FLUXIONS.

INTRODUCTION.

Introduction. THE branch of mathematical analysis which is called in this country the Method of Fluxions, but on the continent the Differential and Integral Calculi, was invented near the end of the 17th century; and Sir Isaac Newton, and Mr Leibnitz, two of the greatest philosophers of that age, have both claimed the discovery.

It will appear very possible that two such men should both fall upon this method of calculation nearly about the same time, if it be considered, that from the beginning of the 17th century its principles were gradually coming into view, in consequence of the united labours and discoveries of a number of mathematicians, such as Napier, Cavallerius, Roberval, Fermat, Barrow, Wallis, and others. And considering the number of men of the first abilities engaged at that time in the study of mathematics, we may reasonably suppose, that the fluxional or differential calculus, would very soon have been found according to the ordinary progress of human knowledge, even although a Newton, or a Leibnitz, had not by the force of superior genius anticipated perhaps by a few years that event. The first intimation that was given of the discovery of the calculus was in the year 1669, when, through the intervention of Dr Barrow, a correspondence was begun between Sir Isaac Newton (then Mr Newton), and Mr Collins, one of the secretaries to the Royal Society. Dr Barrow communicated to the latter a paper by Newton, which had for its title, De analysi per aequationes numero terminorum infinitas. In this paper, besides shewing how to resolve equations by approximation, Newton teaches how to square curves, not only when the expression for the ordinate in terms of the abscissa is a rational quantity,

but also when it involves radical quantities, by first resolving these into an infinite series of rational terms by means of the binomial theorem, a thing which had never before been done. Newton in this paper gives some rather obscure indications of the nature of his calculus, which however serve to shew, beyond all doubt, that he was then in possession of it; and indeed there is good reason to believe that he knew it as early as the year 1665, or even sooner.

These analytical discoveries of Newton were immediately circulated among mathematicians both in this country and abroad, by Dr Barrow, and by Collins and Oldenburg, the two secretaries to the Royal Society.

About the end of the year 1672, Newton communicated to Collins, by letter, a method of drawing tangents to curve lines, illustrated by an example, from which it again plainly appears, that he now possessed his method of fluxions.

In the course of the following year, Leibnitz came to London, and communicated to several members of the Royal Society, some researches relating to the theory of differences. It was however shewn to him, that this subject had been previously treated by Mouton, an astronomer of Lyons; upon this Leibnitz directed his attention to the doctrine of series, which was now considerably advanced, in consequence of the discoveries of the English mathematicians.

The first direct communication that passed between Newton and Leibnitz, was by a letter, which the former addressed to Oldenburg, about the middle of the year 1676. In the beginning of this letter, which was intended to be shewn to Leibnitz, Newton speaks of him with much respect. The letter itself chiefly refers to

the theory of infinite series. In a second letter, written also with a view to its being communicated to Leibnitz, Newton, after bestowing deserved commendation on him, proceeds to explain the steps by which he was led to the discovery of the binomial theorem. He afterwards, among other things, delivers several theorems which have the method of fluxions for their basis; but he does not give their demonstrations, and only observes, that they depend on the solution of a general problem, the enunciation of which he conceals under an anagram of transposed letters, but the meaning of it is this: An equation being given containing any number of flowing quantities, to find their fluxions; and the contrary. This letter affords another proof that Newton was now in full possession of his calculus.

In the end of June 1677, Leibnitz sent to Oldenburg, for the purpose of being communicated to Newton, a letter containing the first essays of his Differential Calculus. The death of Oldenburg, which happened soon after, put an end to the correspondence, and in the year 1684, Leibnitz published his method in the Leipic Act. for the month of October 1684. The title of the memoir which contained it was, Nova methodus pro maximis et minimis, itemque tangentibus, quæ nec fractas, nec irrationales quantitates moratur, et singulare pro illis calculi genus. Thus, in whatever way Leibnitz came by his calculus, whether he discovered it solely by the force of his own genius, or founded it on the method of fluxions, previously invented by Newton, both of which hypotheses are possible, his method was certainly published before Newton's, which, except what transpired in consequence of the circulation of his letters and manuscripts, became only known to the world in general for the first time, by the publication of the Principia in the end of the year 1686.

It seems at first to have been allowed, that Leibnitz had invented his calculus, without having any previous knowledge of what had been done by Newton; for in the first edition of the Principia, Newton says, "In the course of a correspondence which ten years ago I carried on with the very learned geometrician Mr Leibnitz, having intimated to him that I possessed a method of determining maxima and minima, of drawing tangents, and resolving such problems, not only when the equations were rational, but also when they were irrational; and having concealed this method, by transposing the letters of the following sentence—An equation being given, containing any number of flowing quantities, to find their fluxions; and the contrary; this celebrated man answered that he had found a similar method, which he communicated to me, and which differs from mine, only in the enunciation, and in the notation." To this, in the edition of 1714 is added, "and in the manner of conceiving the quantities to be generated *."

There is reason to suppose that Leibnitz might have continued to enjoy undisturbed the honour of being considered as one of the inventors of the fluxional or differential calculus, if he had not manifested a disposition to attribute the invention too exclusively to himself. This called forth some remarks respecting the priority of Newton's claim to the discovery. In particular M. Facio asserted, in a treatise on the Line of swiftest descent, published in 1699, "that he was obliged to own Newton as the first inventor of the diffe-

rential calculus, and the first by many years; and that he left the world to judge whether Leibnitz, the second inventor, had taken any thing from him."

On the other hand, when Newton's treatise on the quadrature of curves, and on the enumeration of lines of the third order was published, which was in 1704, the Leipic journalists insinuated, in a very liberal account which they gave of the work, that Leibnitz was the first inventor, and that Newton had taken his method from Leibnitz's, substituting fluxions for differences.

In consequence of this attack on Newton, Dr John Keill asserted, in the Philosophical Transactions for 1708, that Newton was beyond a doubt the first inventor of the arithmetic of fluxions, and that the same arithmetic having its name and notation changed, was afterwards published by Mr Leibnitz in the Leipic Act. In answer to this, Leibnitz replied, in a letter to Hans Sloane, secretary to the Royal Society, that no one knew better than Newton himself, that the charge against him implied in Keill's assertion was false; and he required Keill to retract what he had said. To this request, however, Keill would by no means accede; but on the contrary, he wrote a long letter to the secretary of the Royal Society, in which he endeavoured to prove, not only that Newton had preceded Leibnitz in the invention, but that he had given to the latter such indications of the nature of his calculus, as made it easy for him to fall upon the same. This letter was sent to Leibnitz, who replied, that Keill, although learned, was too young a man to be fit to judge of what had passed between him and Newton, and he requested the Royal Society to put a stop to Keill's clamours.

The Royal Society being thus appealed to as a judge, appointed a committee to examine all the old letters, papers, and documents, which had passed among the several mathematicians, relating to the question. The judgment of the committee was to the following effect. "That Mr Leibnitz was in London in 1673, and went thence to Paris, where he kept a correspondence with Mr Collins by means of Mr Oldenburg, till about September 1676, and then returned by London and Amsterdam to Hanover; and that Mr Collins was very free in communicating to able mathematicians what he had received from Newton. That it did not appear, that Mr Leibnitz knew any thing of the differential calculus before his letter of the 21st of June 1677, which was a year after a copy of Newton's letter of the 10th of December 1672 had been sent to Paris, to be communicated to him, and above four years after Mr Collins began to communicate that letter to his correspondents; in which letter the method of fluxions was sufficiently described to any intelligent person. That Newton was in possession of his calculus before the year 1669, and that those who had reputed Leibnitz the first inventor, knew little or nothing of his correspondence with Mr Collins, and Mr Oldenburg, long before, nor of Newton's having that method above 15 years before Mr Leibnitz began to publish it in the Leipic Act. That for these reasons, they reckoned Newton the first inventor, and were of opinion, that Mr Keill in asserting the same had been in no ways injurious to Mr Leibnitz."

It is deserving of remark, that the committee delivered no opinion upon the advantage which Leibnitz was accused of having taken of the hints furnished to him in

* Principia
lib. ii. lem.
2. schol.

in the course of his correspondence with Newton; they left the decision of this point to the world in general; and to enable every one to judge for himself, the Royal Society ordered the opinion of the committee to be printed, together with all the documents upon which it was founded. These appeared in 1712 under the title of, Commercium Epistolicum de Analysis promotu. This work was carefully circulated over Europe, to vindicate the title of the English nation to the discovery.

The Commercium Epistolicum having appeared, Leibnitz expressed great dissatisfaction, and threatened to reply in such a manner as to confound his adversaries. There seems no reason however to suppose, that any thing he could have said, would have affected Newton's claim to the honour of being the first inventor; for on this point there cannot be any doubt. With respect, however, to the other question, whether Leibnitz took his calculus from Newton, or found it himself, it is impossible to decide with such certainty. Mr Montucla, in his History of Mathematics*, says, "There are only three places of the Commercium Epistolicum, which treat of the principles of fluxions in so clear a way, as to prove that Newton had found it before Leibnitz, but too obscurely, it seems, to take from the latter the merit of the discovery. One of these is in a letter from Newton to Oldenburg, who had signified to him, that Slufus and Gregory had each found a very simple method of drawing tangents. Newton replied, that he conjectured what the nature of that method was; and he gave an example of it, which shews it to be in effect the same thing as those geometricians had found. He adds, that it is only a particular case, or rather a corollary to a method much more general, which, without a laborious calculation, applies to the finding of tangents to all sorts of curves, geometrical or mechanical, and that without being obliged to free the equation from radicals. He repeats the same thing without explaining himself farther, in another letter, and he conceals the principle of the method under transposed letters. The only place where Newton has allowed any thing of his method to transpire, is in his Analysis per aequationes numero terminis infinitis. He here discloses, in a very concise and obscure manner, his principle of fluxions, but there is no certainty of Leibnitz's having seen this essay. His opponents have never asserted that it was communicated to him by letter, and they have gone no farther than to suspect, that he had obtained a knowledge of it in the interview which he had with Collins, upon his second journey to London. Indeed, this suspicion is not entirely destitute of probability, for Leibnitz admitted, that in this interview, he saw a part of the Epistolary Correspondence of Collins. However I think it would be rash to pronounce upon this circumstance. If Leibnitz had confined himself to a few essays of his new calculus, there might have been some foundation for that suspicion; but the numerous pieces he inserted in the Leipzig Acts, prove the calculus to have received such improvements from him, that probably he owed the invention of it to his genius, and to the efforts he made to discover a method, which put Newton in possession of so many beautiful truths. This is so much the more likely, as, from the method of tangents discovered by Dr Barrow, the transition to the differential calculus was easy, nor was the step too great for such a genius as that with which Leibnitz appears to have been endowed." Such is the opinion of Montucla,

who being a foreigner, cannot be supposed to have been too partial towards Newton, an Englishman. The British mathematicians have hitherto, with few exceptions, entertained an opinion still more decidedly in favour of the claims of their celebrated countryman.

It has been said that Newton took no share in the controversy; this however seems not to have been exactly the case, for besides suppressing in the third edition of his Principia (printed in 1726) the passage we have already quoted, which seems to admit that Leibnitz invented his calculus for himself, he is known to have written the notes which accompany the edition of the Commercium Epistolicum, printed in 1722. Leibnitz had also begun to prepare a Commercium Epistolicum, but he died before it was completed.

Besides the disputes that have happened respecting the inventor of the method of fluxions, the accuracy of the method itself has been the subject of controversy, both in Britain and on the continent. The differential calculus was attacked abroad by Nieuwentijt, a writer of little or no reputation as a mathematician, and by Rolle, who was an expert algebraist, and an indefatigable calculator, but rash, and too confident of the justness of his own opinions, and jealous of the inventions of others. To the first of these writers Leibnitz himself replied, and afterwards Bernoulli and Herman; the attack from Rolle was successfully repelled by Varignon, who was as zealous and intelligent, as his adversary was warm and impetuous.

The very concise manner in which the great inventor of the method of fluxions thought proper to explain its principles, gave occasion to the celebrated Dr Berkley bishop of Cloyne to call in question, not only the logical accuracy of the reasoning employed to establish the theory of fluxions, but also the faith of mathematicians in general, in regard to the truths of religion. The bishop commenced the controversy first in a small work entitled The Minute Philosopher; but its principal attack made its appearance in 1734, under the title of "The Analyst, or A Discourse addressed to an Infidel Mathematician," (understood to be Dr Halley) "wherein it is examined whether the object, principles, and inferences of the modern Analysis are more distinctly conceived than religious mysteries and points of faith." One of the best answers which was made to this work came from the pen of Benjamin Robins, and is entitled, "A discourse concerning the nature and certainty of Sir Isaac Newton's methods of fluxions, and of prime and ultimate ratios." Other mathematicians likewise attempted to defend Newton, and the method of fluxions, against the very cogent and well-directed arguments of the bishop; but the most satisfactory way of removing all objections to the method, was to abandon those obscure and inaccurate modes of expression, of which Berkley had, not without some reason, complained, and to substitute in their place, others more intelligible, and more consonant to the common methods of mathematical reasoning. This was accordingly done by the celebrated Maclaurin, who, in the year 1742, published his Treatise of Fluxions, a work which, although in some respects rather diffuse, placed the principles of the method beyond controversy, by establishing them on the firm basis of geometrical demonstration.

It would extend this introduction to too great a length were we to enter into a detailed account of the various improvements which the calculus has received

from its first invention to the present time. We shall just briefly observe, that among those who contributed the first and the most effectually to its improvement, we may reckon Newton and Leibnitz themselves, the two illustrious rivals for the honour of its discovery; these were followed by the two brothers James and John Bernoulli, by the Marquis de L'Hospital, and many other foreign mathematicians; and in this country we may reckon Craig, Cheyne, Cotes, Taylor, and De Moivre, as among the earliest of its improvers. It is to Cotes in particular that we are indebted for the discovery of the method of finding the fluents of certain rational fractions, a discovery which was extended by De Moivre, so as to form one of the most beautiful and complete branches of the theory of fluxions.

Besides innumerable memoirs on particular branches of the fluxional calculus, which are to be found in academical collections, many distinct treatises have been written on the subject. Some of the most valuable of these are as follow. The Method of Fluxions and Infinite Series, by Sir Isaac Newton. This work was written in Latin, but was not published till the year 1736, when it was translated into English, and given to the world, along with a comment, by Mr Colson. Harmonia Mensurarum, by Cotes, a most valuable and original work, published in 1716. A Treatise on Fluxions, in two books, by Maclaurin, published in 1742. Many parts of the writings of the celebrated Euler have a reference to the theory of fluxions, or the differential and integral calculi. He has, however, three works in particular that relate to that subject; the first is his Introductio in Analysis Infinitorum, the second his Institutiones Calculi Differentialis, and the third his Institutiones Calculi Integralis.

There is a work on this subject which deserves to be particularly mentioned, both on account of its excellence, and the singular circumstance of its being composed by a lady. Its title is, Analytical Institutions, in four books, originally written in Italian, by Donna Maria Gaetana Agnelli. This lady was Professor of Mathematics and Philosophy in the University of Bologna; her work was

originally published in 1748, and has been styled by her countryman Friji, Opus nitidissimum, ingeniosissimum, et certe maximum quod adhuc ex feminae alicujus calamo prodierit. A part of this work has been published in the French language by Bossut. An English translation was prepared for the press many years ago by the late Professor Colson; it remained, however, unpublished, and might still have continued so, but for the liberality of Baron Maseres, who, after satisfying some pecuniary claims upon the manuscript, caused it, in 1801, to be published (we believe at his own expence), in two volumes quarto. The Doctrine and Application of Fluxions, by Thomas Simpson, is a work deservedly in high estimation. The Doctrine of Fluxions, by Emerson, is also very generally read by the British mathematicians. We are sorry, however, to observe, that there is no work in the English language that exhibits a complete view of the theory of fluxions, with all the improvements that have been made upon it to the present time. We cannot at present acquire any tolerable acquaintance with the subject, without consulting the writings of the foreign mathematicians. There are several excellent works in the French language; we may mention in particular a Traité de Calcul Differential et de Calcul Integral, by Cousin, in 2 vols. 4to.; another by Bossut, in 2 vols. 8vo.; and another by La Croix, in 3 vols. 4to. This last deserves particular notice, as the author intended it to comprehend the substance of the various valuable treatises by Euler, as well as of the most important academical memoirs that relate to this subject. The author has also published an abridgment of his work, in one volume octavo. Principiorum Calculi Differentialis et Integralis, by L'Huilier, published in 1795, contains a very clear exposition of the principles of the calculus. The writings of our countrymen Landen and Waring, and of these foreign mathematicians La Grange, Le Gendre, La Place, and many others, abound with improvements in the calculus. Having given this sketch of the history of this very important branch of mathematical science, we proceed to explain its principles.

PART I. THE DIRECT METHOD OF FLUXIONS.

SECT. I. Principles and Definitions.

1. IN the application of algebra to the theory of curve lines, we find that some of the quantities which are the subject of consideration, may be conceived as having always the same magnitude, as the parameter of a parabola, and the axes of an ellipse or hyperbola; while others again are indefinite in respect of magnitude, and may have any number of particular values, such are the co-ordinates at any point in a curve line. This difference in the nature of the quantities which are compared together, has equally place in various other theories, both in pure and mixed mathematics, and it naturally suggests the divisions of all quantities whatever into two kinds, namely, such as are constant, and such as are variable.

2. A constant quantity is that which retains always

the same magnitude, however other quantities with which it is connected may be supposed to change; and a variable quantity is that which is indefinite in respect of magnitude, or which may be supposed to change its value. Thus, in the arithmetic of lines, the radius is a constant quantity, while the cosine, sine, tangent, &c. of an arch, also the arch itself, are variable quantities; and in the conic sections the axes and the parameters of the axes are constant quantities, and any abscissa and its corresponding ordinate are variable quantities.

Constant quantities are usually denoted by the first letters of the alphabet a, b, c, &c. and variable quantities by the last letters x, y, z, &c.

3. Any expression of calculation, containing a variable quantity, along with other constant quantities, is called a Function of that variable quantity. Thus, supposing x to be variable, and the other quantities constant,

Direct Method. any one of these expressions $ax^n$ , $\frac{a+bx^n}{cx^2+dx^3}$ , $a^2$ , $\log. x$ , $\cos. x$ , $\sin. x$ , &c. is a function of $x$ ; and in any such equation as $y = ax + bx + cx^2 + dx^3$ , the quantity $y$ is called a function of $x$ . Even although the variable quantities $x$ and $y$ , should not be separated as in the last example, but should be related to each other as in the following,

ax^3y + by^3 + y^3 = 0,

as, setting aside the consideration of the constant quantities, the value of $y$ depends on that of $x$ , and on the contrary the value of $x$ depends upon that of $y$ , the quantity $y$ is said to be a function of $x$ , and on the other hand $x$ is said to be a function of $y$ .

4. If a variable quantity be supposed to change its value, then a corresponding change will take place in the value of any function of that quantity. Let us examine the nature of this change in the magnitude of a function.

First, let us suppose that, $x$ denoting any variable quantity, the function to be considered is any integer power of that quantity, as $x^1$ , or $x^2$ , or $x^3$ , &c.; then, $x$ being supposed to be increased by an indefinite quantity $h$ , and thus to become $x+h$ , the function will change its value; if it be $x^2$ it will become $(x+h)^2$ , or

x^2 + 2xh + h^2;

and if it be $x^3$ it will become $(x+h)^3$ or

x^3 + 3x^2h + 3xh^2 + h^3;

and if it be $x^4$ it will become $(x+h)^4$ , or

x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4;

and so on, for other integer powers.

If we compare the new value of the function in each of these cases with its former value, it will immediately appear, that the new value may be resolved into two parts, one of which is the original value of the function, and, therefore, the other is the increment which the function has received, in consequence of the change in the value of the variable quantity $x$ . Thus, the function being $x^2$ , we have found its new value to be $x^2 + 2xh + h^2$ , of which expression, the first term $x^2$ is the original value of the function; therefore the other part of the expression viz. $2xh + h^2$ is its increment. In like manner the expression $x^3 + 3x^2h + 3xh^2 + h^3$ , which is the new value of the function $x^3$ , may be resolved into $x^3$ , the original value of the function, and $3x^2h + 3xh^2 + h^3$ its increment; and $x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$ , the new value of the function $x^4$ , may be resolved into $x^4$ , the function itself, and $4x^3h + 6x^2h^2 + 4xh^3 + h^4$ its increment.

5. Having seen that, by conceiving the variable quantity $x$ to receive the indefinite increment $h$ , the functions $x^1$ , $x^2$ , $x^3$ receive the increments

2xh + h^2,

3x^2h + 3xh^2 + h^3,

4x^3h + 6x^2h^2 + 4xh^3 + h^4,

respectively, we next observe that each increment is expressed by a series, the first term of which is the first

power of the indefinite quantity $h$ multiplied by some function of the variable quantity $x$ as a co-efficient. The second term of the series consists of the second power of $h$ , multiplied also by a function of $x$ as a co-efficient; and, in like manner, the third and following terms are composed of the third and higher powers of $h$ (the exponents forming the arithmetical series 1, 2, 3, 4, &c.) each multiplied by a function of $x$ as a co-efficient; and it appears, that the particular form of the function which constitutes the coefficient of any assigned term of the series depends entirely upon the particular form of the original function. Thus, when the original function is $x^2$ , the function which is the coefficient of the first term is $2x$ ; when the original function is $x^3$ , the co-efficient of the first term is $3x^2$ ; when the original function is $x^4$ , the coefficient of the first term is $4x^3$ , and so on. It also appears that the functions of $x$ , which are the coefficients of the powers of $h$ , are composed only of the variable quantity $x$ and given quantities, so that they are entirely independent of the indefinite increment $h$ .

6. These observations may be extended to a function that is any power whatever of a variable quantity, by the application of the binomial theorem. Let $x$ be supposed to become $x+h$ , then $x^n$ will become $(x+h)^n$ ; but by the binomial theorem, (see ALGEBRA, Sect. xvii.) $(x+h)^n$ when expanded into a series, is

x^n + \frac{n}{1} x^{n-1} h + \frac{n(n-1)}{1 \cdot 2} x^{n-2} h^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} x^{n-3} h^3 + \&c.

where it appears, that the first term of the series is the original value of the function, and the following terms are the first, second, and following powers of the increment $h$ , each multiplied by a new function of $x$ , that is independent of the increment. Let us denote the

functions $nx^{n-1}$ , $\frac{n(n-1)}{1 \cdot 2} x^{n-2}$ , and $\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} x^{n-3}$

$x^{n-1}$ , &c. by $p$ , $q$ , $r$ , &c. respectively; and it is to be observed that, in the present case, as well as in the case of any other function of $x$ we may hereafter consider, by the letters $p$ , $q$ , $r$ , &c. or the same letters with accents over them, or lastly the capital letters $P$ , $Q$ , $R$ , &c. we do not mean to denote functions of $x$ of any particular form, but functions of $x$ in general, that consist only of $x$ and given quantities. This being kept in view, it appears that the variable quantity $x$ being supposed to change its value, and to become $x+h$ , the function $x^n$ changes its value, so as to become

x^n + ph + qh^2 + rh^3 + sh^4 + \&c.

a series, the terms of which have the properties already explained in the two preceding sections.

7. Every rational and integer function of a variable quantity $x$ is necessarily of this form,

Ax^a + Bx^b + Cx^c + \&c.

where $A$ , $B$ , $C$ , &c. and $a$ , $b$ , $c$ , are supposed to denote constant quantities.

Let us examine what is the form which the function assumes when the variable quantity $x$ changes its value

Direct Method. to $x+h$ ; and to avoid complicated expressions, let us suppose the function to consist of these two terms $Ax^n + Bx^m$ . We have already found in last section that $x$ being supposed to become $x+h$ , $x^n$ will become

x^n + ph + qh^2 + rh^3 + \&c.

where $p, q, r, \&c.$ denote functions of $x$ independent of $h$ , as explained in the last section, and consequently $Ax^n$ will become

Ax^n + Aph + Aqh^2 + Arh^3 + \&c.

In like manner $Bx^m$ will become

Bx^m + Bp'h + Bq'h^2 + Br'h^3 + \&c.

$p', q', r', \&c.$ denoting also functions of $x$ independent of $h$ ; therefore taking the sum of the two series, it appears that, supposing $x$ to change its value, and to become $x+h$ , the function $Ax^n + Bx^m$ becomes

Ax^n + Bx^m + (Ap + Bp')h + (Aq + Bq')h^2 + (Ar + Br')h^3 + \&c.

now $p$ and $p'$ being functions of $x$ , $Ap + Bp'$ will also be a function of $x$ , and may be denoted more simply by $P$ , and for the same reason $Aq + Bq'$ , $Ar + Br'$ , &c. which are functions of $x$ , may be denoted by $Q, R, \&c.$ thus the expression for the new value of $Ax^n + Bx^m$ is

Ax^n + Bx^m + Ph + Qh^2 + Rh^3 + \&c.

a series, the form and properties of which are in all respects analogous to those of the series that expresses the new value of the function $x^n$ ; and although we have supposed the function to consist of but two terms, yet it is obvious, that whatever be the number of terms, still the form of the series and its properties will be the same; that is, it will consist of two parts, one of which is independent of $h$ , and is the original value of the function, and the other is a series, the terms of which are the successive powers of the increment $h$ , each multiplied by a function of the variable quantity $x$ as a coefficient. This conclusion may be expressed in symbols concisely thus. Let $u$ denote any rational and integer function of a variable quantity $x$ , let $x$ be conceived to change its magnitude, and to become $x+h$ , and let $u'$ denote the new value which the function acquires in consequence of the change in the value of $x$ , then

u' = u + ph + qh^2 + rh^3 + \&c.

where $p, q, r, \&c.$ denote functions of $x$ as already stated.

8. Suppose next the function of $x$ to be of this form:—

(Ax^n + Bx^m + Cx^r + \&c.)^n,

that is, suppose it to be the $n$ th power of a polynomial, consisting of any number of terms whatever. Let the expression between the parentheses be denoted by $v$ , then we are to consider the function $v^n$ . Now when $x$ becomes $x+h$ , we have already found that $v$ becomes

v + ph + qh^2 + rh^3 + \&c.

therefore $v^n$ will become

(v + ph + qh^2 + rh^3 + \&c.)^n,

or, putting $ph + qh^2 + rh^3 + \&c. = M$ ,

v^n \text{ will become } (v + M)^n,

and this expression when expanded into a series by the binomial theorem is

v^n + av^{n-1}M + bv^{n-2}M^2 + cv^{n-3}M^3 + \&c.

where $a, b, c, \&c.$ express numbers.

Now from the form of the series denoted by $M$ , it is manifest that its square, cube, or any power of it whatever, will be a series proceeding by the powers of $h$ , and having for the coefficients of its terms certain combinations of the quantities $p, q, r, \&c.$ which being functions of $x$ , any combinations of them will also be functions of $x$ . Therefore, each of the terms of the above series, expressing the development of $(v + M)^n$ , excepting the first term $v^n$ , will itself be a series proceeding by the powers of $h$ , and having its terms multiplied by functions of $x$ , and consequently their sum will be a series of the same nature. Let us as before denote the functions of $x$ , which are the coefficients of the successive powers of $h$ by $P, Q, R, \&c.$ and we shall have upon the whole

(v + ph + qh^2 + rh^3 + \&c.)^n

expressed by a series of this form

v^n + Ph + Qh^2 + Rh^3 + \&c.

therefore, putting the single letter $u$ for the function $u^n$ , or for

(Ax^n + Bx^m + Cx^r + \&c.)^n

and $u'$ for the new value which $u$ acquires by $x$ changing its value to $x+h$ ,

u' = u + Ph + Qh^2 + Rh^3 + \&c.

a series of the same nature as before.

9. Let us now consider a fractional function of $x$ , and let us suppose it to be

\frac{A'x^{\alpha} + B'x^{\beta} + C'x^{\gamma} + \&c.}{Ax^n + Bx^m + Cx^r + \&c.}

Where $A', A, B', B, \&c.$ also $\alpha, \beta, \gamma, \&c.$ denote constant quantities. Let $v$ denote the numerator of the fraction, and $w$ its denominator, then the function is

\frac{v}{w}, \text{ or } vw^{-1},

now when $x$ becomes $x+h$ , $v$ becomes

v + ph + qh^2 + rh^3 + \&c.

and $w^{-1}$ becomes

w^{-1} + p'h + q'h^2 + r'h^3 + \&c.

and consequently $vw^{-1}$ becomes

(v + ph + qh^2 + \&c.) (w^{-1} + p'h + q'h^2 + \&c.)

and

Direct Method. and the product of these two factors, by actual multiplication is

\left. \begin{array}{l} v w^{-1} + v p' \\ + w^{-1} p \end{array} \right\} \left. \begin{array}{l} + v q' \\ + p p' \\ + w^{-1} q \end{array} \right\} h^1 + \&c.

Now, here as before, it appears that the coefficients of the powers of $h$ are functions of $x$ , therefore, denoting these functions by $P, Q, R, \&c.$ and observing that $v w^{-1}$ is $\frac{v}{w}$ , we have the new value of $\frac{v}{w}$ expressed by the series

\frac{v}{w} + P h + Q h^2 + R h^3 + \&c.

or, substituting the single letter $u$ for $\frac{v}{w}$ , that is, for

\frac{A' x^{\alpha} + B' x^{\beta} + C' x^{\gamma} + \&c.}{A x^{\alpha} + B x^{\beta} + C x^{\gamma} + \&c.}

and putting $u'$ for the value that $u$ acquires when $x$ becomes $x+h$ ,

u' = u + P h + Q h^2 + R h^3 + \&c.

a series in all respects analogous to those already found for the other functions of $x$ .

10. In the functions which we have hitherto considered, the exponents of the powers of $x$ were constant quantities. Let us now consider a function in which the exponent is the variable quantity $x$ itself.

Suppose then the function to be $a^x$ , where $a$ denotes a given number; then, by supposing $x$ to become $x+h$ , the function will become

a^{x+h} = a^x a^h.

Now it has been shown in the article ALGEBRA, § 295, that if $A$ be put to denote the quotient arising from the division of a logarithm of $a$ by the logarithm of $2.7182818\dots$ the exponential quantity $a^h$ is expressed by the series

1 + \frac{A}{1} h + \frac{A^2}{1 \cdot 2} h^2 + \frac{A^3}{1 \cdot 2 \cdot 3} h^3 + \&c.

therefore, $a^{x+h}$ , the new value of the function, is

a^x \left( 1 + \frac{A}{1} h + \frac{A^2}{1 \cdot 2} h^2 + \frac{A^3}{1 \cdot 2 \cdot 3} h^3 + \&c. \right),

this series, by multiplying all its terms by $a^x$ , and putting $p, q, r, \&c.$ for that part of each term which is independent of $h$ , becomes

a^x + p h + q h^2 + r h^3 + \&c.

so that denoting the function $a^x$ by $u$ , and its new value by $u'$ ,

u' = u + p h + q h^2 + r h^3 + \&c.

a series of the same form as the others.

11. From a due consideration of what has been shown relating to the change that takes place in the magnitude of a variable function, corresponding to the

change that takes place in the magnitude of the variable quantity from which the function is formed, we may conclude the truth of the following general proposition to be sufficiently established.

Let $x$ denote a variable quantity, and $u$ any function whatever of that quantity, let $x$ be supposed to receive any increment $h$ , and thus to become $x+h$ , and let $u'$ be the new value which the function acquires by the change in the value of $x$ ; then, the new value of $u$ may in every case be expressed thus:

u' = u + p h + q h^2 + r h^3 + \&c.

where $p, q, r, \&c.$ denote quantities that are quite independent of $h$ , and consequently can only involve the variable quantity $x$ , and given quantities.

12. Having examined what is the general form that any function of a variable quantity acquires by a change in the value of that quantity, and found it to be a series, the first term of which is always the function itself, it is evident that the remaining terms will express the increment that the function receives, in consequence of the change in the magnitude of the variable quantity from which the function is formed. Let us now compare the simultaneous increments of a variable quantity and its function with each other, and that we may at first avoid general reasoning, and fix the mind more completely, let us suppose the functions to have a determinate form, as $x^2, x^3, x^4, \&c.$

Putting $u$ and $u'$ as before to denote the two succeeding values of the function, first let it be supposed that $u = x^2$ , then $x$ being supposed to receive the indefinite increment $h$ , and thus to become $x+h$ , and $u$ to change its value to $u' = (x+h)^2$ , we have

u' = x^2 + 2xh + h^2,

\text{or, } u' = u + 2xh + h^2,

and consequently

u' - u = 2xh + h^2.

Thus it appears, that the simultaneous increments of $x$ and $x^2$ (or $u$ ) are $h$ and $2xh + h^2$ , respectively. Let us now compare these increments, not in respect of their absolute magnitudes, but in respect of their ratio to each other, thus we shall have the increment of $u$ to the increment of $x$ , as $2xh + h^2$ to $h$ , that is, (dividing the two last terms of the proportion by $h$ ) as $2x+h$ to 1. Or, instead of employing any analogy, let us, for the sake of brevity, and in conformity to the algebraic notation, rather express each of these ratios by the quotient arising from the division of the antecedents of the ratio by its consequents, and put the results equal to each other. Then, observing that the symbol $u' - u$ , which expresses the difference between the succeeding values of the function, may be employed to denote its increment, we have

\frac{u' - u}{h} = \frac{2xh + h^2}{h} = 2x + h.

Hence it appears that the expression for the ratio of the increment of the function $u$ to the increment of the variable quantity $x$ is made up of two parts, one of these, viz. $2x$ , is quite independent of $h$ , the increment of $x$ , and the other is in the present case that increment itself. In consequence of this peculiarity in the form

form of the expression for the ratio, it is evident that if the increment $h$ be conceived to be continually diminished, the part of the expression which consists of $h$ will continually diminish, so that the whole expression, viz. $2x+h$ , may become more nearly equal to its first term $2x$ than by any assignable difference; therefore $2x$ may be considered as the limit of the ratio $\frac{u-u'}{h}$ , that is, a quantity to which the ratio may approach nearer than by any assignable difference, but to which it cannot be considered as becoming absolutely equal.

Let us next suppose that $u=x^3$ , then $x$ being supposed to become $x+h$ , we have $u'=(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ , or

u' = u + 3x^2h + 3xh^2 + h^3,

and consequently

u' - u = 3x^2h + 3xh^2 + h^3,

\text{and } \frac{u' - u}{h} = 3x^2 + 3xh + h^2.

Here it is evident, as in the former case, that the expression for the ratio $\frac{u' - u}{h}$ is composed of two parts, one of these, viz. the first term $3x^2$ , is a function of $x$ that is independent of the increment $h$ , but the other, viz. $3xh + h^2$ , or $h(3x+h)$ , is the product of two factors, one of which is the increment itself. From the particular form of this latter part of the expression for the ratio, it is plain, that $h$ being supposed to be continually diminished, that part will also diminish, and may become less than any assignable quantity. Therefore in this case, as well as in the former, the ratio $\frac{u' - u}{h}$ has a limit, and that limit is the first term of the general expression for the ratio, namely the quantity $3x^2$ .

Suppose, next, that $u=x^4$ , and consequently

u' = (x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4,

or,

u' - u = 4x^3h + 6x^2h^2 + 4xh^3 + h^4,

\text{and } \frac{u' - u}{h} = 4x^3 + h(6x^2 + 4xh + h^3).

Here, as in the two former cases, we have only to inspect the general expression for the ratio $\frac{u' - u}{h}$ to discover, that by supposing $h$ to be continually diminished, the latter part of the expression, viz. $h(6x^2 + 4xh + h^3)$ , and which comprises all the terms except the first, will become smaller than any assignable quantity; and consequently, that the first term $4x^3$ is the limit of the ratio.

13. It is easy to see that the property which we have found to belong to the ratio of the simultaneous increments of a variable quantity, and its function in these three particular cases, is an immediate consequence of the form of the expression for the increment of the function, so that it is not peculiar to the functions $x^2$ , $x^3$ , and $x^4$ , but belongs equally to all functions whatever.

For we have found, § 11, that $u$ being supposed to

denote any function of a variable quantity $x$ , as for example, $ax^n$ , or $ax^n + bx^m + \&c.$ or

\frac{ax^n + bx^m + \&c.}{a'x^n + b'x^m + \&c.}

and $u'$ being put for the new value which the function acquires when $x$ becomes $x+h$ ,

u' = u + ph + qh^2 + rh^3 + \&c.

where $p, q, r, \&c.$ denote functions of $x$ that are independent of $h$ , therefore,

u' - u = ph + qh^2 + rh^3 + \&c.

and

\frac{u' - u}{h} = p + qh + rh^2 + \&c.

or,

\frac{u' - u}{h} = p + h(q + rh + \&c.)

Thus it appears, that whatever be the form of the function, the ratio $\frac{u' - u}{h}$ is always expressed by a quantity which may be resolved into two parts; one of these, viz. $p$ , is independent of the increment $h$ , and the other, viz. $h(q + rh + \&c.)$ , is the product of $h$ by a series, the first term of which is a function of $x$ , and the remaining terms also functions of $x$ multiplied by the first, second, third, and higher powers of $h$ . Now from the particular form of this last part of the general expression for the ratio, it is manifest, that $h$ being conceived to be continually diminished, the quantity $h(q + rh + \&c.)$ will also be continually diminished, and may become less than any assignable quantity; therefore, the limit of the ratio $\frac{u' - u}{h}$ is simply $p$ , that is, the function of $x$ , which is the coefficient of the first or simple power of $h$ in the general expression for the increment.

14. From what has been just shewn, we may infer the truth of the following general proposition relative to the simultaneous changes that take place in a variable quantity and its function.

Let $x$ denote a variable quantity and $u$ any function of that quantity, let $x$ be conceived to change its value and become $x+h$ , where $h$ denotes an arbitrary increment, and let $u'$ denote the new value that the function acquires, in consequence of the change in the magnitude of $x$ . Then, observing that $h$ and $u' - u$ are the simultaneous increments of the variable quantity and its function, if $h$ be conceived to be continually diminished, the ratio $\frac{u' - u}{h}$ will continually approach to a certain Limit, which will be different for different functions, but always the same for the same function, and in every case quite independent of the magnitude of the increments.

The ratio which is the limit of the ratio of the increments, when these increments are conceived to be continually diminished, may be called the limiting ratio of the increments.

15. The analytical fact contained in the preceding proposition, affords the foundation for a mathematical theory of great extent, and which may be divided into

Direct Method. into two distinct branches; one having for its object the resolution of the following problem. Having given the relation of any number of variable quantities to each other, to determine the limiting ratios of their increments; and the other the converse of that problem, namely, Having given the limiting ratios, to determine the relations of the quantities themselves.

The theory to which we have alluded constitutes the METHOD of FLUXIONS, and in explaining the foundation of the method, we have endeavoured to show, that it rests upon a principle purely analytical, namely the theory of limiting ratios; and this being the case, the subject may be treated as a branch of pure mathematics, without having occasion to introduce any ideas foreign to geometry.

16. Sir Isaac Newton, however, in first delivering the principles of the method, thought proper to employ considerations drawn from the theory of motion. But he appears to have done this chiefly for the purpose of illustration, for he immediately has recourse to the theory of limiting ratios; and it has been the opinion of several mathematicians of great eminence (A) that the consideration of motion was introduced into the method of fluxions at first without necessity, and that succeeding writers on the subject ought to have established the theory upon principles purely mathematical, independent of the ideas of time and velocity, both of which seem foreign to investigations relating to abstract quantity.

17. That we may conform to the usual method of treating this subject, we proceed to show how the theory of motion is commonly applied to the illustration of the nature of variable quantities, and of the relations that result from their being conceived to change their value.

As quantities of every kind, if we abstract from their position, figure, and such affections, and consider their magnitude only, may be represented by lines, we may consider a variable quantity $x$ , and $u$ any function of that quantity, to be represented by two lines $AP$ , $BQ$ , which have $A$ , $B$ , one extremity of each given, and which vary by the points $P$ , $Q$ , their other extremities, moving in the directions $AC$ , $BD$ , while the equation expressing the relation between $x$ and $u$ , or their representative lines $AP$ , $BQ$ , remains always the same.

Diagram showing two horizontal lines, AP and BQ, representing variable quantities. Line AP starts at point A on the left and ends at point P in the middle. Line BQ starts at point B on the left and ends at point Q in the middle. Points P and Q are vertically aligned. To the right of P is point C, and to the right of Q is point D. The lines extend horizontally to the right.

18. The lines $AP$ , $BQ$ being thus conceived to vary, the relation that is supposed always to subsist between them, in respect of their magnitudes, will necessarily give rise to another relation, namely, that which will constantly subsist between the velocities of the moving points $P$ and $Q$ , by which the lines are generated.

With a reference to this particular mode of conceiving variable quantities to exist, the quantities themselves have been called flowing quantities or fluents, and the

measures of the velocities with which the variable quantities increase have been called their fluxions.

19. To simplify the hypothesis, we may suppose that the point which generates the line $AP$ , or $x$ , moves uniformly; thus the measure of its velocity, or the fluxion of $x$ , will be a given quantity, with which the measure of the velocity of the point $Q$ , or the fluxion of $u$ , may be continually compared. To determine then the fluxions, or rather the ratio of the fluxions of $x$ a variable quantity, and $u$ , any function of that quantity, is in effect to resolve the following problem.

Having given an equation expressing the relation at every instant between the spaces passed over by two points, one of which moves with an uniform velocity: It is required to find an expression for the ratio that the measures of the velocities have at every instant to each other.

20. Now it is a fundamental principle in the theory of motion, resulting indeed from the very nature of a variable velocity, that when two velocities are compared together, whether they be both variable, or one of them uniform, and the other variable, the measures of their velocities are any quantities having to each other the ratio that is the limit of the ratio of the spaces described in the same time, when those spaces are conceived to be continually diminished. And hence it follows that the ratio of the fluxions of two variable quantities is no other than the limiting ratio of their simultaneous increments.

That the theory of motion may be applied to the generation of variable algebraic quantities, we have supposed them to be represented by lines; this, however, is not necessary, if the variable quantities are themselves geometrical magnitudes; for like as a line is conceived to be generated by the motion of a point, so a surface may be considered as generated by the motion of a line, a solid by the motion of a surface, and an angle by the rotation of one of the lines which contain it; and the fluxions of those quantities at any instant, or position, will be the measures of the velocities, or degrees of swiftness, according to which they increase at that instant or position.

But in every case the ratio of the rates of increase, or fluxions of two homogeneous magnitudes, will be the limiting ratio of their simultaneous increments.

21. Having thus found that by conceiving variable quantities as generated by motion, and taking their velocities, or rates of increase, as an object for the mind to contemplate and reason on, we are in the end led to the consideration of the limiting ratios of their increments, a subject which is purely mathematical, and independent of the ideas of time or velocity, we shall exchange the definition of a fluxion given in § 18, which involves those ideas, for another that rests entirely upon the existence of limiting ratios.

By the fluxions then of two variable quantities having any assigned relation to each other, we are in the following treatise always to be understood to mean any

(A) Such as Lagrange, Confin, La Croix, &c. abroad, and Landen in this country.

indefinite quantities which have to each other the limiting ratio of their simultaneous increments (B).

In conformity to this definition of fluxions, it is evident that we are to consider them, not as absolute, but as relative quantities, which derive their origin from the comparison of variable quantities with each other in respect of their simultaneous variations of magnitude.

22. Sir Isaac Newton employed different symbols at different times to denote the fluxions of variable quantities. It is now however common in Britain to denote them by the same letters employed to express the quantities themselves, and each having a dot over it. Thus $\dot{x}$ denotes the fluxion of the variable quantity expressed by $x$ , and in the like manner $\dot{u}$ , $\dot{v}$ , $\dot{z}$ , denote the fluxions of the variable quantities $u$ , $v$ , $z$ , respectively.

23. Suppose now that $u$ is any function of a variable quantity $x$ , and that the limiting ratio of the simultaneous increments of $u$ and $x$ is the ratio of $p$ to 1, where $p$ denotes some other function of $x$ ; then, from the definition just given of fluxions, we have

\frac{\dot{u}}{\dot{x}} = p, \text{ and } u = p \cdot x.

Hence it follows as a consequence of the preceding definition, that the fluxion of $u$ , any function of a variable quantity $x$ , is the product arising from the multiplication of that function of $x$ which is the expression for the limiting ratio of the increments by the fluxion of the variable quantity $x$ itself.

SECT. II. Investigation of the Rules of the Direct Method of Fluxions.

24. The method of fluxions naturally resolves itself into two parts, as we have already observed, § 15. We proceed to explain the first of these, which is called the Direct Method, and which treats of finding the ratio of the fluxions of variable quantities, having given the relations of those quantities to each other.

25. We shall begin with investigating the ratio of the fluxions of two variable quantities in that particular case, when one of them is in the power of the other.

Let us suppose then, that $u$ is such a function of a variable quantity $x$ , that $u = x^n$ , where $n$ denotes any number whatever, it is required to determine the ratio of the fluxions of $u$ and $x$ .

If we recur to the definition which has been given of the fluxions of variable quantities in § 21, it will appear that we have in effect resolved the problem just proposed in three particular cases, when treating of the limiting ratios of the increments of variable quantities. For it has been shown in § 12. that when $u = x^2$ , and $x$ is supposed to change its value, so as to become $x+h$ , where

$h$ denotes an indefinite increment, and in consequence of this change in the magnitude of $x$ , $u$ also changes its value, and becomes $u'$ , then, observing that $u' - u$ and $h$ are the simultaneous increments of $u$ and $x$ , the limiting ratio of $\frac{u' - u}{h}$ is $2x$ . Let this expression for the limiting ratio be put equal to the ratio of the fluxions of $u$ and $x$ , that is to $\frac{\dot{u}}{\dot{x}}$ , thus we have $\frac{\dot{u}}{\dot{x}} = 2x$ , and $\dot{u} = 2x \cdot \dot{x}$ .

Hence it appears, that whatever be the magnitude of the quantity that expresses the fluxion of $x$ , the fluxion of $u$ or $x^2$ will be expressed by the fluxion of $x$ multiplied by $2x$ .

Again, when $u = x^3$ , and $u'$ , $u' - u$ and $h$ denote the same as before, it has been shown, § 12, that the limiting ratio of $\frac{u' - u}{h}$ is $3x^2$ , therefore, § 21, $\frac{\dot{u}}{\dot{x}} = 3x^2$ , and

$\dot{u} = 3x^2 \cdot \dot{x}$ ; that is, the fluxion of $x^3$ is expressed by the fluxion of $x$ multiplied by $3x^2$ .

And when $u = x^4$ , it has been shown, § 12, that the limiting ratio of $\frac{u' - u}{h}$ is $4x^3$ . Therefore, § 21,

\frac{\dot{u}}{\dot{x}} = 4x^3 \text{ and } \dot{u} = 4x^3 \cdot \dot{x}.

To resolve the problem generally, or when $u = x^n$ , let us suppose $x$ to become $x+h$ , and $u$ to become $u'$ , then $u' = (x+h)^n$ . But this last quantity, when expanded into a series by the binomial theorem (ALGEBRA, SECT. XVII.) is

x^n + p \cdot h + q \cdot h^2 + r \cdot h^3 + \&c.
\text{or } u + p \cdot h + h^2 (q + r \cdot h + \&c.)

where $p$ , $q$ , $r$ , &c. denote functions of $x$ , independent of $h$ . Therefore,

u' - u = p \cdot h + h^2 (q + r \cdot h + \&c.),
\text{and } \frac{u' - u}{h} = p + h (q + r \cdot h + \&c.).

Therefore, supposing $h$ to be continually diminished, the limit of $\frac{u' - u}{h}$ is $p$ ; but, whatever be the nature of the exponent $n$ , $p$ is always $n x^{n-1}$ , (ALGEBRA, § 267.), therefore, the limit of $\frac{u' - u}{h}$ is $n x^{n-1}$ , and consequent-

\frac{\dot{u}}{\dot{x}} = n x^{n-1}, \text{ and } \dot{u} = n x^{n-1} \cdot \dot{x}.

26. As we shall have frequently occasion to employ the result of this investigation, it will be proper to express it in the form of a practical rule, thus.

To find the fluxion of any power of a variable quantity. Multiply the fluxion of the variable quantity itself by the exponent of the power, and by a power of the quantity whose exponent is less by unity than the given exponent, and the product will be the fluxion required.

27.

(B) We are here to be understood to mean the ratio of the numerical values of the increments, which may always be compared with each other, whether the variable quantities be of the same kind, as both lines, or both surfaces, &c. or of different kinds, as the one a line, and the other a surface.

27. In determining the limit of the ratio of the simultaneous increments of $x$ and $x^n$ , we have referred to the binomial theorem; but the only application we have had occasion to make of that theorem was to determine the numeral coefficient of the second term of the development of $(x+h)^n$ , when $n$ is supposed to be any number whatever, which is an inquiry of a more simple nature than the general investigation of the theorem. We shall now show how that coefficient may be deduced from the first principles of algebra. Thus the investigation of the fluxion of $x^n$ will be rendered independent of the general demonstration of the binomial theorem; and we shall hereafter show that the theorem itself is easily investigated by the direct method of fluxions.

28. Since $x+h$ is equal to $x(1+\frac{h}{x})$ it follows that $(x+h)^n = x^n(1+\frac{h}{x})^n$ , thus the development of $(x+h)^n$ is reduced to that of $(1+\frac{h}{x})^n$ , or putting $\frac{h}{x} = v$ , to

$(1+v)^n$ . Now if we give particular values to $n$ , and suppose it to be 1, 2, 3, &c. or -1, -2, &c. or lastly $\frac{1}{2}$ , $\frac{1}{3}$ , &c. we can find the series that expresses the powers of $1+v$ , whose exponents are those numbers, by the operations of multiplication, division, and evolution, and in every particular case we shall find, that the powers of $1+v$ are expressed by a series of this form,

1 + Av + Bv^2 + Cv^3 + Dv^4 + \dots

Where $A, B, C, D, \dots$ denote numeral coefficients which depend for their value only on $n$ the exponent of the power, and not on the quantity $v$ ; and as the form of the series will be found to be the same whatever particular values we may give to the exponent, we may conclude that it is the same, whether the exponent be positive or negative, whole or fractional.

29. First let us suppose that the exponent is a whole positive number, then, because

(1+v)^n = 1 + Av + Bv^2 + \dots

if we multiply both sides of this equation by $1+v$ , and collect together the like powers of $v$ , it will appear that

(1+v)^{n+1} = 1 + A \begin{Bmatrix} v \\ + 1 \end{Bmatrix} + B \begin{Bmatrix} v^2 \\ + A \end{Bmatrix} + \dots

Hence it appears that the coefficient of the second term of any power of $1+v$ exceeds that of the next less power by unity. Now in the case of the first power of $1+v$ , the coefficient of the second term is obviously 1, therefore, in the second power it is 2, in the third power 3, and universally, in the $n$ th power it is $n$ ; so that $n$ being a whole positive number,

(1+v)^n = 1 + nv + \dots

30. Let us next suppose the exponent to be a fraction denoted by $\frac{m}{n}$ , so that

(1+v)^{\frac{m}{n}} = 1 + Av + Bv^2 + \dots

Let both sides of this equation be raised to the power $n$ , having first substituted $Mv$ for $Av + Bv^2 + \dots$ then

(1+v)^m = (1+Mv)^n

Now as we have just found that $m$ and $n$ being integers,

(1+v)^m = 1 + mv + \dots

\text{and } (1+Mv)^n = 1 + nMv + \dots

Here we stop at the second term, that being the only one whose coefficient is required. Substitute now for $Mv$ its value $Av + \dots$ then, stopping again at the second term, we get

(1+Mv)^n = 1 + nAv + \dots

therefore,

1 + mv + \dots = 1 + nAv + \dots

and making the coefficients of $v$ in each series equal to each other,

m = nA, \text{ and } A = \frac{m}{n}.

31. In the last place, let us suppose that the exponent is a negative quantity either whole or fractional, so that

(1+v)^{-\frac{m}{n}} = 1 + A'v + B'v^2 + \dots

\text{or } \frac{1}{(1+v)^{\frac{m}{n}}} = 1 + A'v + B'v^2 + \dots

(1+v)^{\frac{m}{n}} = \frac{1}{1 + A'v + B'v^2 + \dots}

then, multiplying both sides by $(1+v)^{\frac{m}{n}}$ we get

1 = (1+v)^{\frac{m}{n}} (1 + A'v + B'v^2 + \dots)

or, substituting $1 + Av + Bv^2 + \dots$ for $(1+v)^{\frac{m}{n}}$ and actually multiplying the two series,

1 = 1 + A \begin{Bmatrix} v \\ + A' \end{Bmatrix} + B \begin{Bmatrix} v^2 \\ + AA' \end{Bmatrix} + B' \begin{Bmatrix} v^3 \\ + B' \end{Bmatrix} + \dots

Now that this equation may subsist, whatever be the value of $v$ , it is necessary that

A + A' = 0,

B + AA' + B' = 0,

\dots

and by these equations we may determine $A', B', C', \dots$ . It is, however, only required at present to find the first of these, viz. $A'$ , now we have $A' = -A$ ; but $A$ being the coefficient of the second term of the series expressing $(1+v)^{\frac{m}{n}}$ , the exponent of which is positive, we have already found it to be $\frac{m}{n}$ , therefore, $A' = -\frac{m}{n}$ .

32. As we have found that the coefficients of the second term of the developments of $(1+v)^n$ , $(1+v)^{\frac{m}{n}}$ and $(1+v)^{-\frac{m}{n}}$ are $m$ , $\frac{m}{n}$ , and $-\frac{m}{n}$ , respectively, it appears that whatever be the number denoted by $n$ , the two first terms of the series expressing $(1+v)^{\frac{m}{n}}$ are, $1 + nv$ , and therefore, substituting for $v$ its value $\frac{h}{x}$ , and multiplying by $x^n$ , the two first terms of the series expressing $(x+h)^n$ are $x^n + nx^{n-1}h$ , agreeing with what we have

have assumed in § 25. as given by the binomial theorem (c).

33. The mode of reasoning employed to determine the ratio of the fluxions of $u$ and $x$ , when the former is a function of the latter of the form $x^n$ , will apply equally when the function has any other assigned form. But instead of investigating in this manner the fluxion of every particular function, it is better to consider a complex function as the sum, or difference, or product or quotient, &c. of other simple functions, and to investigate rules for each of these cases, supposing that the fluxions of the simple functions are previously known.

34. Let us first suppose that $u$ , a function of a variable quantity $x$ , is equal to the sum of $v$ and $w$ , two other functions of $x$ . It is required to find the fluxion of $u$ , having given the fluxions of $v$ and $w$ .

Let $x$ be conceived to change its value, and to become $x+h$ , then, as $v$ and $w$ will also change their values, § 11, the one to

v' = v + p h + q h^2 + r h^3 + \dots

and the other to

w' = w + p' h + q' h^2 + r' h^3 + \dots

if $u'$ as usual denote the corresponding new value of $u$ , we have

\begin{aligned} u &= v + w \\ u' &= v + p h + q h^2 + \dots \\ &+ w + p' h + q' h^2 + \dots \\ u' - u &= (p + p') h + (q + q') h^2 + \dots \\ \frac{u' - u}{h} &= p + p' + (q + q') h + \dots \end{aligned}

If we now conceive $h$ to be continually diminished, we shall have the limit of $\frac{u' - u}{h}$ expressed by $p + p'$ . But

$p$ is the limit of $\frac{v' - v}{h}$ , § 14, and in like manner $p'$ is the limit of $\frac{w' - w}{h}$ , therefore,

\text{limit of } \frac{u' - u}{h} = \text{limit of } \frac{v' - v}{h} + \text{limit of } \frac{w' - w}{h}.

Substitute now the ratio of the fluxions instead of the limited ratios, and we have

\frac{\dot{u}}{\dot{x}} = \frac{\dot{v}}{\dot{x}} + \frac{\dot{w}}{\dot{x}};

\text{therefore, } \dot{u} = \dot{v} + \dot{w}.

35. If we suppose $s$ to be a function of $x$ , and $u$ , $v$ , $w$ , &c. other functions of $x$ , such that

s = a u + b v + c w + \dots

where $a$ , $b$ , $c$ , denote any given numbers, positive or negative, then, by reasoning as above, it is evident that

s' = a \dot{u} + b \dot{v} + c \dot{w} + \dots

Therefore, to find the fluxion of the sum of any number of functions, each multiplied by a constant quantity. Multiply the fluxion of each function by its constant coefficient, and the sum of the products is the fluxion required.

36. If $c$ denote a constant quantity, and $u$ , $v$ , be functions of $x$ , such, that $u = c + v$ ; then $x$ being supposed to become $x+h$ , and consequently $v$ to become $v'$ , or $v + p h + q h^2 + \dots$ and $c+v$ to become $c+v' + p h + q h^2 + \dots$ we have

\begin{aligned} u &= c + v, \\ \text{and } u' &= c + v + p h + q h^2 + \dots \\ \text{and hence } u' - u &= p h + q h^2 + \dots \\ \text{or } u' - u &= v' - v, \end{aligned}

\text{therefore, } \frac{u' - u}{h} = \frac{v' - v}{h},

and these ratios being always equal, their limits must also be equal; therefore, substituting for the limiting ratios those of the fluxions, we have $\frac{\dot{u}}{\dot{x}} = \frac{\dot{v}}{\dot{x}}$ and $\dot{u} = \dot{v}$ ; that

is to say, the fluxion of $c+v$ is $\dot{v}$ , from which it appears, That the fluxion of any variable function is the very same as the fluxion of the same function, increased or diminished by any constant quantity. This is a remark of great importance in the theory of fluxions, as will appear hereafter.

37. Let us now suppose that $u$ , $v$ , $w$ , are functions of $x$ , such, that $u = v w$ , it is required to find the fluxion of $u$ , supposing the fluxions of $v$ and $w$ to be given.

By supposing that $x$ , $u$ , $v$ , and $w$ , change their values as usual, we have

\begin{aligned} u &= v w \\ u' &= (v + p h + q h^2 + \dots)(w + p' h + q' h^2 + \dots) \end{aligned}

and this last expression by multiplication becomes

u' = v w + v p' h + v q' h^2 + \dots + w p h + w q h^2 + \dots

therefore

u' - u = (v p' + w p) h + (v q' + w q) h^2 + \dots

dividing now by $h$ , and taking the limit of $\frac{u' - u}{h}$ , we have that limit expressed by $v p' + w p$ ; but $p'$ is the limit of $\frac{v' - v}{h}$ § 14, and in like manner $p$ is the limit of $\frac{w' - w}{h}$ ; therefore

$\frac{u' - u}{h}$ ; therefore

\frac{u' - u}{h}; \text{ therefore}

the

(c) In this investigation we have supposed $n$ to be a rational number. If, however, it were irrational, still the result would be the same; for, corresponding to every such number, two rational numbers, one greater, and the other less than it, may be found, which shall differ from each other by less than any assignable quantity. Therefore, the general properties of these numbers must belong also to the irrational number which is their limit.

\text{the limit of } \frac{u' - u}{h} = \begin{cases} v \times \text{limit of } \frac{v' - v}{h} \\ + w \times \text{limit of } \frac{w' - w}{h} \end{cases}

Hence, by substituting for the limiting ratios the ratios of the fluxions, we have

\frac{\dot{u}}{u} = \frac{v \dot{v}}{v} + \frac{w \dot{w}}{w} \text{ and } \dot{u} = v \dot{v} + w \dot{w}.

Therefore, to find the fluxion of the product of any two functions, multiply the fluxion of each function by the other function, and the sum of these products is the fluxion required.

38. We have just now seen that when

u = v w

\text{then } \dot{u} = v \dot{w} + w \dot{v}.

Let each side of the latter equation be divided by the corresponding side of the former, thus we get

\frac{\dot{u}}{u} = \frac{\dot{v}}{v} + \frac{v \dot{w}}{w};

suppose now that the function $w$ is the product of two other functions $s, t$ , so that

w = s t,

then, because $w = s t$ , from what has been shewn it follows that $\frac{\dot{w}}{w} = \frac{\dot{s}}{s} + \frac{\dot{t}}{t}$ ; therefore, substituting this value

of $\frac{\dot{w}}{w}$ in the equation $\frac{\dot{u}}{u} = \frac{\dot{v}}{v} + \frac{v \dot{w}}{w}$ , it becomes

\frac{\dot{u}}{u} = \frac{\dot{v}}{v} + \frac{\dot{s}}{s} + \frac{\dot{t}}{t}.

In general, if we suppose that

u = v s t r \&c.

by reasoning as above it will be found that

\frac{\dot{u}}{u} = \frac{\dot{v}}{v} + \frac{\dot{s}}{s} + \frac{\dot{t}}{t} + \frac{\dot{r}}{r} + \&c.

whatever be the number of factors.

Suppose the number of factors to be three, so that

u = v s t

\text{and } \frac{\dot{u}}{u} = \frac{\dot{v}}{v} + \frac{\dot{s}}{s} + \frac{\dot{t}}{t}

then substituting $v s t$ for $u$ in this last equation, and taking away the denominators, we find

\dot{u} = s t \dot{v} + v s \dot{t} + v t \dot{s}.

And as a similar result will be found, whatever be the number of factors, we may conclude that The fluxion of the product of any number of functions is equal to the sum of the products of the fluxion of each function by all the other functions.

39. Let us in the last place suppose that $u = \frac{v}{w}$ , and

that it is required to find the fluxion of $u$ , having given the fluxions of $v$ and $w$ .

From the given equation we have $v = w u$ , and there-

fore (§ 37.) $\dot{v} = w \dot{u} + u \dot{w}$ , let $\frac{\dot{v}}{w}$ be substituted for $u$

in this equation, and it becomes $v = w \dot{u} + \frac{v \dot{w}}{w}$ , from which we easily obtain

\dot{u} = \frac{w \dot{v} - v \dot{w}}{w^2}.

Hence we have the following rule for finding the fluxion of a fraction.

Multiply the fluxion of the numerator by the denominator, and from the product subtract the fluxion of the denominator multiplied by the numerator, and divide the remainder by the square of the denominator; the result is the fluxion required.

40. It will now be proper to shew the application of these general rules for determining the fluxions of variable functions to some particular examples.

Example 1. Suppose $u = a + b \sqrt{x} = \frac{c}{x}$ . Required the fluxion of $u$ .

Here $a$ being a constant quantity, the fluxion of $a + b \sqrt{x} = \frac{c}{x}$ is the same as the fluxion of $b \sqrt{x} = \frac{c}{x}$ , § 36, or $b x^{\frac{1}{2}} = c x^{-1}$ . Now, by § 26, the fluxion of $x^{\frac{1}{2}}$ is $\frac{1}{2} x^{\frac{1}{2}-1} \dot{x}$ , which expression is equivalent to $\frac{1}{2} x^{-\frac{1}{2}} \dot{x}$ , or to $\frac{\dot{x}}{2 \sqrt{x}}$ , and in like manner the fluxion

of $x^{-1}$ is $-x^{-1-1} \dot{x}$ or $-x^{-2} \dot{x}$ , or $\frac{-\dot{x}}{x^2}$ , therefore, multiplying the fluxion of $x^{\frac{1}{2}}$ by $b$ , and the fluxion of $x^{-1}$ by $c$ , and taking the sum of the products, agreeably to the rule in § 35, we have

\dot{u} = \frac{b \dot{x}}{2 \sqrt{x}} + \frac{c \dot{x}}{x^2}

\text{or } \dot{u} = \left( \frac{b}{2 \sqrt{x}} + \frac{c}{x^2} \right) \dot{x}.

Ex. 2. Suppose $u = a + \frac{b}{\sqrt[3]{x^3}} = \frac{c}{x^3 \sqrt[3]{x}} + \frac{d}{x^2}$ .

By writing the function thus

u = a + b x^{-\frac{1}{3}} = c x^{-\frac{4}{3}} + d x^{-2}

the application of the same rules employed in the last example give us

\dot{u} = -\frac{1}{3} b x^{-\frac{4}{3}} \dot{x} + \frac{4}{3} c x^{-\frac{7}{3}} \dot{x} - 2 d x^{-3} \dot{x}

or, exchanging the fractional indices for the radical sign, and otherwise reducing,

\dot{u} = \frac{-2 b \dot{x}}{2 x^{\frac{1}{3}} \sqrt[3]{x^3}} + \frac{4 c \dot{x}}{3 x^{\frac{1}{3}} \sqrt[3]{x^3}} - \frac{2 d \dot{x}}{x^3}

Ex. 3.

Ex. 3. Suppose $u = (a + bx^m)^n$ .
In order to find the fluxion of this function by the rules already laid down, it will be necessary to consider it first as a function of a variable quantity that is itself a function of $x$ . Let us then put $a + bx^m$ equal to $v$ , and thus the proposed equation becomes $u = v^n$ ; then, $u$ being considered as a function of $v$ , we have by § 26. $\dot{u} = n v^{n-1} \dot{v}$ . Again, $v$ being considered as a function of $x$ , from the equation $v = a + bx^m$ , we find by § 36. and § 26. $\dot{v} = m b x^{m-1} \dot{x}$ . Let this value of $\dot{v}$ be substituted in the expression for $\dot{u}$ , and it becomes

\dot{u} = m n b v^{n-1} x^{m-1} \dot{x},

which, by substituting for $v$ its value $a + bx^m$ , is also

\dot{u} = m n b x^{m-1} (a + bx^m)^{n-1} \dot{x}.

Ex. 4. Suppose $u = \sqrt{(a^2 - x^2)}$ .
Here we are to proceed as in the last example, and first put $a^2 - x^2 = v$ , then $u = \sqrt{v} = v^{\frac{1}{2}}$ ; and therefore (§ 26) $\dot{u} = \frac{1}{2} v^{\frac{1}{2}-1} \dot{v} = \frac{\dot{v}}{2\sqrt{v}} = \frac{\dot{v}}{2u}$ . Again, since $v = a^2 - x^2$ , by § 26. we find $\dot{v} = -2x \dot{x}$ . Substitute this value of $\dot{v}$ in the expression for $\dot{u}$ , and we have

\dot{u} = -\frac{x \dot{x}}{\sqrt{a^2 - x^2}},

which, by restoring $\sqrt{(a^2 - x^2)}$ for $v$ , and leaving out the number common to the numerator and denominator, becomes

\dot{u} = \frac{-x \dot{x}}{\sqrt{(a^2 - x^2)}}.

Ex. 5. Suppose $u = \sqrt{(a + bx + cx^2)}$ .
By proceeding in the same manner as in last example, we find

\dot{u} = \frac{b + 2cx \dot{x}}{\sqrt{(a + bx + cx^2)}}.

Ex. 6. Suppose $u = x(a^2 + x^2) \sqrt{(a^2 - x^2)}$ .
Here the proposed function is the product of these three functions, viz. $x$ , $a^2 + x^2$ , and $\sqrt{(a^2 - x^2)}$ . Therefore, its fluxion will be found by proceeding according to the rule in § 38.

Now the fluxion of $x$ is $\dot{x}$ , and the fluxion of $a^2 + x^2$ is $2x \dot{x}$ , and the fluxion of $\sqrt{(a^2 - x^2)}$ has been found in last example to be $\frac{-x \dot{x}}{\sqrt{(a^2 - x^2)}}$ . Therefore, multi-

plying the fluxion of each function by the product of the other two functions, and taking the sum of all these products, we find

\dot{u} = \begin{cases} (a^2 + x^2) \sqrt{a^2 - x^2} \dot{x} \\ + 2x^2 \sqrt{a^2 - x^2} \dot{x} \\ - \frac{x^2 (a^2 + x^2) \dot{x}}{\sqrt{(a^2 - x^2)}} \end{cases}

and this equation, by reducing all the terms on the latter side of it to a common denominator, is more simply expressed thus,

\dot{u} = \frac{(a^4 + a^2 x^2 - 4x^4) \dot{x}}{\sqrt{(a^2 - x^2)}}.

Ex. 7. Suppose $u = \frac{a + x}{a^2 + x^2}$ .

Here we employ the rule given in § 39. for finding the fluxion of a fractional function; thus we find

\dot{u} = \frac{(a^2 + x^2) \dot{x} - 2x (a + x) \dot{x}}{(a^2 + x^2)^2},

which when reduced, is

\dot{u} = \frac{(a^2 - 2ax - x^2) \dot{x}}{(a^2 + x^2)^2}.

Ex. 8. Suppose

u = \sqrt{\left\{ (a - \frac{b}{\sqrt{x}} + \sqrt{(c^2 - x^2)})^2 \right\}}

To simplify this expression, we put

\frac{b}{\sqrt{x}} = y, \quad \sqrt{(c^2 - x^2)} = z.

Thus we have

u = \sqrt{(a - y + z)^2} = (a - y + z).

Now as the fluxion of $a - y + z$ , is $-\dot{y} + \dot{z}$ (§ 35), we find from § 26, that by considering $a - y + z$ as a single variable quantity,

\begin{aligned} \dot{u} &= \frac{1}{2} (a - y + z)^{\frac{1}{2}-1} (-\dot{y} + \dot{z}) \\ &= \frac{1}{2} (a - y + z)^{-\frac{1}{2}} (-\dot{y} + \dot{z}) \\ &= \frac{1}{2} \sqrt{a - y + z} (-\dot{y} + \dot{z}) \end{aligned}

but, since $y = \frac{b}{\sqrt{x}} = b x^{-\frac{1}{2}}$ , we have, § 26,

\dot{y} = -\frac{1}{2} b x^{-\frac{3}{2}} \dot{x} = -\frac{b \dot{x}}{2x \sqrt{x}};

and since $z = \sqrt{(c^2 - x^2)} = (c^2 - x^2)^{\frac{1}{2}}$ , by considering $c^2 - x^2$ as a single variable quantity, and observing that its fluxion is $-2x \dot{x}$ , we find by § 26, that

\dot{z} = \frac{1}{2} (c^2 - x^2)^{-\frac{1}{2}} \times (-2x \dot{x}) = \frac{-x \dot{x}}{\sqrt{(c^2 - x^2)}}.

Instead of $y$ , $z$ , $\dot{y}$ , $\dot{z}$ , substitute now their values in the expression for the fluxion of $u$ , thus it becomes

\begin{aligned} \dot{u} &= \frac{1}{2} \sqrt{a - \frac{b}{\sqrt{x}} + \sqrt{(c^2 - x^2)}} \\ &\times \left( \frac{b \dot{x}}{2x \sqrt{x}} - \frac{x \dot{x}}{\sqrt{(c^2 - x^2)}} \right). \end{aligned}

Ex. 9. Suppose $u = a v^m y^n$ , where $v$ and $y$ denote any functions of a variable quantity.

Then, § 37,

\dot{u} = \left\{ \begin{array}{l} a y^n \times \text{fluxion of } v^m \\ + a v^m \times \text{fluxion of } y^n \end{array} \right\}

But

Direct Method. But fluxion of $v^n = n v^{n-1} \dot{v}$ , § 26,
and fluxion of $y^n = n y^{n-1} \dot{y}$ ;

\begin{aligned} \dot{u} &= a m y^n v^{n-1} \dot{v} + a n v^n y^{n-1} \dot{y}, \\ &= a v^{n-1} y^{n-1} (m y \dot{v} + n v \dot{y}). \end{aligned}

Ex. 10. Suppose $u = \frac{v+x}{y^3}$ , where $v$ , $x$ and $y$ denote any function of a variable quantity. Then, because fluxion $(v+x) = \dot{v} + \dot{x}$ , § 34, and fluxion $y^3 = 3 y^2 \dot{y}$ , § 26, we have, § 39,

\begin{aligned} \dot{u} &= \frac{y^3(\dot{v} + \dot{x}) - 3(v+x)y^2 \dot{y}}{y^6}, \\ &= \frac{y(\dot{v} + \dot{x}) - 3(v+x)\dot{y}}{y^4}. \end{aligned}

41. As when $u$ denotes that particular function of $x$ which is $x^n$ , we have (§ 25.)

\frac{\dot{u}}{x} = n x^{n-1};

so, in general, whatever be the form of the function denoted by $u$ , we have always

\frac{\dot{u}}{x} = p,

where $p$ denotes a new function of $x$ , resulting from the analytical process employed to find the fluxion of the function $u$ , and depending for its form upon the particular form of that function: just as in involution, or any of the other operations of algebra, a result is obtained depending upon the particular nature of the operation, and the quantities operated upon.

Let us put $p$ to denote the particular function $n x^{n-1}$ , or the expression for $\frac{\dot{u}}{x}$ the ratio of the fluxion of $u$ to the fluxion of $x$ when $u = x^n$ , then, supposing that $n-1$ is not equal to 0, (for in that case $n x^{n-1}$ would be simply $n$ , a given number,) we may reason concerning the ratio of the fluxions of the variable quantities $p$ and $x$ , in all respects as concerning the ratio of the fluxions of $u$ and $x$ ; and accordingly, from the equation

p = n x^{n-1},

we get, by taking the fluxions,

\frac{\dot{p}}{x} = n(n-1)x^{n-2},

or, considering $p$ as denoting generally the function of $x$ that results from the operation of finding the fluxion of the original function $u$ , whatever be the form of that function, we have

\frac{\dot{p}}{x} = q,

where $q$ denotes a new function of $x$ , derived from $p$ , the former function, by the same kind of operation as that by which $p$ was deduced from $u$ .

Suppose now $q$ to denote the particular function $n(n-1)x^{n-2}$ , then,

\begin{aligned} \frac{\dot{q}}{x} &= n(n-1)(n-2)x^{n-3}, \\ &\text{or } \frac{\dot{q}}{x} = r, \end{aligned}

where $r$ denotes a function of $x$ derived from $q$ , as $q$ was derived from $p$ , or $p$ from the original function $u$ . And it is evident that we may proceed in this manner as far as we please, unless it happen that in finding the series of functions $p$ , $q$ , $r$ , &c. we at last arrive at a result that is a constant quantity, and then the series of operations will terminate. Thus if the function was $a x^4$ , we should have

\begin{aligned} u &= a x^4, \\ \frac{\dot{u}}{x} &= 4 a x^3 = p, \\ \frac{\dot{p}}{x} &= 4 \cdot 3 a x^2 = q, \\ \frac{\dot{q}}{x} &= 4 \cdot 3 \cdot 2 a x = r, \\ \frac{\dot{r}}{x} &= 4 \cdot 3 \cdot 2 \cdot 1 \cdot a = 24 a. \end{aligned}

Here the expression for $\frac{\dot{r}}{x}$ is a constant quantity, which has no fluxion.

Hence it appears, that relatively to any function of a variable quantity, there exists a series of limiting ratios, deducible from that function, and from each other, by a repetition of the operation of finding the fluxion of a variable function.

42. In treating of the fluxion of a function, we have hitherto regarded the fluxion of the variable quantity $x$ , from which the function is formed, merely as one of the terms of a ratio, without considering whether it was a constant or a variable quantity.

Now as we may assume any hypothesis respecting the nature of the fluxion of $x$ , that is not inconsistent with what has been already delivered, we shall suppose it to be constant. This assumption, if we consider the fluxions of variable quantities as the measures of their respective velocities, or rates of increase, is in effect the same thing as to suppose that the variable quantity $x$ increases uniformly. Then, as in the expressions

\frac{\dot{u}}{x} = p, \frac{\dot{p}}{x} = q, \frac{\dot{q}}{x} = r, \&c.

or these others, which follow from them,

\dot{u} = p x, \dot{p} = q x, \dot{q} = r x, \&c.

the symbol $\dot{x}$ is to be understood as denoting a constant quantity, it follows that if $p$ be variable, then $p x$ , or $\dot{u}$ will be variable; and if $q$ be variable, then $q x$ , or $\dot{p}$ , will

$\dot{p}$ , will be variable; and if $\dot{r}$ be variable, then $\dot{r}\dot{x}$ , or $\dot{q}$ will be variable, and so on.

43. Let us now recur to the relation in which the succeeding functions $p, q, r, \dots$ stand to the original function $u$ .

By performing that particular analytical operation upon the function $u$ , which consists in finding its fluxion, we obtain $\dot{p}\dot{x}$ as the expression for its fluxion, that is, we get $\dot{u} = \dot{p}\dot{x}$ ; and by repeating the operation on the function $p$ , we get $\dot{p} = \dot{q}\dot{x}$ ; and therefore $\dot{p}\dot{x} = \dot{q}\dot{x}^2$ ; but, $\dot{x}$ being regarded as a constant quantity, $\dot{p}\dot{x}$ is deduced from $p\dot{x}$ , considered as a function of $x$ , just in the same manner as $\dot{p}\dot{x}$ is derived from the original function $u$ ; therefore the expression $\dot{q}\dot{x}^2$ is deduced from the function $u$ by performing the operation of taking the fluxion twice; that is, first upon the function $u$ itself, and then upon $\dot{u}$ , or $\dot{p}\dot{x}$ , the expression for its fluxion; and in this second operation $\dot{x}$ (or the fluxion of the quantity from which the function is formed) is considered as a constant quantity.

The expression $\dot{q}\dot{x}^2$ , obtained in this manner from the function $u$ , is called the second fluxion of the function; and to express its relation to the function $u$ , it is denoted by $\ddot{u}$ , that is, by the letter denoting the function itself with two dots over it. Thus, like as $\dot{u} = \dot{p}\dot{x}$ , we have

\ddot{u} = \dot{q}\dot{x}^2 \text{ and } \frac{\ddot{u}}{\dot{x}^2} = \dot{q}.

Again, since $\dot{q} = \dot{r}\dot{x}$ , it follows that $\dot{q}\dot{x}^2 = \dot{r}\dot{x}^3$ ; but, as $\dot{x}$ is constant, $\dot{q}\dot{x}^2$ is derived from $\dot{q}\dot{x}^2$ , by the operation of finding its fluxion, considering it as a function of $x$ , just in the same manner as $\dot{q}\dot{x}^2$ , or $\ddot{u}$ , is derived from $p\dot{x}$ , or $\dot{u}$ , and in the same manner as $\dot{u}$ is derived from the original function $u$ ; therefore, like as $\dot{p}\dot{x}$ or $\dot{u}$ is the first fluxion of the function, and $\dot{q}\dot{x}^2$ or $\ddot{u}$ is its second fluxion, so $\dot{r}\dot{x}^3$ is called its third fluxion, and is denoted by $\dot{r}$ , that is, by the letter expressing the function itself, having three dots placed over it, so that

\dot{r}\dot{x}^3 \text{ and } \frac{\dot{r}\dot{x}^3}{\dot{x}^3} = \dot{r}.

The fourth fluxion of a variable function $u$ is denoted by $\dot{u}$ , that is, by the letter $u$ with four dots over it, and is derived from the third fluxion, in the same manner as the third is derived from the second, or the second from the first, or the first fluxion from the variable function itself; observing, that in repeating the operation of taking the fluxions, the symbol $\dot{x}$ (or the fluxion of the variable quantity from which the function is formed) is considered as a constant quantity. And the same mode of notation and deduction is to be understood as applying to a fluxion of any order whatever of a variable function.

44. To illustrate what has been said respecting the

second and higher orders of fluxions of a function, let us suppose $u$ to denote the particular function $ax^n$ ; then, proceeding agreeably to what has been laid down in last section, we obtain, by the rule for finding the fluxion of any power of a variable quantity (§ 26.)

\begin{aligned} \dot{u} &= nax^{n-1}\dot{x}, \\ \ddot{u} &= n(n-1)ax^{n-2}\dot{x}^2, \\ \dots & \\ \dot{u} &= n(n-1)(n-2)(n-3)ax^{n-4}\dot{x}^4, \&c. \end{aligned}

Here we have exhibited the first, second, third, and fourth fluxions of the function $ax^n$ ; the law of continuation is obvious, and it appears that when $n$ is any positive integer, the function $ax^n$ will have as many orders of fluxions, as there are units in $n$ , and no more; for if $n$ were supposed $=3$ , then, as the fourth fluxion, and all the subsequent ones, are multiplied by $n-3$ , or in that case by $3-3=0$ , they consequently would vanish, and a similar observation may be made when $n$ is any other whole positive number.

45. That we might be able to apply the rules of § 26, § 34, &c. to the determination of the fluxion of a complex function of a variable quantity, we have found it convenient in some cases to consider such a function as composed of other more simple functions of the same quantity, and we have expressed its fluxion by means of the fluxions of those other functions. In finding the fluxion of any higher order than the first of such a complex function by those rules, we must keep in mind, that it is only the fluxion of $x$ , the variable quantity from which the functions are all formed, that is to be considered as constant, and that the fluxions of the functions themselves are in general variable quantities; so that each of them may have a second, third, &c. fluxion, as well as the function which is composed of them.

Let us suppose, for example, that

u = \sqrt{a^2 + x^2};

then, considering $a^2 + x^2$ as a function of $x$ , and putting $v$ to denote it, we have $u = \sqrt{v} = v^{\frac{1}{2}}$ , and $\dot{u} = \frac{1}{2}v^{-\frac{1}{2}}\dot{v}$ $= \frac{\dot{v}}{2\sqrt{v}}$ ; but since $v = a^2 + x^2$ , it follows that $\dot{v} = 2x\dot{x}$ ;

therefore, substituting for $v$ and $\dot{v}$ their respective values, we have

\dot{u} = \frac{x\dot{x}}{\sqrt{a^2 + x^2}}.

Now, to find the second fluxion of $u$ , we may either take the fluxion of this last expression, viz. $\frac{x\dot{x}}{\sqrt{a^2 + x^2}}$ ,

and consider the symbol $\dot{x}$ , which is found in it, as denoting a constant quantity; or we may recur to the

equation $\dot{u} = \frac{\dot{v}}{2\sqrt{v}}$ , and take the fluxion of this other expression for $\dot{u}$ ; and in this case, we must consider that both $v$ and $\dot{v}$ denote variable functions of $x$ , and therefore that the fluxion of $\frac{\dot{v}}{\sqrt{v}}$ may be found by the rule

for

Direct Method. for finding the fluxion of a fraction; observing that $\ddot{v}$ is to be substituted as the fluxion of $\dot{v}$ . Accordingly, proceeding by this last method, and considering that the fluxion of $\sqrt{v}$ , the denominator of the fraction is $\frac{\dot{v}}{\sqrt{v}}$ , we find

\dot{u} = \left\{ \begin{array}{l} \frac{\frac{1}{2}\sqrt{v}\ddot{v} - \frac{1}{2}\dot{v}\dot{v}}{\sqrt{v}} \\ = \frac{2v\ddot{v} - \dot{v}^2}{4v\sqrt{v}} \end{array} \right.

Now from the equation $v = a^2 + x^2$ , we have $\dot{v} = 2x\dot{x}$ , and (observing that $\dot{x}$ is constant) $\ddot{v} = 2\dot{x}^2$ . Let these values of $v$ , $\dot{v}$ , and $\ddot{v}$ , be now substituted in the expression for $\dot{u}$ , and it becomes

\dot{u} = \left\{ \begin{array}{l} \frac{4(a^2 + x^2)\dot{x}^2 - 4x^2\dot{x}^2}{4(a^2 + x^2)\sqrt{a^2 + x^2}} \\ = \frac{a^2 + x^2}{(a^2 + x^2)^{\frac{3}{2}}} \end{array} \right.

The very same expression for $\dot{u}$ would have been found if we had employed the other method.

By proceeding as in this last example, the rules already delivered for finding the first fluxion of any function of a variable quantity will apply to the finding of the fluxion of any higher order.

Thus if we had $u = vt$ , where $v$ and $t$ denote each a function of another variable quantity $x$ , and it were required to find the different orders of fluxions of $u$ , considered also as a function of $x$ ; then, by the rule of § 37, we have

\dot{u} = \dot{t}v + t\dot{v},

and $\ddot{u} = \text{fluxion of } \dot{t}v + \text{fluxion of } t\dot{v}$ ;

but $v$ and $t$ being variable functions of $x$ , we may consider $\dot{v}$ and $\dot{t}$ as denoting also variable functions of $x$ , the fluxions of which are to be denoted by $\ddot{v}$ and $\ddot{t}$ respectively; now by the rule in § 37, we have

\text{fluxion of } \dot{t}v = \dot{v}\dot{t} + t\ddot{v},

and fluxion of $t\dot{v} = \dot{t}v + v\dot{t}$ ;

\text{therefore, } \ddot{u} = 2\dot{v}\dot{t} + v\ddot{t}.

By considering $v$ , $\dot{v}$ , $\ddot{v}$ , also $t$ , $\dot{t}$ , as denoting each a distinct function of $x$ , we may find the third fluxion of $u$ from the second, in the same manner as the second has been found from the first, and so on for the other orders of fluxions of $u$ . If it be now required to express the successive orders of fluxions of $u$ in terms of $x$ and its fluxion, we must find the values of $v$ , $\dot{v}$ , &c. also of $t$ , $\dot{t}$ , &c. in terms of $x$ and its fluxion, and these values, also the particular functions of $x$ denoted by $v$ and $t$ , being substituted in the expressions found for $\dot{u}$ , $\ddot{u}$ , &c. will give to these expressions the form required.

If for example we suppose that

VOL. VIII. PART II.

\begin{array}{l} v = a + b x^n, \\ \text{then } \dot{v} = m b x^{n-1} \dot{x} \end{array} \quad \begin{array}{l} t = c + d x^m, \\ \dot{t} = n d x^{m-1} \dot{x}, \end{array}

and considering $\dot{x}$ as constant,

\begin{array}{l} \ddot{v} = m(m-1) b x^{n-2} \dot{x}^2, \\ \text{&c.} \end{array} \quad \begin{array}{l} \ddot{t} = n(n-1) d x^{m-2} \dot{x}^2 \end{array}

these values of $v$ , $t$ , $\dot{v}$ , $\dot{t}$ , &c. being substituted in the expressions of $\dot{u}$ , $\ddot{u}$ , &c. will give the successive fluxions of $u$ in terms of $x$ and $\dot{x}$ only.

46. If the fluxion of a variable quantity be considered as the measure of its rate of increase, if that rate be uniform, then its measure will be a constant quantity; but if it be variable, then its measure will be a variable quantity, which will also have a certain rate of increase or decrease; and the measure of this rate will be its fluxion, or will be the fluxion of the fluxion of the original variable quantity; that is, it will be the second fluxion of the original variable quantity. And if this second fluxion is not a constant quantity, then the measure of its rate of variation will be its fluxion, or will be the third fluxion of the original variable quantity, and so on. Thus a quantity will have a successive order of fluxions till some one fluxion become constant, and then it will have no more.

47. We have hitherto supposed the equation expressing the relation between a variable quantity, and a function of that quantity, to be of such a form, that the function was found alone, and of the first degree on one side of the equation, and some power, or combination of powers, of the variable quantity on the other; as in these examples,

u = a x^n, \quad u = \frac{a + b x^m}{c + d x^k}.

In such cases as these, $u$ is said to be an explicit function of $x$ . We are now to consider how the ratio of the fluxions is to be found when the relation between the variable quantity and its function is expressed by an equation, the terms of which involve different powers, both of the function, and the variable quantity; as in the following example,

y^2 - a x y + b x^2 - c = 0,

where we are to consider $y$ as a function of $x$ ; but from the particular manner in which its relation to $x$ is expressed, it is said to be an implicit function of that quantity.

Now in this example, by the resolution of a quadratic equation, we find

y = \frac{a x \pm \sqrt{\{(a^2 - 4b) x^2 + 4c\}}}{2},

and as $y$ is here an explicit function of $x$ , its fluxion or the ratio of its fluxion to that of $x$ , might be determined by the rules already laid down. But it is to be observed that it is only in the particular case of the proposed equation being of the second degree that we can effect the solution generally in this manner. If it

Direct Method. were of a higher order, this particular mode of solution would be often impracticable, for want of a general method of resolving equations.

48. We may however in all cases resolve the problem, without a previous resolution of the equation, by reasoning as follows.

Whatever be the degree of the equation, by giving particular values to $x$ , we can, by the theory of equations, obtain corresponding particular values of $y$ ; therefore, we may be assured that in every case $y$ is expressible by means of $x$ in some way or other, if not in finite terms, at least in the form of a series, the terms of which shall involve powers of $x$ . Hence we may infer, as in the case of explicit functions, that when $x$ changes its value, and becomes $x+h$ , $y$ will also change its value, and become

y+ph+qh^2+\&c.

where $p, q, \&c.$ denote functions of $x$ , that are independent of the arbitrary quantity $h$ . Let us denote $ph+qh^2+\&c.$ the increment of $y$ , by the single letter $k$ ; then $y+k$ is the new value of $y$ , corresponding to $x+h$ , the new value of $x$ . Let these new values be substituted instead of $x$ and $y$ in the proposed equation.

y^3-axy+b x^3-c=0,

and as the result must still be $=0$ , we have

(y+k)^3-a(x+h)(y+k)+b(x+h)^3-c=0;

which equation, by actually involving its terms, substituting for $k$ its value $ph+qh^2+\&c.$ and arranging the result in the form of a series proceeding by the powers of $h$ , becomes

\left. \begin{aligned} & y^3-axy+b x^3-c \\ & +(2py-a(px+y)+2bx)h \\ & +Qh^2+Rh^3+\&c. \end{aligned} \right\} =0

Here $Q, R, \&c.$ denote quantities independent of $h$ , and involving $x, y, p, q, \&c.$ that is to say, $x$ , and functions of $x$ , and therefore $Q, R, \&c.$ are also functions of $x$ . Now as this equation must subsist whatever $h$ may be, which is a quantity quite arbitrary and independent of the coefficients by which its powers are multiplied, it follows (as has been observed when treating of the method of indeterminate coefficients, ALGEBRA, § 261.) that the coefficients of the different powers of $h$ must be each equal to 0.

Therefore,

\left. \begin{aligned} & y^3-axy+b x^3-c=0 \\ & 2py-a(px+y)+2bx=0, \\ & \&c. \end{aligned} \right\}

From the first of these equations we can infer nothing, as it is no other than the proposed equation itself; but from the second we find

p = \frac{ay-2bx}{2y-ax}.

Now $h$ , and $k=ph+qh^2+\&c.$ being the simultaneous increments of $x$ and $y$ , we have $\frac{k}{h}=p+qh+\&c.$ therefore, supposing $h$ to be continually diminished, and

putting $\frac{y}{x}$ equal to the limit of $\frac{k}{h}$ , we have $\frac{y}{x}=p$ , therefore

\frac{y}{x} = \frac{ay-2bx}{2y-ax};

thus we have obtained an expression for the ratio of the fluxions of $y$ and $x$ , from which we find

2y\dot{y}-a(x\dot{y}+\dot{y}x)+2bx\dot{x}=0,

and this is precisely the expression we should have obtained, had we taken the fluxion of each term of

y^3-axy+b x^3-c=0,

the proposed equation, and put the result equal to 0.

49. But to see that this will always be the case, whatever be the degree of the equation, we have only to observe, that, by the very same process employed to deduce from the original equation

y^3-axy+b x^3-c=0,

these two others

2yp-a(xp+y)+2bx=0,

2y\dot{y}-a(x\dot{y}+\dot{y}x)+2bx\dot{x}=0;

if we suppose the equation to be generally expressed thus,

y^l+a y^m x^n \dots + x^r+c=0,

where the exponents $l, m, n$ , and $r$ denote constant quantities, we shall obtain

\left. \begin{aligned} & ly^{l-1}p+a(m y^{m-1}x^n p+n y^m x^{n-1}) \\ & \dots + r x^{r-1} \end{aligned} \right\} =0,

and hence, by substituting for $p$ its value $\frac{\dot{y}}{x}$ , and bringing $\dot{x}$ from the denominator,

\left. \begin{aligned} & ly^{l-1}\dot{y}+a(m y^{m-1}x^n \dot{y}+n y^m x^{n-1} \dot{x}) \\ & \dots + r x^{r-1} \dot{x} \end{aligned} \right\} =0.

From which it appears that, when the relation between $x$ , a variable quantity, and $y$ , a function of that quantity, is expressed by an equation, the terms of which are brought all to one side, so as to produce an expression $=0$ ; the relation of the fluxions will be found, by taking the fluxion of each term of the equation (considering $y$ as a function of $x$ ), and putting the sum of these fluxions equal to 0.

50. Having from the equation

y^3-axy+b x^3-c=0,

found that

\dot{y} = \frac{(ay-2bx)\dot{x}}{2y-ax},

if it be required to find the second fluxion of $y$ , we have only to take the fluxion of the latter side of this equation,

Direct Method. equation, considering $x$ as constant, and $y$ as a function of $x$ ; thus we have

\dot{y} = \frac{\left\{ (2y-ax)(ay-2bx)\dot{x} \right\}}{(2y-ax)^2}

an equation which abbreviates to

\dot{y} = \frac{(4b-a^2)(ay-yx)\dot{x}}{(2y-ax)^2},

and from which we may exterminate $\dot{y}$ by means of the equation.

\dot{y} = \frac{(ay-2bx)\dot{x}}{2y-ax}.

By the same mode of proceeding we may determine the third or any higher fluxion of the function $y$ .

51. As far as we have yet gone in explaining the principles of fluxions, we have had continually occasion to employ the rule for finding the fluxion of the particular function $x^n$ , where $x$ denotes a variable quantity, and $n$ any constant number; and we may therefore, in respect of other functions, consider $x^n$ as a simple function. Besides the function $x^n$ , writers on Analysis have considered each of the following as also constituting a simple analytic function of a variable quantity; viz.

$a^x$ , where $a$ is constant, and $x$ is variable.

$\text{Log. } x$ , that is the logarithm of $x$ , a variable number.

$\text{Sin. } x$ , that is the sine of $x$ , a variable arch of a circle, radius being unity.

$\text{Cos. } x$ , that is the cosine of $x$ , a variable arch of a circle, radius being as before unity.

52. We have already found the fluxion of $x^n$ , and we proceed to find the fluxions of the other simple functions of $x$ ; and, as in the investigation of these we shall have occasion to employ the binomial theorem, it will be proper to show how that theorem may be deduced from the principles already explained. We are then to find the series that expresses $(a+x)^n$ , when $n$ is any number whatever. Or, since $(a+x)^n$ is equal to $a^n(1+v)^n$ ,

where $v$ denotes the fraction $\frac{x}{a}$ , we may leave the quantity $a^n$ out of consideration, as has been formerly observed, § 28, and seek the series that expresses $(1+v)^n$ .

As we have already pointed out (§ 28) the process of induction by which we may find the general form of the series, we shall not here repeat it, but assume

(1+v)^n = 1 + Av + Bv^2 + Cv^3 + Dv^4 + \&c.

where $A, B, C, D, \&c.$ denote numbers that are independent of $v$ .

Now, as the fluxion of a variable function must be the same, whether that function be expressed by one term, or developed into a series of terms; by performing the operation of taking the fluxion on each side of the above equation, the results must be equal, that is, § 26.

n(1+v)^{n-1}v = Av + 2Bv^2 + 3Cv^3 + 4Dv^4 + \&c.

or leaving out the quantity $v$ , common to each term,

n(1+v)^{n-1} = A + 2Bv + 3Cv^2 + 4Dv^3 + \&c.

Let both sides of this equation be multiplied by $1+v$ , and divided by $n$ , thus we shall have

(1+v)^n = \frac{1}{n} \left\{ A + 2Bv + 3Cv^2 + 4Dv^3 + \&c. \right\} + \frac{1}{n} \left\{ A + 2Bv + 3Cv^2 + 4Dv^3 + \&c. \right\}

Thus, by performing on the quantities the analytical process of taking their fluxions, we have obtained a new expression for $(1+v)^n$ . Let the quantities that are independent of $v$ in each expression be put equal to each other, and also the coefficients of like powers of $v$ ; thus we obtain

1 = \frac{A}{n}, \text{ and hence } A = n

A = \frac{A+2B}{n}, B = \frac{n-1}{2} A

B = \frac{2B+3C}{n}, C = \frac{n-2}{3} B

C = \frac{3C+4D}{n}, D = \frac{n-3}{4} C

\&c. \quad \&c.

Or, substituting successively the expression for each coefficient in that which follows it,

A = n,

B = \frac{n(n-1)}{2},

C = \frac{n(n-1)(n-2)}{2 \cdot 3},

D = \frac{n(n-1)(n-2)(n-3)}{2 \cdot 3 \cdot 4},

\&c.

Hence it appears that

(1+v)^n = 1 + nv + \frac{n(n-1)}{2}v^2 + \frac{n(n-1)(n-2)}{2 \cdot 3}v^3 + \frac{n(n-1)(n-2)(n-3)}{2 \cdot 3 \cdot 4}v^4 + \&c.

and therefore, substituting $\frac{x}{a}$ for $v$ , and multiplying by $a^n$ ,

(a+x)^n = a^n + na^{n-1}x + \frac{n(n-1)}{2}a^{n-2}x^2 + \frac{n(n-1)(n-2)}{2 \cdot 3}a^{n-3}x^3 + \&c.

where the law of continuation is evident.

53. We now proceed to investigate the fluxion of the function $u = a^x$ , $a$ being supposed constant, and $x$ the variable quantity, to which the function is referred.

4 X 2 Let

Direct Method. Let $x$ be supposed, as formerly, to change its value, and to become $x+h$ , and put $u'$ for the new value that the function acquires by this change in the magnitude of $x$ , then we have

u' = a^{x+h} = a^x \times a^h,

and, taking the difference between the two succeeding values,

u' - u = a^x \times a^h - a^x = a^x(a^h - 1).

We must now develop the expression $a^h - 1$ into a series, the terms of which are arranged according to the successive powers of the increment $h$ . To effect this, let us put $b = a - 1$ , so that $a = 1 + b$ , and $a^h = (1 + b)^h$ ; but by the binomial theorem, this last expression may be expanded into the following series:

1 + hb + \frac{h(h-1)}{2} b^2 + \frac{h(h-1)(h-2)}{2 \cdot 3} b^3 + \dots

Therefore,

a^h = 1 + hb + \frac{h(h-1)}{2} b^2 + \frac{h(h-1)(h-2)}{2 \cdot 3} b^3 + \dots

As the terms of this series are not arranged according to the powers of $h$ , but according to the powers of $b$ , it is necessary that we transform it into another having the required form; now this may be effected by actually multiplying together all the factors that constitute each term, and arranging the series anew in such a manner, that each of its terms may be a power of $h$ , multiplied by a coefficient composed only of the powers of $b$ , and given numbers;

Accordingly we have

hb = b h,

\frac{h(h-1)}{2} b^2 = -\frac{b^2}{2} h + \frac{b^2}{2} h^2,

\frac{h(h-1)(h-2)}{2 \cdot 3} b^3 = \frac{b^3}{3} h - \frac{b^3}{2} h^2 + \frac{b^3}{6} h^3,

&c.

Therefore, by taking the sum of all the quantities on each side of these equations, we get the series,

1 + hb + \frac{h(h-1)}{2} b^2 + \frac{h(h-1)(h-2)}{2 \cdot 3} b^3 + \dots

otherwise expressed thus,

1 + Ah + Bh^2 + Ch^3 + \dots

where $A$ is equal to the infinite series $b - \frac{b^2}{2} + \frac{b^3}{3}$

- \frac{b^4}{4} + \dots

that is, to

\frac{a-1}{1} - \frac{(a-1)^2}{2} + \frac{(a-1)^3}{3} - \frac{(a-1)^4}{4} + \dots

and $B, C, \dots$ are also quantities composed of the powers of $b$ , and consequently are independent of $h$ ; but as these all disappear in the course of the investigation, it is not necessary to express them otherwise,

than by a general symbol. Therefore, we have now got

a^h = 1 + Ah + Bh^2 + Ch^3 + \dots

and consequently,

u' - u = a^x(a^h - 1) = a^x(Ah + Bh^2 + Ch^3 + \dots)

and

\frac{u' - u}{h} = Aa^x + Ba^x h + Ca^x h^2 + \dots

Hence, when $h$ is conceived to be continually diminished, we have the limit of $\frac{u' - u}{h}$ expressed by $Aa^x$ , and therefore, § 21,

\frac{\dot{u}}{\dot{x}} = Aa^x, \text{ and } \dot{u} = Aa^x \dot{x}.

54. In the preceding investigation, we have had occasion to develop the exponential expression $a^h$ into a series of this form,

1 + Ah + Bh^2 + Ch^3 + \dots

that is, a series the terms of which are the successive powers of the exponent, each multiplied by a coefficient, which is independent of the exponent.

We have however only determined the coefficients of the first two terms of the series, these being the only ones we had occasion to employ.

The result of the investigation however may be applied to determine all the coefficients by the very same kind of process as that which we have employed in § 53, to determine the coefficients of the terms of the series which constitutes the other expansion of $a^h$ .

Instead of denoting the exponent by $h$ , let us consider it as a variable quantity, and express it by $x$ , then, from what has been shewn it appears that

a^x = 1 + Ax + Bx^2 + Cx^3 + Dx^4 + \dots

where $A, B, C, \dots$ express constant quantities. Let the operation of taking the fluxions be now performed on both sides of this equation, (observing that the fluxion of $a^x$ is $Aa^x \dot{x}$ ) and let all the terms be divided by $\dot{x}$ , which is common to each, thus we obtain

Aa^x = A + 2Bx + 3Cx^2 + 4Dx^3 + \dots

and, dividing by $A$ ,

a^x = 1 + 2 \frac{B}{A} x + 3 \frac{C}{A} x^2 + 4 \frac{D}{A} x^3 + \dots

Let the coefficients of the same powers of $x$ in each of the two series expressing $a^x$ be put equal to each other, then,

\begin{aligned} \frac{2B}{A} &= A, \text{ hence, } B = \frac{A^2}{2}, \\ \frac{3C}{A} &= B, \dots C = \frac{A^3}{2 \cdot 3}, \\ \frac{4D}{A} &= C, \dots D = \frac{A^4}{2 \cdot 3 \cdot 4}, \\ \&c. & \quad \&c. \end{aligned}

Therefore,

Direct Method. Therefore, substituting these values of B, C, D, &c. in the original series, we have

a^x = 1 + Ax + \frac{A^2x^2}{2} + \frac{A^3x^3}{2 \cdot 3} + \frac{A^4x^4}{2 \cdot 3 \cdot 4} + \&c.

the same result as has been found in the article ALGEBRA, § 293. by proceeding in a different manner.

55. If we suppose $x=1$ , then the preceding equation becomes

a = 1 + A + \frac{A^2}{2} + \frac{A^3}{2 \cdot 3} + \frac{A^4}{2 \cdot 3 \cdot 4} + \&c.

and if we suppose $x = \frac{1}{A}$ , it becomes

a^{\frac{1}{A}} = 1 + 1 + \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \&c.

thus the quantity $a^{\frac{1}{A}}$ is equal to a constant number, which is the value of $a$ when $A=1$ , and which, by taking the sum of the first ten terms of the series, is found to be 2.7182818, or, by taking the sum of a greater number of terms, more accurately

2.718281828459045 \dots

We shall, in the remainder of this treatise, denote this

number always by $e$ , then $a^{\frac{1}{A}} = e$ , and $a = e^A$ , and taking the logarithms, $\log. a = A \times \log. e$ , hence

A = \frac{\log. a}{\log. e}.

56. If we now substitute this value of $A$ in the expression for the fluxion of $u=a^x$ , found in § 53, it becomes

\dot{u} = \frac{\log. a}{\log. e} a^x \dot{x}.

Hence it appears, that the fluxion of the function $a^x$ is equal to the fluxion of $x$ multiplied by the function itself, and by the quotient arising from the division of the logarithm of $a$ by the logarithm of $e$ , where $e$ denotes

$1 + \frac{1}{1} + \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \&c.$ a series whose sum is 2.7182818 nearly.

57. Let us now consider the third simple function of $x$ , namely $u = \log. x$ . Let $a$ be the radical number of the particular system, in which $u$ is a logarithm, and $x$ the corresponding number; then from the nature of logarithms (see ALGEBRA, § 277.) we have $a^x = x$ . Now, whether we consider $u$ as a function of $x$ , or $x$ as a function of $u$ , the limiting ratio of their simultaneous increments, and consequently the ratio of their fluxions will be the very same. But by considering $x$ as a function of $u$ , we have immediately, from what has been shown in § 53, and § 55,

\frac{\dot{x}}{\dot{u}} = A a^u = A x,

and therefore, $u = \frac{1}{A} \frac{\dot{x}}{x} = \frac{\log. e \dot{x}}{\log. a x}$ , but as $a$ is the radical number of the system, $\log. a = 1$ , therefore

\dot{u} = \log. e \frac{\dot{x}}{x}.

The number which we have denoted by $e$ occurs very often in analytical investigations; it is the radical number of the system of logarithms first invented by Baron Napier, and called by some writers Hyperbolic logarithms, but by others, with more propriety, Napierian logarithms.

The expression $\frac{\log. e}{\log. a}$ is called the modulus of the system of logarithms whose radical number is $a$ . In the Napierian system $\frac{\log. e}{\log. a} = \frac{\log. e}{\log. e} = 1$ , that is, the modulus is unity; but in the common system, or that in which $a=10$ , the modulus $\frac{\log. e}{\log. a} = 0.434294482$ . The rule for finding the fluxion of the logarithm of a variable quantity may now be expressed thus:

Multiply the fluxion of the variable quantity by the modulus of the system, and divide the product by the variable quantity itself, the result is the fluxion required.

58. By the application of the rule for finding the fluxion of the logarithm of a variable quantity, we may readily find the fluxions of exponential functions in general. Thus, for example, if $u=x^y$ , $x$ and $y$ being both functions of any variable quantity $x$ , then $\log. u = y \times \log. x$ ; and taking the fluxions (considering $y \times \log. x$ as the product of two functions $y$ and $\log. x$ , and proceeding by the rules of § 37 and last),

\frac{\dot{u}}{u} = \dot{y} \log. x + y \frac{\dot{x}}{x},

and hence

\dot{u} = u (\dot{y} \log. x + y \frac{\dot{x}}{x}),

= x^y (\dot{y} \log. x + y \frac{\dot{x}}{x}).

59. We are next to consider the functions $u = \sin. x$ , and $u = \cos. x$ .

Suppose $x$ to change its value, and to become $x+h$ , and $u$ to become $u'$ , then, since

u = \sin. x, \text{ and } u' = \sin. (x+h),

u' - u = \sin. (x+h) - \sin. x;

but by the arithmetic of sines (see ALGEBRA, § 353), $\sin. (x+h) = \sin. x \cos. h + \cos. x \sin. h$ , therefore,

\begin{aligned} u' - u &= \sin. x \cos. h + \cos. x \sin. h - \sin. x \\ &= \cos. x \sin. h - \sin. x (1 - \cos. h). \end{aligned}

In this case, as when treating formerly of other functions, we might consider the above expression for $u' - u$ , as resolvable into a series $p h + q h^2 + \&c.$ proceeding by the powers of the increment, and thence we might

might find the limit of $\frac{u'-u}{h}$ as before. But we may discover the limit otherwise, by proceeding as follows: Because

\sin^2 h = 1 - \cos^2 h = (1 + \cos h)(1 - \cos h)

\therefore 1 - \cos h = \frac{\sin^2 h}{1 + \cos h};

Let this value of $1 - \cos h$ be substituted in the expression for $u' - u$ , and it becomes

u' - u = \cos x \sin h - \frac{\sin x \sin^2 h}{1 + \cos h};

And hence, dividing by $h$ , and arranging the terms so as to exhibit the ratio $\frac{\sin h}{h}$ , we get

\frac{u' - u}{h} = \frac{\sin h}{h} \left\{ \cos x - \frac{\sin h \sin x}{1 + \cos h} \right\}.

Conceive now $h$ to be continually diminished, and we shall have the limit of $\frac{u' - u}{h}$ equal to the limit of $\frac{\sin h}{h}$ multiplied by the limit of the following expression

\cos x - \frac{\sin h \sin x}{1 + \cos h}.

Now, the sine of an arch being less than the arch itself, we have $\frac{\sin h}{h} < 1$ . Again, the arch being less than its tangent, $\frac{\sin h}{h} > \frac{\sin h}{\tan h}$ ; but $\tan h = \frac{\sin h}{\cos h}$ , and therefore $\frac{\sin h}{\tan h} = \cos h$ ; consequently $\frac{\sin h}{h} > \cos h$ . Hence it appears, that the expression for the ratio $\frac{\sin h}{h}$ is less than 1, or radius, but greater than $\cos h$ . But $h$ being conceived to be continually diminished, $\cos h$ continually approaches to 1, and may come nearer to it than by any assignable difference; therefore, the limit of $\frac{\sin h}{h}$ is 1. As to the other expression, $\cos x - \frac{\sin h \sin x}{1 + \cos h}$ ; when $h$ is supposed to be continually diminished, its second term, to wit, $\frac{\sin h \sin x}{1 + \cos h}$ may become less than any assignable quantity; therefore the limit of the expression is simply $\cos x$ : thus, upon the whole we have found that the limit of $\frac{u' - u}{h}$ is $\cos x$ , and therefore

\frac{u'}{x} = \cos x, \text{ and } \dot{u} = \dot{x} \cos x.

The fluxion of the other function, $u = \cos x$ , is easily deduced from that which we have just found, by proceeding thus:

Put $c$ to denote a quadrant, then $\cos x = \sin(c - x)$ , and therefore

u = \sin(c - x).

Now, it has been just shewn that

flux. of $\sin(c - x) = \cos(c - x) \times \text{flux. of } (c - x)$ but $\cos(c - x) = \sin x$ , and the fluxion of $c - x$ is $-\dot{x}$ , therefore

\dot{u} = -\dot{x} \sin x.

Thus it appears, that the fluxion of the sine of a variable arch is equal to the fluxion of the arch multiplied by its cosine; and that the fluxion of the cosine is equal to the fluxion of the arch (taken with a negative sign) multiplied by the sine.

60. We can now very readily find the fluxion of any other function of an arch of a circle. Thus, suppose $u = \tan x$ ; then, because $\tan x = \frac{\sin x}{\cos x}$ , we have $u = \frac{\sin x}{\cos x}$ . This expression being considered as a fractional function of $x$ , we have, by § 39, and what has been just now shewn,

\begin{aligned} \dot{u} &= \frac{\dot{x} \cos^2 x + x \sin^2 x}{\cos^2 x} \\ &= \frac{\dot{x} (\cos^2 x + \sin^2 x)}{\cos^2 x}; \end{aligned}

or, since $\cos^2 x + \sin^2 x = 1$ , and $\frac{1}{\cos^2 x} = \sec^2 x$ ,

\dot{u} = \frac{\dot{x}}{\cos^2 x} = \dot{x} \sec^2 x.

\text{Hence also we have } \dot{x} = \frac{\dot{u}}{\sec^2 x} = \frac{\dot{u}}{1 + u^2}.

In like manner, if we suppose $u = \sec x$ , then, because $\sec x = \frac{1}{\cos x}$ , we have $u = \frac{1}{\cos x}$ , and

\dot{u} = \frac{\dot{x} \sin x}{\cos^2 x},

or, since $\frac{\sin x}{\cos x} = \tan x$ , and $\frac{1}{\cos^2 x} = \sec^2 x$ ,

\dot{u} = \dot{x} \tan x \sec x.

Proceeding in this manner, we find that when $u = \cotan x$ , then

\dot{u} = \frac{-\dot{x}}{\tan^2 x \cos^2 x} = \frac{-\dot{x}}{\sin^2 x};

And when $u = \text{cosec } x$ , then

\dot{u} = \frac{-\dot{x} \cos x}{\sin^2 x} = -\dot{x} \cotan x \text{ cosec } x.

61. Let us now consider the fluxions of geometrical magnitudes: And first let it be required to find the expression for the fluxion of BDPC the area bounded by CP, a curve line, and by CB, PD, the ordinates at its extremities, and BD, the portion of AE, the line of the abscissas, which lies between those ordinates. Let the numerical measures of AD and PD, the co-ordinates at the point P, be denoted by $x$ and $y$ , and the numerical measure of the area BDPC by $s$ ; then

Direct Method. then $y$ and $s$ may both be considered as functions of the abscissa $x$ .

Let $x$ , or $AD$ , be supposed to change its value, and to become $AD'$ , and let $D'P'$ , and $BD'P'C$ be the corresponding new values of $y$ and $s$ ; then $DD'$ , and $DD'P'P$ will be the geometrical expressions for the simultaneous increments of the abscissa and area. But, as one of these quantities is a line, and the other a space, they cannot be compared in respect of their ratio. Therefore, let us consider $a$ as denoting a line whose numerical value is unity, and then the numerical values of the increments of the abscissa and area may be considered as analogous to the geometrical quantities $DD' \times a$ , and the area $DD'P'P$ respectively, which quantities being homogeneous may now be compared with each other. We are now to investigate the limit of $\frac{\text{area } DD'P'P}{a \times DD'}$ , the general expression for the ratio of the increments of $s$ and $x$ . Draw $PM$ and $P'N$ parallel to $AE$ , meeting the ordinates in $M$ and $N$ . The curvilinear area $DD'P'P$ is greater than the rectangle $DD'MP$ , that is, greater than $PD \times DD'$ ; but less than the rectangle $DD'PN$ , that is, less than $P'D' \times DD'$ , therefore

\frac{\text{area } DD'P'P}{a \times DD'} > \frac{PD \times DD'}{a \times DD'} > \frac{PD}{a},

\text{and } \frac{\text{area } DD'P'P}{a \times DD'} < \frac{P'D' \times DD'}{a \times DD'} < \frac{P'D'}{a}.

But the increments being supposed to be continually diminished, $\frac{PD}{a}$ is the limit of $\frac{P'D'}{a}$ , therefore $\frac{PD}{a}$ is also the limit of $\frac{\text{area } DD'P'P}{a \times DD'}$ , and hence (§ 21.) $\frac{\dot{s}}{\dot{x}} = \frac{PD}{a} = \frac{y}{1} = y$ , and $\dot{s} = y\dot{x}$ .

That is, the fluxion of a curvilinear area is equal to the product of the ordinate, and the fluxion of the abscissa.

62. Before we proceed to investigate the expression for the fluxion of an arch of a curve, it is necessary that we should inquire what is the limited ratio of an arch of a curve to its chord.

Let $APB$ be any curve line, all the parts of which are concave towards its chord $AP$ . Let $AQ, QP$ be tangents at the extremities of the arch, and let $apq$ be a triangle similar to $APQ$ , but having its base $ap$ of a given magnitude, then

AQ + QP : AP :: ap + qp : ap.

Suppose now the point $P$ to approach to $A$ , then the angles at $A$ and $P$ , and consequently the angles at $a$ and $p$ , which are equal to them, will decrease, and may become less than any assignable angles; therefore, the limit of the ratio of $ap + qp$ to $ap$ is evidently a ratio of equality; hence also the limit of the ratio of $AQ + QP$ to $AP$ is the ratio of equality; and since the arch $AP$ is less than $AQ + QP$ , but greater than its chord $AP$ , the limit of the arch $AP$ to its chord $AP$ must also be the ratio of equality.

63. We proceed now to find the fluxion of an arch of a curve. Let $APP'$ be a curve line of any kind, and $AB, BP$ any two co-ordinates at a point $P$ in the curve. Put $x$ for $AB$ , the abscissa, $y$ for $BP$ , the ordinate, and $z$ for the curve line $AP$ , then $z$ and $y$ may be considered as each a function of $x$ . Draw $P'B'$ another ordinate, and draw $PM$ parallel to $AB$ , meeting $P'P'$ in $M$ , and draw the chord $PP'$ ; then $PM, MP'$ , and the arch $PP'$ are the simultaneous increments of $x, y$ , and $z$ respectively. Now we have

\frac{\text{arch } PP'}{PM} = \frac{\text{arch } PP'}{\text{chord } PP'} \times \frac{\text{chord } PP'}{PM}

But $\text{chord } PP' = \sqrt{(PM^2 + MP'^2)} = PM \sqrt{(1 + \frac{MP'^2}{PM^2})}$ ; therefore,

\frac{\text{arch } PP'}{PM} = \frac{\text{arch } PP'}{\text{chord } PP'} \times \sqrt{(1 + \frac{MP'^2}{PM^2})}.

Suppose now the increments to be continually diminished, then, as $\frac{\dot{z}}{\dot{x}} = \text{limit of } \frac{\text{arch } PP'}{PM}$ , and $\frac{\dot{y}^2}{\dot{x}^2} = \text{limit of } \frac{MP'^2}{PM^2}$ (§ 21.), and $1 = \text{limit of } \frac{\text{arch } PP'}{\text{chord } PP'}$ (last §) we have

\frac{\dot{z}}{\dot{x}} = \sqrt{(1 + \frac{\dot{y}^2}{\dot{x}^2})}, \text{ and } \dot{z} = \sqrt{(\dot{x}^2 + \dot{y}^2)}.

Hence it appears that the square of the fluxion of a curve line of any kind is equal to the sum of the squares of the fluxions of the co-ordinates.

64. The expression for the fluxion of a solid may be found by the same mode of reasoning as that which we have employed, § 61, to find the fluxion of a curvilinear area. Let $APQp$ be a portion of a solid generated by the revolution of $APB$ , a curve line, about $AC$ , a line taken in the plane of the curve, as an axis. Let $PD, p, P'D', p'$ be the lines in which $BA, b$ , a plane passing along the axis $AC$ , meets $PQ, p, P'Q', p'$ , the planes of two circles formed by sections of the solid perpendicular to its axis. Draw $PM$ and $P'N$ parallel to $AD$ . Put $AD = x, DP = y$ , let $s$ denote the solid $APQp$ , having $y$ for the radius of its circular base, and $x$ for its altitude; put $\pi$ for the number $3.14159\dots$ , viz. the circumference of a circle having its diameter $= 1$ , and let $a$ denote an area, having its numerical measure expressed by unity; then $DD'$ , or $a \times DD'$ being considered as the increment of $x$ , the portion of the solid comprehended between the parallel planes $PQ, p, P'Q', p'$ will be the corresponding increment of $s$ , which we are

to consider as a function of $x$ ; hence (§ 21.) $\frac{\dot{s}}{\dot{x}}$ is equal to the limiting ratio of the portion of the solid, comprehended between the planes $PQ, p$ , and $P'Q', p'$ to the solid $a \times DD'$ . But the former of these solids being evidently greater than a cylinder $Pm$ , having the circle $PA, p$ for its base, and $DD'$ for its altitude, that is greater than $\pi PD^2 \times DD'$ , and less than a cylinder $Np'$ , having the circle $P'Q', p'$ for its base, and $DD'$ for its altitude, that is less than $\pi P'D'^2 \times DD'$ ; it

Direct Method. it follows, that as long as $DD'$ has an assignable magnitude,

\begin{aligned} \frac{\dot{s}}{x} & > \pi PD' \times DD' \times \frac{1}{a \times DD'} \\ & > \frac{\pi PD'}{a}; \\ \text{and } \frac{\dot{s}}{x} & < \pi P'D' \times DD' \times \frac{1}{a \times DD'} \\ & < \frac{\pi P'D'}{a}; \end{aligned}

but the increment $DD'$ being continually diminished, $\frac{\pi P'D'}{a}$ , the greater limit of $\frac{\dot{s}}{x}$ , approaches continually to its lesser limit $\frac{\pi PD'}{a} = \frac{\pi y^2}{a} = (\text{because } a=1) \pi y^2$ , so as to come nearer to it than by any assignable difference, therefore $\frac{\dot{s}}{x} = \pi y^2$ , and $\dot{s} = \pi y^2 x$ . Now, if we observe that $\pi y^2$ is the area of the circle $PQp$ , it will appear, that the fluxion of a solid generated by the revolution of a curve about its axis is equal to the fluxion of the axis multiplied by the general expression for the area of a circle formed by supposing the solid to be cut by a plane perpendicular to its axis.

65. To find the fluxion of the surface of the solid, let us denote that surface by $s$ , and let $x$ and $y$ denote as before; then the surface contained between the circles $PQp$ and $P'Q'p'$ will be the increment of $s$ , corresponding to $DD'$ the increment of $x$ . Draw the chord $PP'$ ; then the curve line $PP'$ being supposed to revolve about the axis $AC$ , and thus to generate the increment of the surface of the solid, the chord $PP'$ will generate at the same time the convex surface of a frustum of a cone; now the limiting ratio of the curve line $PP'$ to its chord $PP'$ being the ratio of equality, the limiting ratio of the surfaces generated by the revolution of those lines will also be the ratio of equality;

therefore $\frac{\dot{s}}{x}$ , which is equal to the limit of

\frac{\text{surf. gener. by arch } PP'}{x \times DD'}

will also be equal to the limit of

\frac{\text{surf. gener. by chord } PP'}{DD'};

but the convex surface of a frustum of a cone is equal to the product of its slant side into half the sum of the circumferences of its two bases (see GEOMETRY), and in the present case these circumferences are equal to $2 PD \times \pi$ , and $2 P'D' \times \pi$ , therefore $\frac{\dot{s}}{x}$ is equal to the limit

\frac{\pi(PD + P'D') PP'}{DD'} = \pi(PD + P'D') \frac{PP'}{DD'};

but the point $D'$ being supposed to approach to $D$ , the

limit of $PD + P'D'$ , will manifestly be $2 PD = 2y$ ; and since $\frac{PP'}{DD'} = \frac{(\sqrt{DD'^2 + P'M^2})}{DD'} = \sqrt{1 + \frac{P'M^2}{DD'^2}}$ , the limit of this expression (if we consider that $P'M$ and $DD'$ are the simultaneous increments of $y$ and $x$ ) is evidently equal to $\sqrt{1 + \frac{y^2}{x^2}}$ , therefore

\frac{\dot{s}}{x} = 2\pi y \sqrt{1 + \frac{y^2}{x^2}},

and consequently

\dot{s} = 2\pi y \sqrt{(x^2 + y^2)}.

If we now observe that $2\pi y$ is the circumference of the circle $PQp$ , and $\sqrt{(x^2 + y^2)}$ is the fluxion of the curve line $AP$ , § 63, it will appear, that the fluxion of the surface of a solid generated by the revolution of a curve about its axis is equal to the fluxion of the curve line multiplied by the general expression for the circumference of a circle formed by supposing the curve to be cut by a plane perpendicular to its axis.

SECT. III. The Application of the Direct Method of Fluxions.

HAVING explained the principles of the direct method of fluxions at as great a length as we think suitable to the work of which this treatise forms a part, we proceed to shew how the calculus may be applied to the resolution of some general problems in Analysis and Geometry.

Investigation of a general formula for expanding a Function into a Series.

66. In treating of the principles of the method of fluxions, we have, from an examination of particular functions, inferred by induction, that $u$ being any function of a variable quantity $x$ , which was either actually expressed, or capable of being expressed by a combination of the powers of $x$ , then, $x$ being supposed to change its value, and to become $x+h$ , the new value which the function $u$ will acquire when $x+h$ is substituted in it instead of $x$ will always be capable of being expanded into a series of this form,

u + p h + q h^2 + r h^3 + \&c.

where $p, q, \&c.$ denote functions of $x$ that are quite independent of $h$ .

We have shewn that, from the particular form of this development, it happens that the ratio of $p h + q h^2 + r h^3 + \&c.$ the increment of the function, to $h$ the increment of the variable quantity $x$ itself, admits of a limit, which is always expressed by $p$ , the coefficient of its second term; and as we have defined this limit to be the expression for the ratio of the fluxions of $u$ and $x$ , so that $p = \frac{\dot{u}}{\dot{x}}$ , the new value of the function may also be expressed thus

u + \frac{\dot{u}}{\dot{x}} h + q h^2 + r h^3 + \&c.

And

Direct Method. And this expression may be considered as indicating not only the general form of the series, but also the particular relation subsisting between $u$ , the original function, and $p$ , the coefficient of the second term of the series, the latter being in every case that function of $x$ which results from the operation of taking the fluxion of the former, and dividing by $x$ .

We are now to investigate the relation that subsists between each of the remaining coefficients and the original function.

67. First let us suppose the function $u$ to have the particular form $x^n$ , $n$ being a constant number. Then $x$ changing its value to $x+h$ , $u$ changes to $u'=(x+h)^n$ , therefore, by the binomial theorem (§ 52.)

u' = x^n + n x^{n-1} h + \frac{n(n-1)}{2} x^{n-2} h^2 + \frac{n(n-1)(n-2)}{2 \cdot 3} x^{n-3} h^3 + \&c.

But since $u=x^n$ , by taking the successive fluxions of $u$ , and considering $x$ as constant, we have

\begin{aligned} \dot{u} &= n x^{n-1}, \\ \ddot{u} &= n(n-1) x^{n-2}, \\ \ddot{\dot{u}} &= n(n-1)(n-2) x^{n-3}, \\ \ddot{\dot{\dot{u}}} &= n(n-1)(n-2)(n-3) x^{n-4}, \\ &\&c. \end{aligned}

Let $u$ , $\dot{u}$ , $\ddot{u}$ , $\ddot{\dot{u}}$ , $\ddot{\dot{\dot{u}}}$ , $\&c.$ be now substituted for $x^n$ , $n x^{n-1}$ , $n(n-1) x^{n-2}$ , $\&c.$ , respectively, in the series for $u'$ , and we have

u' = u + \dot{u} h + \frac{\ddot{u}}{x^2} \frac{h^2}{2} + \frac{\ddot{\dot{u}}}{x^3} \frac{h^3}{2 \cdot 3} + \frac{\ddot{\dot{\dot{u}}}}{x^4} \frac{h^4}{2 \cdot 3 \cdot 4} + \&c.

68. This manner of expressing the development of $u'$ , or $(x+h)^n$ , indicates directly the relation that each of the coefficients of the successive powers of $h$ has to the original function.

The first term of the series is the original function $u$ , or $x^n$ , itself, or it is what the function $(x+h)^n$ becomes upon the supposition that $h=0$ . The second term is $h$ , or $\frac{h}{x}$ , multiplied by the coefficient $\dot{u}$ , which coefficient is a function of $x$ derived from the original function by the operation of taking its fluxion, and dividing the result by $x$ . The third term is $\frac{h^2}{1 \cdot 2}$ multiplied by the coefficient $\frac{\ddot{u}}{x^2}$ , that is, by a function of $x$ derived from the preceding coefficient $\frac{\dot{u}}{x}$ by the same

operation as that coefficient was derived from the original function, namely by taking the fluxion of $\frac{\dot{u}}{x}$ , considering $x$ as constant, and dividing by $x$ . The fourth term is $\frac{h^3}{1 \cdot 2 \cdot 3}$ multiplied by $\frac{\ddot{\dot{u}}}{x^3}$ , that is, by a function of $x$ deduced from the third coefficient by the very same operation as that by which the third was derived from the second, or the second from the first. And so on with respect to all the other terms of the series, the $n$ th term being the product of $\frac{h^{n-1}}{1 \cdot 2 \cdot 3 \dots (n-1)}$ , and the $(n-1)$ th fluxion of the function $u$ divided by $x^{n-1}$ .

69. Let us now suppose that $u$ denotes any other function of $x$ , then, whatever be its nature, it may always be conceived as capable of being expressed by a series, the terms of which are powers of $x$ , in this manner;

A x^a + B x^b + C x^c + D x^d + \&c.

where $A, B, C, \&c.$ $a, b, c, \&c.$ denote constant numbers. Thus we have

u = A x^a + B x^b + C x^c + \&c.

Then, $x$ being supposed to become $x+h$ , and (in consequence of the change in the value of $x$ ) $u$ to become $u'$ , we have

u' = A(x+h)^a + B(x+h)^b + C(x+h)^c + \&c.

Let us now denote $A x^a$ by $P$ , $B x^b$ by $Q$ , $C x^c$ by $R$ , $\&c.$ then by last §

\begin{aligned} A(x+h)^a &= P + \dot{P} h + \frac{\ddot{P}}{x^2} \frac{h^2}{2} + \frac{\ddot{\dot{P}}}{x^3} \frac{h^3}{2 \cdot 3} + \&c. \\ B(x+h)^b &= Q + \dot{Q} h + \frac{\ddot{Q}}{x^2} \frac{h^2}{2} + \frac{\ddot{\dot{Q}}}{x^3} \frac{h^3}{2 \cdot 3} + \&c. \\ C(x+h)^c &= R + \dot{R} h + \frac{\ddot{R}}{x^2} \frac{h^2}{2} + \frac{\ddot{\dot{R}}}{x^3} \frac{h^3}{2 \cdot 3} + \&c. \\ &\&c. \end{aligned}

Therefore, substituting these developments in the series expressing $u'$ ,

u' = \begin{cases} P + Q + R + \&c. \\ + \left( \dot{P} + \dot{Q} + \dot{R} + \&c. \right) h \\ + \left( \frac{\ddot{P}}{x^2} + \frac{\ddot{Q}}{x^2} + \frac{\ddot{R}}{x^2} + \&c. \right) \frac{h^2}{2} \\ + \left( \frac{\ddot{\dot{P}}}{x^3} + \frac{\ddot{\dot{Q}}}{x^3} + \frac{\ddot{\dot{R}}}{x^3} + \&c. \right) \frac{h^3}{2 \cdot 3} \\ + \&c. \end{cases}

u = P + Q + R + \dots

\frac{\dot{u}}{x} = \frac{\dot{P}}{x} + \frac{\dot{Q}}{x} + \frac{\dot{R}}{x} + \dots

\frac{\ddot{u}}{x^2} = \frac{\ddot{P}}{x^2} + \frac{\ddot{Q}}{x^2} + \frac{\ddot{R}}{x^2} + \dots

&c.

Therefore, substituting $u$ , $\frac{\dot{u}}{x}$ , &c. for the series to which they are respectively equal,

u' = u + \frac{\dot{u}}{x} \frac{h}{1} + \frac{\ddot{u}}{x^2} \frac{h^2}{1.2} + \frac{\ddot{u}}{x^3} \frac{h^3}{1.2.3} + \dots

Hence it appears that $u$ being any function of $x$ whatever, if $x+h$ be substituted in that function instead of $x$ , the series expressing the development of this new value of the function will have the general properties which have been shewn, in last §, to belong to it in the case of the function having the particular value $x^n$ .

The very general theorem which we have just now investigated is one of the most elegant and important in analysis. It was first published by Dr Brooke Taylor in a work entitled Methodus Incrementorum, which made its appearance about the year 1716. The theorem itself is generally known by the name of Taylor's theorem. It is more general than the celebrated Binomial theorem, inasmuch as this last, and innumerable others, are comprehended in it as particular cases.

70. We shall now give some examples to shew the manner of applying Taylor's theorem, as well as its great utility as an instrument of analysis.

Example 1. Suppose $u = a^x$ , $a$ being constant and $x$ variable. Then $x$ becoming $x+h$ , $u$ becomes $u' = a^{x+h}$ . Now from the equation $u = a^x$ we derive (§ 56.)

\frac{\dot{u}}{x} = A a^x, \quad (\text{here } A \text{ denotes } \frac{\log a}{\log e}).

Again, considering $x$ as constant, and repeating the operation of taking the fluxion on $\frac{\dot{u}}{x} = A a^x$ , we get $\frac{\ddot{u}}{x^2} = A^2 a^x$ , and hence

\text{again } \frac{\ddot{u}}{x^3} = A^3 a^x, \quad \&c. \quad \text{Therefore, substituting for}

u', u, \frac{\dot{u}}{x}, \frac{\ddot{u}}{x^2}, \&c. \text{ their values in the general theorem}

u' = u + \frac{\dot{u}}{x} h + \frac{\ddot{u}}{x^2} \frac{h^2}{2} + \dots \text{ it becomes}

a^{x+h} = a^x \left( 1 + A h + \frac{A^2}{2} h^2 + \frac{A^3}{2.3} h^3 + \dots \right)

Suppose now, that $x=0$ , then, as in this case $a^x=1$ , we have

a^h = 1 + A h + \frac{A^2}{2} h^2 + \frac{A^3}{2.3} h^3 + \dots

or, exchanging $h$ for $x$ ,

a^x = 1 + A x + \frac{A^2 x^2}{2} + \frac{A^3 x^3}{2.3} + \dots

the same result as we formerly obtained in § 54.

Ex. 2. Suppose $u = \log x$ . Then, $x$ becoming $x+h$ , $u$ becomes $u' = \log(x+h)$ . Now from the equation $u = \log x$ , we find (by § 57.) $\frac{\dot{u}}{x} = \frac{M}{x}$ .

Here $M$ denotes the modulus of the system. Again, supposing $\dot{x}$ constant, we find by § 26, $\frac{\ddot{u}}{x^2} = -\frac{M}{x^2}$ , $\frac{\ddot{u}}{x^3} = -\frac{M}{x^3}$ .

$\therefore \frac{\ddot{u}}{x^4} = -\frac{2.3M}{x^4}$ &c. Therefore, substituting as before the values of $u'$ , $u$ , $\frac{\dot{u}}{x}$ , &c. in the general for-

\text{mula } u' = u + \frac{\dot{u}}{x} h + \frac{\ddot{u}}{x^2} \frac{h^2}{2} + \dots \text{ it becomes}

\log(x+h) = \log x + \frac{M}{x} h - \frac{M}{2x^2} h^2 + \frac{M}{3x^3} h^3 - \dots

If we suppose $x=1$ , and change $h$ into $y$ , we have, because $\log x = \log 1 = 0$ ,

\log(1+y) = M \left( y - \frac{y^2}{2} + \frac{y^3}{3} - \dots \right)

For the particular method of applying these two series to the calculation of logarithms, see ALGEBRA, § 285 to § 291. See also LOGARITHMS.

Ex. 3. Suppose now $u = \sin x$ . Then $u' = \sin(x+h)$ . From $u = \sin x$ , by the application of the rule in § 59, we deduce $\frac{\dot{u}}{x} = \cos x$ , $\frac{\ddot{u}}{x^2} = -\sin x$ , $\frac{\ddot{u}}{x^3} = -\cos x$ , $\frac{\ddot{u}}{x^4} = \sin x$ , &c.

Therefore, substituting for $u'$ , $u$ , $\frac{\dot{u}}{x}$ , &c. their values in the general formula as before, we have

\begin{aligned} \sin(x+h) &= \sin x + \cos x \frac{h}{1} - \sin x \frac{h^2}{1.2} \\ &\quad - \cos x \frac{h^3}{1.2.3} + \sin x \frac{h^4}{1.2.3.4} + \dots \end{aligned}

or $\sin(x+h)$ is equal to

\begin{aligned} \sin x \left( 1 - \frac{h^2}{2} + \frac{h^4}{1.2.3.4} - \dots \right) \\ + \cos x \left( h - \frac{h^3}{1.2.3} + \frac{h^5}{1.2.3.4.5} - \dots \right) \end{aligned}

If we suppose $x=0$ , then, as in that case $\sin x=0$ , the preceding formula becomes

$\sin$ ,

\text{fin. } h = h - \frac{h^3}{1 \cdot 2 \cdot 3} + \frac{h^5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} - \&c.

or, substituting

x

instead of

h

\text{fin. } x = x - \frac{x^3}{1 \cdot 2 \cdot 3} + \frac{x^5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} - \&c.

Ex. 4. Suppose $u = \cos. x$ , then $u' = \cos. (x+h)$ , and since $u = \cos. x$ , by § 59, $\frac{u}{x} = -\sin. x$ , $\frac{\dot{u}}{x^2} = -\cos. x$ , $\frac{\ddot{u}}{x^3} = \sin. x$ , &c. Therefore, substituting as before these values in the general formula $u' = u + \frac{\dot{u}}{x} h + \frac{\ddot{u}}{x^2} \frac{h^2}{2} + \frac{\ddot{u}}{x^3} \frac{h^3}{6} + \&c.$ it becomes

\begin{aligned} \cos. (x+h) &= \cos. x - \sin. x \frac{h}{1} - \cos. x \frac{h^2}{2} \\ &\quad + \sin. x \frac{h^3}{6} + \&c. \end{aligned}

\cos. (x+h)

is equal to

\begin{aligned} \cos. x \left( -\frac{h^2}{1 \cdot 2} + \frac{h^4}{1 \cdot 2 \cdot 3 \cdot 4} - \&c. \right) \\ \text{fin. } x \left( h - \frac{h^3}{1 \cdot 2 \cdot 3} + \frac{h^5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} - \&c. \right) \end{aligned}

which expression, when $x=0$ , and therefore $\cos. x=1$ , $\sin. x=0$ , becomes simply

\cos. h = 1 - \frac{h^2}{1 \cdot 2} + \frac{h^4}{1 \cdot 2 \cdot 3 \cdot 4} - \&c.

or substituting

x

for

h

\cos. x = 1 - \frac{x^2}{1 \cdot 2} + \frac{x^4}{1 \cdot 2 \cdot 3 \cdot 4} - \&c.

71. It may be remarked that in each of these examples, from the development of $u'$ the new value of the function of $u$ , we have been able to deduce a development of $u$ the function itself. But it is easy to see, that by proceeding in the same manner with the general formula as we have done in these particular examples, we shall obtain a general expression for the development of any function whatever.

The general formula is

u' = u + \frac{\dot{u}}{x} \frac{h}{1} + \frac{\ddot{u}}{x^2} \frac{h^2}{2} + \frac{\ddot{u}}{x^3} \frac{h^3}{6} + \&c.

Now, $u'$ being the value that $u$ assumes when $x+h$ is substituted in it instead of $x$ , if we suppose $x=0$ , then $u'$ becomes the very same function of $h$ , that $u$ is of $x$ . Let us denote the values which each of the functions $u, \frac{\dot{u}}{x}, \frac{\ddot{u}}{x^2}, \&c.$ acquire, when $x=0$ , by $U, U', U'', \&c.$ respectively.

Then ( $F$ ) $u'$ (considered as the same function of $h$ that $u$ is of $x$ ) is equal to

U + U' \frac{h}{1} + U'' \frac{h^2}{1 \cdot 2} + U''' \frac{h^3}{1 \cdot 2 \cdot 3} + \&c.

Let $x$ be now supposed to be substituted both in $u'$ , and the series which is its development instead of $h$ , then $u'$ becomes $u$ , and we have

u = U + U' \frac{x}{1} + U'' \frac{x^2}{1 \cdot 2} + U''' \frac{x^3}{1 \cdot 2 \cdot 3} + \&c.

and in this formula it is to be considered, as already stated, that $U, U', U'', \&c.$ denote the particular values

which the functions $u, \frac{\dot{u}}{x}, \frac{\ddot{u}}{x^2}, \&c.$ acquire respectively, by supposing that in each of them $x$ is taken $=0$ .

72. As an example of the application of this series let us resume the equation $u = a^x$ , then $\frac{\dot{u}}{x} = A a^x$

(\text{§ 56.}) \frac{\ddot{u}}{x^2} = A^2 a^x, \frac{\ddot{u}}{x^3} = A^3 a^x, \&c.

Suppose now that $x=0$ , then $u$ , or $a^x$ becomes $a^0=1$ , $\frac{\dot{u}}{x} = A a^x$ becomes $A$ , $\frac{\ddot{u}}{x^2} = A^2 a^x$ becomes $A^2$ , &c. so that $U=1, U'=A, U''=A^2, \&c.$ substituting therefore these values in the general formula, it becomes

a^x = 1 + A \frac{x}{1} + A^2 \frac{x^2}{1 \cdot 2} + A^3 \frac{x^3}{1 \cdot 2 \cdot 3} + \&c.

Let us next suppose that $u$ is an arch of a circle of which the fine is $x$ (radius being unity), then $x = \sin. u$ . Now the ratio of the fluxion of $u$ to the fluxion of $x$ will be the very same whether we consider $u$ as a function of $x$ , or $x$ as a function of $u$ ; therefore (§ 59.)

$\dot{x} = \dot{u} \cos. u$ , and $\frac{\dot{u}}{x} = \frac{1}{\cos. u}$ , but since $\sin. u = x$ , $\cos. u = \sqrt{1-x^2}$ , therefore,

\frac{\dot{u}}{x} = \frac{1}{\sqrt{1-x^2}}.

4 Y 2

Taking

(F) For the sake of illustration let us take a particular example. Suppose $u = (a+x)^n$ , then $\frac{\dot{u}}{x} = n(a+x)^{n-1}$ , $\frac{\ddot{u}}{x^2} = n(n-1)(a+x)^{n-2}$ , &c. Suppose now that $x=0$ , then $u$ becomes $a^n$ , $\frac{\dot{u}}{x}$ becomes $n a^{n-1}$ , $\frac{\ddot{u}}{x^2}$ becomes $n(n-1) a^{n-2}$ , &c. so that in this particular case we have $U=a^n, U'=n a^{n-1}, U''=n(n-1) a^{n-2}, \&c.$

Taking now the fluxion of $\frac{1}{\sqrt{(1-x^2)}}$ , and the fluxion of the result, &c. we have

\begin{aligned}\frac{\dot{u}}{x^2} &= \frac{x}{(1-x^2)^{\frac{1}{2}}} \\ \frac{\dot{u}}{x^3} &= \frac{1}{(1-x^2)^{\frac{1}{2}}} + \frac{3x^2}{(1-x^2)^{\frac{3}{2}}} \\ \frac{\dot{u}}{x^4} &= \frac{3 \cdot 3x}{(1-x^2)^{\frac{1}{2}}} + \frac{3 \cdot 5x^3}{(1-x^2)^{\frac{3}{2}}} \\ &\&c.\end{aligned}

Suppose now that $x=0$ , then $u$ becomes 0, $\frac{\dot{u}}{x}$ becomes

\frac{\dot{u}}{x^2} \text{ becomes } 0, \quad \frac{\dot{u}}{x^3} \text{ becomes } 1, \quad \frac{\dot{u}}{x^4} \text{ becomes } 0, \quad \&c. \text{ so}

that $U=0$ , $U'=1$ , $U''=0$ , $U'''=1$ , $U''''=0$ , &c. Therefore, substituting in the general formula, we find

u = x + \frac{x^3}{1 \cdot 2 \cdot 3} + \&c.

By prosecuting the computations farther, we may find

u = x + \frac{1^2 x^3}{2 \cdot 3} + \frac{3^2 x^5}{2 \cdot 3 \cdot 4 \cdot 5} + \frac{3^2 5^2 x^7}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} + \&c.

Application of the Method of Fluxions to the Drawing of Tangents.

73. The theory of tangents to curve lines furnishes a good illustration of the truth of the principle which we have considered as the foundation of the method of fluxions, namely, that whatever be the form of a function, the ratio of its increment to the increment of the variable quantity from which the function is formed, is in every case susceptible of a limit.

Let AB, the abscissa of a curve, be the geometrical expression of a variable quantity $x$ , and let BP the corresponding ordinate, be the expression for $y$ , any function of $x$ ; then the curve line itself is the locus of the equation expressing the relation between $x$ and $y$ . Let PT, a tangent to the curve at P, meet AB the abscissa in T; through P draw any straight line meeting the abscissa in D, and the curve in $p$ ; draw the ordinate $p b$ , and from P draw Pn parallel to the abscissa, meeting the ordinate $b p$ in $n$ .

The triangles DBP, Pnp are similar; therefore

pn : nP :: PB : BD.

Now $pn$ , and $nP$ , or $Bb$ , are the increments of PB and BA, or of $y$ and $x$ respectively, therefore the ratio of the simultaneous increments of PB and BA, or $y$ and $x$ , whatever be their magnitudes, is equal to the ratio of PB to BD. Conceive now the point $p$ to approach continually to P, then the angle contained by the straight line $pPD$ , and the tangent PT, will decrease, and the point D will approach to T; at the same time $np$ , and $nP$ , the increments of $y$ and $x$ , will be continually diminished; still, however, they will have

to each other the ratio of PB to BD, but this ratio approaches continually to the ratio of PB to BT, and becomes at last more nearly equal to it than any assignable ratio; therefore the ratio of PB to BT is the limit of the ratio of PB to BD, and consequently is also the limit of the ratio of $p n$ , the increment of $y$ , to $n P$ , the increment of $x$ . And as this conclusion does not depend upon the particular nature of the curve, or upon any particular relation supposed to subsist between $x$ and $y$ , we may conclude, that whatever be the form of the function, the ratio of the simultaneous increments of the function, and the variable quantity from which it is formed, has a limit to which it approaches when the increments are conceived to be continually diminished.

It is now easy to see how the method of fluxions may be applied to the determination of tangents to curves, for since the ratio of the ordinate PB to the subtangent BT is always the limiting ratio of the increments of the ordinate and abscissa, it is equal to the ratio of their fluxions, that is

\dot{y} : \dot{x} :: y : \text{subtan. BT.}

Hence in any curve whatever, referred to an axis, the subtangent, (that is, the segment of the abscissa between the ordinate and tangent) is equal to $\frac{\dot{x}}{\dot{y}} y$ where $x$ denotes the abscissa, and $y$ the ordinate at the point of contact; and the subtangent being found, the position of the tangent is thereby determined.

Let us apply the above general formula to some examples.

Example 1. Let the proposed curve be a circle. It Fig. 7. is required to determine the position of PT, a tangent at any point P in its circumference.

Put $2a$ for AE the diameter, also $x$ for AB the abscissa, and $y$ for BP the ordinate at the point of contact.

From the nature of the curve, we have

AB \times BE = BP^2, \text{ that is} \\ x(2a-x) = y^2.

Hence taking the relations of the fluxions of $x$ and $y$ , we have

\begin{aligned}2a\dot{x} - 2x\dot{x} &= 2y\dot{y}, \\ \text{therefore } \frac{\dot{x}}{\dot{y}} &= \frac{y}{a-x}, \\ \text{and BT} &= \frac{\dot{x}}{\dot{y}} y = \frac{y^2}{a-x};\end{aligned}

from which it appears that BT the subtangent is a third proportional to $a-x$ and $y$ , that is, to CB the distance of the ordinate from the centre, and BP the ordinate, agreeing with what is known from the elements of geometry.

Ex. 2. Let the curve be a parabola, required the Fig. 8. same as before.

Put $x$ for AB, the abscissa, and $y$ for BP the ordinate at P the point of contact; also $a$ for the parameter; then, from the nature of the curve

PB^2 = a \times AB, \text{ that is} \\ ax = y^2

therefore,

Direct Method. therefore, taking the fluxions, we get $\dot{a}\dot{x} = 2\dot{y}\dot{y}$ , and

\frac{\dot{x}}{\dot{y}} = \frac{2y}{a}, \text{ and}

BT = \frac{\dot{x}}{\dot{y}} y = \frac{2y^2}{a} = \frac{2ax}{a} = 2x;

from which it appears that the sub-tangent BT is double the abscissa BA.

Fig. 9. Ex. 3. Let the curve be an ellipse. Put $AB = x$ , $BP = y$ , $AC$ the semi-transverse axis $= a$ , $CH$ the semi-conjugate axis $= b$ .

The nature of the curve is such, that

AC^2 : CH^2 :: AB \times BE : BP^2

\text{or } a^2 : b^2 :: (2a-x)x : y^2

\text{Hence } a^2 y^2 = b^2 (2a-x)x,

and taking the relation of the fluxions,

a^2 y \dot{y} = b^2 (a-x) \dot{x};

\text{Therefore } \frac{\dot{x}}{\dot{y}} = \frac{a^2 y}{b^2 (a-x)}, \text{ and}

BT = \frac{\dot{x}}{\dot{y}} y = \frac{a^2 y^2}{b^2 (a-x)}.

But from the preceding equation expressing the nature

of the curve, $\frac{a^2 y^2}{b^2} = (2a-x)x$ ; therefore

BT = \frac{\dot{x}}{\dot{y}} y = \frac{(2a-x)x}{a-x}.

To this expression for BT, let $BC = a-x$ be added, and we have

CT = a-x + \frac{(2a-x)x}{a-x} = \frac{a^2}{a-x};

from which it appears that $CB : CA :: CA : CT$ .

Fig. 10. Ex. 4. Suppose the curve to be a hyperbola, and let it be required to find, as in the preceding examples.

Put $a$ and $b$ , as in the case of the ellipse, to express the semitransverse and semiconjugate axes, then the equation of the curve is

a^2 y^2 = b^2 (2a+x)x,

and taking the relation of the fluxions,

a^2 y \dot{y} = b^2 (a+x) \dot{x},

from which we find

BT = \frac{\dot{x}}{\dot{y}} y = \frac{a^2 y^2}{b^2 (a+x)}

or, substituting $(2a+x)x$ for $\frac{a^2 y^2}{b^2}$ ,

BT = \frac{\dot{x}}{\dot{y}} y = \frac{(2a+x)x}{a+x}.

Let this expression for BT be subtracted from the expression for CB, that is from $a+x$ , and we have

CT = a+x - \frac{2ax+x^2}{a+x} = \frac{a^2}{a+x},

therefore $CB : CA :: CA : CT$ .

Ex. 5. Suppose the curve APD to be a cycloid, of Fig. 11. which AE is the axis, and AQE a semicircle described on the axis as a diameter. Suppose AC, the radius, to be unity; put $AB = x$ , $BP = y$ , and the arch $AQ = v$ ; then, $AB = 1 - \cos. v$ and $BQ = \sin. v$ . Now, from the nature of the curve, $PB = \text{arch } AQ + BQ$ ; hence we have

x = 1 - \cos. v, \quad y = v + \sin. v,

and taking the fluxions, by § 59.

\dot{x} = \dot{v} \sin. v, \quad \dot{y} = \dot{v} + \dot{v} \cos. v;

therefore,

BT = \frac{\dot{x}}{\dot{y}} = \frac{\dot{v} y \sin. v}{\dot{v} + \dot{v} \cos. v} = \frac{y \sin. v}{1 + \cos. v} \\ = \frac{PB \times BQ}{EB};

but from the nature of the circle $\frac{BQ}{EB} = \frac{AB}{BQ}$ , there-

fore, $BT = \frac{PB \times AB}{BQ}$ , and consequently $BQ : BA ::$

$BP : BT$ , from which it appears that if the chord AQ be drawn, the tangent PT is parallel to the chord QA.

74. If PT (Fig. 6.) be a tangent to the curve AP at Fig. 6. the point P, and PC be drawn perpendicular to the tangent, meeting AC the axis of the curve in C, then the line PC is called a normal to the curve at the point P; and BC, the distance between the ordinate and the extremity of the normal, is called the sub-normal.

The triangles TBP, BPC being similar, we have $TB : BP :: BP : BC$ ; or, since $TB : BP :: x : y$ , (§ 73) $x : y :: y : BC$ , hence in any curve, BC the sub-normal is equal to $\frac{y^2}{x}$ ; and from this expression, we may find

the sub-normal in the same way as we have found the sub-tangent in the examples of last §.

75. As by plane trigonometry

TB : BP :: \text{rad.} : \text{tangent of } T,

and from § 73, $TB : BP :: x : y$ ,

\text{therefore } x : y :: 1 : \tan. T,

hence it appears that $\frac{y}{x}$ expresses the numeral tangent of the angle T, that is, the angle contained by a tangent to the curve, and the axis of the curve. In like manner we have

y : x :: BP : BD :: CB : BP :: 1 : \tan. C,

therefore $\frac{x}{y}$ expresses the tangent of the angle C, that is, the angle contained by a normal to the curve and its axis.

Application

Application of Fluxions to Problems relating to MAXIMA and MINIMA.

76. If a variable quantity be supposed to change its magnitude, then any function of that quantity will also change its magnitude. When the variable quantity is supposed to increase continually so as to acquire successively all degrees of magnitude, there are some functions of such a form, that they either increase continually, or decrease continually; but there are others again which either increase to a certain limit, after which they decrease; or else they decrease to a certain limit, after which they increase.

If, in consequence of the continual increase of a variable quantity, a function of that quantity first increases to a certain limit, and afterwards decreases, when it arrives at that limit it is then said to be a maximum. Or if it decrease to a certain limit, and afterwards increase, when it arrives at that limit, it is then said to be a minimum.

77. Let us consider the function $y = b - (x - a)^2$ . If we suppose $x = 0$ , then $y = b - a^2$ . Suppose now $x$ to be at first very small, and to increase; then as $(x - a)^2$ will decrease, $y$ will also increase, till $x$ become $= a$ , and then $y$ will become $= b$ , when it is a maximum; for $x$ being supposed to become greater than $a$ , $y$ will be less than $b$ . By supposing $x$ to increase till $(x - a)^2$ become equal to $b$ , then $y$ will decrease to $0$ ; and $x$ being still supposed to increase, $y$ will become negative.

Let us next suppose $y = b + (x - a)^2$ . In this case when $x = 0$ , $y = b + a^2$ ; as $x$ increases, $(x - a)^2$ decreases, and consequently $y$ decreases, till $x = a$ , and then $y = b$ , a minimum; for $x$ becoming greater than $a$ , $y$ becomes greater than $b$ .

78. Every function that either increases or decreases continually has neither maximum nor minimum; for whatever value such a function may acquire, in the one case it may always have a greater, and in the other a less value.

The characteristic property of a maximum value of a function, by which it is made the object of analytical inquiry, consists in its being greater than the values immediately preceding, and also greater than the values immediately following it; and that of a minimum consists in its being less than the values immediately preceding, and also less than the values immediately following it.

In some cases a function may increase to a certain limit and then decrease, and afterwards increase again indefinitely; or the contrary. Hence it may happen that such a function may have values greater than its maximum or less than its minimum as they have been here defined. And indeed it is easy to conceive that a function may increase and decrease alternately several times; in such a case it must be considered as having several maxima and minima.

79. Since $y$ any function of a variable quantity $x$ may be considered as the ordinate of a curve, of which $x$ is the abscissa, it is evident that to determine the greatest or least value of such a function, we have only to seek the greatest or least ordinate of the curve which

is the locus of the equation expressing the relation between $x$ and $y$ . Let us suppose this curve to be DPE, and that AB is the value of $x$ , corresponding to BP the maximum or minimum value of the ordinate $y$ ; it is evident, that in the case of a maximum, the curve must be concave towards AC, at least to a certain extent, on each side of the point P, as in fig. 12. but that in the case of a minimum it must be convex towards AC, as in fig. 13.; and also, that in either case, if a straight line be drawn through P parallel to AC, the curve must be wholly on one side of that line, to a certain extent on each side of the point P, and therefore, that the line PQ must be a tangent to the curve at the point P.

Now when PQ a tangent to a curve at P (fig. 8.) meets the axis in T, it has been shewn, § 75, that $\frac{y}{x}$ is the expression for the tangent of the angle T, radius being unity; but this angle vanishes, when PQ, instead of meeting AC, is parallel to it, as in fig. 12. and fig. 13.; therefore, as in this case the tangent of the angle is $= 0$ ,

we have $\frac{y}{x} = 0$ .

Hence it appears, that to determine the maximum or minimum of $y$ , a function of $x$ , we must find the fluxion of the function, and divide it by $x$ , and put the result equal to $0$ .

80. We proceed to illustrate this rule by some examples.

Ex. 1. To divide a given number $a$ into two such parts, that their products may be the greatest possible.

Let $x$ denote the one part, then $a - x$ will be the other part, and $x(a - x)$ the product of the two parts. Therefore, by the question

y = x(a - x) = ax - x^2, \text{ a maximum,}

hence, taking the fluxion of the function,

\dot{y} = a - 2x, \text{ and } \frac{\dot{y}}{x} = a - 2x,

therefore, $a - 2x = 0$ , and $x = \frac{1}{2}a$ .

Thus it appears that the product of the parts will be the greatest possible, when each is half the given number.

Ex. 2. To find the fraction which shall exceed its cube by the greatest quantity possible.

Let $x$ denote the fraction, then its cube is $x^3$ , so that we have

y = x - x^3, \text{ a maximum;}

therefore, taking the fluxion of the function,

\dot{y} = 1 - 3x^2, \text{ and } \frac{\dot{y}}{x} = 1 - 3x^2 = 0,

hence $3x^2 = 1$ , and $x = \sqrt{\frac{1}{3}}$ , the fraction required.

Ex. 3. To determine the greatest rectangle that can be inscribed in a given triangle.

Put the base AC of the triangle $= b$ , and its altitude BD $= a$ , and let Bn, the altitude of the rectangle $p'q'r's$ , considered as variable, be denoted by $x$ , then, because

Direct Method. because of the parallel lines AC, pq, it will be, as BD : AC :: Dn : pq, that is a : b :: a - x : pq, hence pq = $\frac{b(a-x)}{a}$ , and the area of the rectangle, or pq x Bn, will be = $\frac{b(a-x)x}{a}$ , therefore y = $\frac{b(a-x)x}{a}$ must be a maximum, and hence $\dot{y} = \frac{ba\dot{x} - 2bx\dot{x}}{a}$ ; thus we have

\frac{\dot{y}}{x} = \frac{ba - 2bx}{a} = 0,

and $ba - 2bx = 0$ , and $x = \frac{1}{2}a$ ; hence it appears that the greatest inscribed rectangle is that whose altitude is half the altitude of the triangle.

81. It is proper to observe that the value of a quantity, when a maximum or minimum, may often be determined more readily by considering that any given multiple, or part of the function, likewise any power or root of it, must then be also a maximum or minimum. Thus in the preceding example, in which the function to be a maximum is $\frac{b(a-x)x}{a}$ , we may reject the constant multiplier $\frac{b}{a}$ , and then the function to be a maximum is $y = (a-x)x = ax - x^2$ , the fluxion of which being taken, we have $\dot{y} = a\dot{x} - 2x\dot{x}$ and $\frac{\dot{y}}{x} = a - 2x = 0$ , hence $x = \frac{1}{2}a$ , the same value as before.

82. Ex. 4. Of all the right-angled plane triangles having the same given hypotenuse, to find that whose area is greatest.

Fig. 15.

Let ABC be the triangle; put AB = x, and AC, the given hypotenuse, = a, then BC = $\sqrt{a^2 - x^2}$ , and consequently the area of the triangle is $\frac{x}{2} \sqrt{a^2 - x^2}$ , which being a maximum, its square $\frac{x^2(a^2 - x^2)}{4}$ , also four times that square, or $x^2(a^2 - x^2)$ , will likewise be a maximum, therefore, $y = x^2(a^2 - x^2) = a^2x^2 - x^4$ , a max. hence $\dot{y} = 2a^2x - 4x^3$ , and $\frac{\dot{y}}{x} = 2a^2 - 4x^2 = 0$ , hence $2x^2 = a^2$ , and $x = a\sqrt{\frac{1}{2}}$ .

Fig. 16.

Ex. 5. To determine the greatest cylinder that can be inscribed in a given cone. Put AC the base of the cone = b, BH its altitude = a, and DF, the diameter of the end of the cylinder DIGF inscribed in the cone, = x. From the similar triangles BAC, BDF, we have AC : BH :: DF : BE, that is b : a :: x : BE, hence BE = $\frac{ax}{b}$ , and EH = BH - BE = $a - \frac{ax}{b} = \frac{ab - ax}{b}$ . Put c for the number 78539, then the area of the base of the cylinder is (by the Elements of Geometry) $c \times DF^2$ , and

its solid content $c \times DF^2 \times EH = \frac{ca^2x^2(b-x)}{b}$ , hence $\frac{ca^2x^2(b-x)}{b}$ is to be a maximum, therefore, leaving out the constant multiplier $\frac{ca}{b}$ , we have $y = x^2(b-x) = bx^2 - x^3$ , a maximum; taking now the fluxions, we get $\dot{y} = 2bx - 3x^2$ , and $\frac{\dot{y}}{x} = 2b - 3x^2 = 0$ , hence $3x^2 = 2bx$ , and $x = \frac{2}{3}b$ , and consequently HF = $\frac{1}{3}BH$ .

Ex. 6. To find the sun's place in the ecliptic, when that part of the equation of time which arises from the obliquity of the ecliptic is a maximum. Let EQ be the equator, EC the ecliptic, S the sun's place, and SA his declination, then this part of the equation of time is the difference of the sun's longitude ES, and right ascension EA, turned into time. Put the arch ES = x, the arch EA = v, which is to be considered as a function of x, and put a for the cosine of the angle E = 23° 28'. Then, by Spherical Trigonometry, in the right-angled spherical triangle EAS, we have tan. EA = cos. E x tan. ES; therefore, to determine x we have tan. v = a tan. x, and y = x - v, a maximum.

From the second of these equations we get $\dot{y} = \dot{x} - \dot{v}$ , and $\frac{\dot{y}}{x} = 1 - \frac{\dot{v}}{x} = 0$ , and from the first, by § 60, $\dot{v} \sec^2 v = a \dot{x} \sec^2 x$ , and $\frac{\dot{v}}{x} = \frac{a \sec^2 x}{\sec^2 v}$ , therefore, $1 - \frac{a \sec^2 x}{\sec^2 v} = 0$ , and $\sec^2 v = a \sec^2 x$ , or $1 + \tan^2 v = a + a \tan^2 x$ , but from the first equation $\tan^2 v = a^2 \tan^2 x$ , therefore, $1 + a^2 \tan^2 x = a + a \tan^2 x$ , or $a(a-1) \tan^2 x = a-1$ ,

hence $\tan x = \frac{1}{\sqrt{a}} = 1.04416$ , the tangent of 46° 14', the sun's longitude when this part of the equation is a maximum.

83. We have deduced the rule of § 79. for determining when a function is a maximum, or minimum, from the consideration of curve lines. The whole theory of maxima and minima may however be explained in a manner purely analytical, as follows: Let us suppose that y is any function whatever of x, and that x has acquired the value that produces the maximum or minimum of the function; then, if we suppose x - h and x + h to be substituted successively in the function instead of x, the two resulting values ought to be both less than the maximum or both greater than the minimum value. Let

Let us denote the value of the function that results from the substitution of $x-h$ by $y$ and that which results from the substitution of $x+h$ by $y'$ , then by the theorem given in § 69,

y = y - \frac{\dot{y}}{x} h + \frac{\ddot{y}}{x^2} \frac{h^2}{2} - \frac{\ddot{y}}{x^3} \frac{h^3}{2 \cdot 3} + \&c.

y' = y + \frac{\dot{y}}{x} h + \frac{\ddot{y}}{x^2} \frac{h^2}{2} + \frac{\ddot{y}}{x^3} \frac{h^3}{2 \cdot 3} + \&c.

In each of these values, $\frac{\dot{y}}{x}$ , the coefficient of $h$ , must either be equal to some quantity, positive or negative, or else it must $\equiv 0$ . Let us suppose, if possible, that it is equal to some quantity, positive or negative; now as $h$ may be conceived to be so small that the term $\frac{\dot{y}}{x} h$ , or any other term, shall exceed the sum of all the terms that follow it in each series, (D) if we suppose $h$ to have such a value, then, because of the term $\frac{\dot{y}}{x} h$ having the sign — in the one series, and + in the other, it follows that the one value of the function is greater, and the other less than $y$ , the maximum or minimum value: But this conclusion is inconsistent with the nature of a maximum or minimum, therefore $\frac{\dot{y}}{x}$ cannot in the case of a maximum or minimum be equal to any positive or negative quantity whatever.

If however we assume $\frac{\dot{y}}{x} \equiv 0$ , so that the two values are,

y = y + \frac{\ddot{y}}{x^2} \frac{h^2}{2} - \frac{\ddot{y}}{x^3} \frac{h^3}{2 \cdot 3} + \&c.

y' = y + \frac{\ddot{y}}{x^2} \frac{h^2}{2} + \frac{\ddot{y}}{x^3} \frac{h^3}{2 \cdot 3} + \&c.

then as the second term $\frac{\ddot{y}}{x^2} \frac{h^2}{2}$ has the same sign in both, when that term is greater than all the terms that follow

it, we shall have both values greater than $y$ when $\frac{\ddot{y}}{x^2}$ is positive, and both less when it is negative, the first case corresponding to a maximum, and the second to a minimum.

Hence, to determine the maximum or minimum of the function $y$ , it appears that we must take the fluxion of $y$ , and divide it by $x$ , and put the result equal to 0, which agrees with what was shown in § 79.

84. Although in the case of a function admitting of a maximum or minimum we have always $\frac{\dot{y}}{x} \equiv 0$ , yet we must not conclude that conversely the one or the other of these has place every time that $\frac{\dot{y}}{x} \equiv 0$ . For if it so happens that the value of $x$ that renders $\frac{\dot{y}}{x} \equiv 0$ , causes also $\frac{\ddot{y}}{x^2}$ to vanish, without at the same time making $\frac{\ddot{y}}{x^3}$ to disappear, then we have

y = y - \frac{\ddot{y}}{x^3} \frac{h^3}{2 \cdot 3} + \frac{\ddot{y}}{x^4} \frac{h^4}{2 \cdot 3 \cdot 4} - \&c.

y' = y + \frac{\ddot{y}}{x^3} \frac{h^3}{2 \cdot 3} + \frac{\ddot{y}}{x^4} \frac{h^4}{2 \cdot 3 \cdot 4} + \&c.

and as by giving a proper value to $h$ , $\frac{\ddot{y}}{x^3} \frac{h^3}{2 \cdot 3}$ may be rendered greater than the sum of all the following terms in each series, it follows, that $\frac{\ddot{y}}{x^3}$ being supposed to be any quantity either positive or negative, because of its sign being different in the two values, the one of them will be greater, and the other less than $y$ , the maximum or minimum value, which result is inconsistent with the nature of a maximum or minimum. If however $\frac{\ddot{y}}{x^3}$ be assumed $\equiv 0$ , then

y =

(D) If this should not appear sufficiently obvious, let

A h + B h^2 + C h^3 + D h^4 + \&c.

be such a series, where $A, B, C, D, \&c.$ denote quantities either positive or negative, but which are independent of $h$ . Then, writing the series thus,

h(A + B h + C h^2 + D h^3 + \&c.)

it is obvious that if $h$ be conceived to be continually diminished, and at last to become $\equiv 0$ , the part

B h + C h^2 + D h^3 + \&c.

will also become $\equiv 0$ , therefore before it vanishes it will be less than $A$ , or any other assignable quantity, therefore $B h + C h^2 + D h^3 + \&c.$ may become less than $A h$ .

\therefore y = y + \frac{\dot{y}}{x^4} \frac{h^4}{2 \cdot 3 \cdot 4} + \dots

\therefore y' = y + \frac{\dot{y}}{x^4} \frac{h^4}{2 \cdot 3 \cdot 4} + \dots

here again, the coefficient of the second term having in both values the same sign, the conditions of the maxi-

mum or minimum are fulfilled, and the sign of $\frac{\dot{y}}{x^4}$ shows when the one or the other is to have place.

It must now be sufficiently evident, without proceeding any further, that a function can only admit of a maximum or a minimum when the first of its fluxions that does not vanish, is of an even order (or is its second or fourth fluxion, &c.), and that the sign of that fluxion is negative in the case of a maximum, but positive in the case of a minimum.

85. We shall conclude this theory by applying it to an example. Let $y$ be such a function of $x$ that

y^3 - 2mxy + x^3 - a^3 = 0,

(y - mx)\dot{y} - (my - x)\dot{x} = 0,

\text{and hence } \frac{\dot{y}}{x} = \frac{my - x}{y - mx} = 0,

\text{therefore } my - x = 0, \text{ and } y = \frac{x}{m}.

To find the value of $x$ , let this value of $y$ be substituted in the original equation, it thus becomes

\frac{x^3}{m^3} - x^3 - a^3 = 0,

x = \frac{ma}{\sqrt{(1-m^3)}}, \text{ and } y = \frac{a}{\sqrt{(1-m^3)}}.

We must now examine what is the nature of the expression for $\frac{\dot{y}}{x^3}$ . Taking the fluxion of the equation

\frac{\dot{y}}{x} = \frac{my - x}{y - mx}, \text{ and considering that } x \text{ is constant, we have}

\frac{\dot{y}}{x} = \frac{(1-m^3)(x\dot{y} - y\dot{x})}{(y-mx)^2},

therefore, dividing by $x$ ,

\frac{\dot{y}}{x^2} = \frac{(1-m^3)}{(y-mx)^2} \left\{ \frac{\dot{y}}{x} x - y \right\},

but as in the present case $\frac{\dot{y}}{x} = 0$ , and $y = \frac{x}{m}$ , this expression becomes simply

VOL. VIII. Part II.

\therefore \frac{\dot{y}}{x^2} = \frac{-m}{x\sqrt{(1-m^3)}},

which equation, by putting instead of $x$ its value $\frac{ma}{\sqrt{(1-m^3)}}$ , becomes

\therefore \frac{\dot{y}}{x} = -\frac{1}{a\sqrt{(1-m^3)}},

and as this result is negative, we conclude that the value which we have found for $y$ is a maximum.

Of the values of fractions, the numerators and denominators of which vanish at the same time.

86. There are some fractional functions of such a nature, that by giving a particular value to the variable quantity, both the numerator and denominator of the fraction vanish, and thus the fraction is reduced to this form $\frac{0}{0}$ , an expression from which nothing can be concluded.

We have an example of this in the fraction $\frac{x-a}{x^3-a^3}$ , which, by supposing $x=a$ becomes $\frac{a-a}{a^3-a^3} = \frac{0}{0}$ ; we must not however conclude that the fraction has no determinate value in this particular case, for if we consider that its numerator and denominator have a common divisor, viz. $x-a$ , it is evident that by taking this divisor out of both, the fraction $\frac{x-a}{x^3-a^3} = \frac{x-a}{(x-a)(x+a)}$ , becomes $\frac{1}{x+a}$ , an expression, which in the case of $x=a$ is equal to $\frac{1}{2a}$ .

87. In general, if we make $x=a$ in an expression of this form $\frac{P(x-a)^m}{Q(x-a)^n}$ , it becomes $\frac{0}{0}$ ; however its true value is either nothing, or finite, or infinite, according as $m > n$ , or $m = n$ , or $m < n$ ; for by taking out the factors common to the numerator and denominator, the fraction becomes $\frac{P(x-a)^{m-n}}{Q}$ in the first case, $\frac{P}{Q}$ in the second, and $\frac{P}{Q(x-a)^{n-m}}$ in the third; here we suppose that $P$ and $Q$ are such functions as neither become nothing, nor infinite, by the supposition of $x=a$ .

88. Therefore, when by giving a particular value to $x$ a function of that quantity assumes the form $\frac{0}{0}$ , to discover the true value of the function in this particular case, we must disengage the factors which are common to the numerator and denominator. This may be done in most cases by finding their common measure (ALGEBRA, § 49.) but the direct method of fluxions furnishes us with another method.

In the expression $P(x-a)$ , where $P$ denotes any function of $x$ that is independent of $x-a$ , if we suppose $x=a$ , then the expression vanishes; the fluxion however of the expression, viz. $(x-a)\dot{P} + P\dot{x}$ , is a quantity which does not vanish when $x=a$ , but is then reduced to its last term, that is to $P\dot{x}$ .

Again,

Again, the function $P(x-a)^n$ vanishes by supposing $x=a$ , but if we take its fluxion, viz. $(x-a)^n \dot{P} + 2(x-a) \dot{P} x$ , and again the fluxion of this quantity, we get

(x-a)^n \ddot{P} + 4(x-a) \dot{P} \dot{x} + 1 \cdot 2 P x^2,

an expression which does not vanish upon the hypothesis of $x=a$ , but is reduced to its last term, viz. $1 \cdot 2 P x^2$ . By proceeding in this manner, it is easy to see that by taking the fluxion of a function of the form $P(x-a)^m$ $m$ times successively ( $m$ being a whole number) we shall finally obtain an expression, all the terms of which, except the last, vanish by supposing that $x=a$ ; and that the last term will be $1 \cdot 2 \cdot 3 \dots m P x^m$ , an expression free from the factor $(x-a)^m$ , and involving only the function $P$ .

89. It is not necessary that we should know the number $n$ , nor that we should exhibit the factor $(x-a)^n$ , in order to determine when the expression $P(x-a)^n$ is freed from that factor. We have only to ascertain after each operation of taking the fluxion, whether the result vanishes or not, when we substitute $a$ instead of $x$ ; for in the last case the operation is finished, and the result is the quantity $1 \cdot 2 \cdot 3 \dots P x^m$ . Suppose for example the function to be $x^3 - a x^2 - a^2 x + a^3$ , which vanishes when $x=a$ , its first fluxion also vanishes when $x=a$ , but not its second fluxion, which is $(6x-2a)x^2$ , hence we may conclude that the function has the form $P(x-a)^2$ , which is besides obvious, because

x^3 - a x^2 - a^2 x + a^3 = (x+a)(x-a)^2.

90. In applying these observations to the fraction $\frac{P(x-a)^m}{Q(x-a)^n}$ it appears, that by repeating the operation of taking the fluxions of its numerator and denominator, they will be freed at once from the factor $x-a$ , if $m=n$ . If a result, which does not vanish, be obtained first from the numerator, then we may be assured, that the factor, $(x-a)$ is found in the numerator raised to a less power than in the denominator, and in this case the fraction is infinite when $x=a$ . If on the contrary the first result that does not vanish is found from the denominator, then the numerator contains a higher power of $(x-a)$ than the denominator, and in this case, when $x=a$ , the fraction vanishes.

The rule for finding the value of a function which becomes $\frac{0}{0}$ by giving a particular value to $x$ , may therefore be expressed thus. Take the successive fluxions of both the numerator and denominator until a result which does not vanish be obtained from either the one or the other, or from both at the same time; in the first case the function is infinite, in the second it is equal to 0, and in the last case its value is finite.

91. We proceed to illustrate this rule by a few examples.

Ex. 1. The value of the function $\frac{x^3 - 1}{x^2 - 1}$ is required when $x=1$ .

The fluxion of the numerator is $3x^2 \dot{x}$ , and that of the

denominator is $2x \dot{x}$ , neither of which quantities vanish when $x=1$ , therefore, in this particular case, the value of the fraction is $\frac{3x^2}{2x} = \frac{3}{2}$ .

Ex. 2. Suppose the fraction to be $\frac{a x^3 - 2 a c x + a c^2}{b x^2 - 2 b c x + b c^2}$ which vanishes when $x=c$ .

By taking the fluxions of the numerator and denominator we obtain $\frac{2 a x^2 - 2 a c \dot{x}}{2 b x - 2 b c \dot{x}} = \frac{a x - a c}{b x - b c}$ , a fraction, the numerator and denominator of which still vanish upon the hypothesis of $x=c$ , we therefore take the fluxions a second time, and get $\frac{2 a x^2}{2 b x^2} = \frac{a}{b}$ for the value of the proposed fraction in the particular case of $x=c$ .

Ex. 3. Suppose the fraction to be

\frac{x^3 - a x^2 - a^2 x + a^3}{x^2 - a^2},

which vanishes when $x=a$ . In this example, by taking the fluxions of the numerator and denominator once, we get

\frac{3x^2 x - 2 a x \dot{x} - a^2 \dot{x}}{2 x \dot{x}} = \frac{3x^2 - 2 a x - a^2}{2 x},

an expression, of which only the numerator vanishes upon the supposition of $x=a$ ; hence we may conclude the true value of the fraction in this case to be 0.

The contrary happens in the fraction

\frac{a x - x^2}{a^2 - 2 a^2 x + 2 a x^2 - x^4};

we may therefore conclude that when $x=a$ , this last fraction becomes infinite.

92. The rule § 90 can only be applied when the factors common to the numerator and denominator are integer powers of $x-a$ , for as by taking the fluxions, the index of $(x-a)^m$ is diminished by an unit at each operation; when $m$ is a fraction we shall at last arrive at a result containing negative powers of $x-a$ , which therefore, when $x=a$ , will become infinite. The following mode of proceeding will however apply to all cases whatever.

Let $\frac{X}{X'}$ be a fraction of which the numerator and denominator both vanish when $x=a$ ; by substituting in it $a+h$ instead of $x$ , the functions $X$ and $X'$ may be expanded into a series of this form,

A h^m + B h^{\beta} + \dots \quad A' h^{\alpha} + B' h^{\beta} + \dots

which are ascending, that is, having the exponents of the powers positive and increasing; because the series must become 0, upon the hypothesis that $h=0$ . We have therefore

\frac{A h^m + B h^{\beta} + \dots}{A' h^{\alpha} + B' h^{\beta} + \dots}

instead of the proposed fraction.

Direct Method. Now if $a = a'$ , by dividing the numerator and denominator of this expression by the factor $h^a$ , which is common to all the terms of each, it becomes

\frac{A h^{a-a'} + B h^{\beta-a'} + \&c.}{A' + B h^{\beta-a'} + \&c.}

a quantity which, by supposing $h=0$ , is reduced to $\frac{0}{A'}$ , that is to 0. If again $a = a'$ , the expression for the fraction, after dividing the numerator and denominator by $h^a$ , is

\frac{A + B h^{\beta-a} + \&c.}{A' + B h^{\beta-a} + \&c.}

which, by supposing $h$ to be $\infty$ , becomes simply $\frac{A}{A'}$ , a finite quantity. If, however, $a < a'$ , then the expression for the fraction is

\frac{A + B h^{\beta-a} + \&c.}{A h^{a-a'} + B h^{\beta-a} + \&c.}

which, when $h = \infty$ , becomes $\frac{A}{0}$ , an expression which may be considered as infinite. Thus it appears that in each case the true value of the fraction depends only on $A$ and $A'$ , the first terms of the series.

The following rule is applicable to every function that can appear under the indeterminate form $\frac{0}{0}$ .

Find the first term of each of the ascending series which express the developments of the numerator and denominator when $a+h$ is substituted in them instead of $x$ . Reduce the new function formed of these first terms to its most simple form, and make $h = \infty$ ; the results shall be the different values of the proposed function when $x$ is made equal to $a$ .

Example. Suppose the function to be

\frac{\sqrt{x} - \sqrt{a} + \sqrt{(x-a)}}{\sqrt{(x^2-a^2)}},

which, when $x=a$ , becomes $\frac{0}{0}$ . By substituting $a+h$ instead of $x$ , and developing the results into series, the numerator becomes $h^{\frac{1}{2}} + \frac{h}{2\sqrt{a}} + \&c.$ and the denominator $\sqrt{2a} h^{\frac{1}{2}} + \frac{h^{\frac{3}{2}}}{2\sqrt{2a}} + \&c.$ Taking now the first

term of each series, we have $\frac{h^{\frac{1}{2}}}{2\sqrt{2a} h^{\frac{1}{2}}} = \frac{1}{2\sqrt{2a}}$ , an expression in which $h$ is not found; therefore the value of the function is $\frac{1}{2\sqrt{2a}}$ , when $x=a$ .

Of the Radii of Curvature.

93. Let HC/CF represent a material curve, or mould.

Let a thread be fastened to it at H, and made to pass along the curve, so as to coincide with it in its whole extent from H to F. Let the thread be now unrolled or evolved from the curve, then its extremity F will describe another curve line FAPP'. The curve HCF is called the EVOLUTE of the curve FAP; and the curve FAP is called the INVOLUTE of the curve HCF.

94. From this mode of conceiving the curve to be generated, we may draw the following conclusions.

1st. Suppose PC to be a portion of the thread detached from the evolute, then PC will be a tangent to the evolute at C.

2dly. The line PC will be perpendicular to a tangent to the curve FAP at the point P, or will be $\propto$ normal to the curve at that point. For the point P may be considered as describing at the same time an element of the curve FAP, and an element of a circle $q P q'$ , whose momentary centre is C, and which has PC for its radius.

3dly. That part of the curve between F and P, which is described with radii all of which are shorter than CP, is more incurved than a circle described on C as a centre, with a radius equal to CP. And in like manner PP', the part of the curve on the other side of P, which is described with radii greater than PC, is less incurved than that circle.

4thly. The circle $q P q'$ has the same curvature as the curve APP' itself has at P: hence it is called an EQUICURVE circle, and its radius PC is called the RADIUS of CURVATURE at the point C.

95. We are now to investigate how the radius of curvature at any point in FAP any proposed curve may be found.

Let AB and BP be the co-ordinates at P any point in the curve, and PC its radius of curvature; and let PC meet AB in E. Put the abscissa AB = $x$ , the ordinate BP = $y$ , the arch AP = $s$ , the angle AEP (that is, the arch which measures that angle, radius being unity) = $v$ , the radius of curvature PC = $r$ . Take P' another point in the curve, and let P'C' be the radius of curvature at that point. Let P'C' meet AB in E', and PC in D, and on D as a centre, with a radius = 1, describe an arch of a circle, meeting the radii PC, P'C' in $m$ and $n$ . Then the arch PP' will be the increment of $s$ ; and since the angle PDP' is the difference of the angles PEA, P'E'A, the arch $mn$ will be the corresponding increment of $v$ .

Suppose now the point P' to approach continually to P, then the points C' and D will approach to C, and the ratio of the arch PP', the increment of $s$ , to the arch $mn$ the increment of $v$ , will approach to the ratio of CP to Cm, that is to the ratio of $r$ to 1; therefore the ratio of $r$ to 1 is the limit of the ratio of PP' to $mn$ , or $r = \lim \frac{PP'}{mn}$ , and passing to the ratio of the

fluxions, $r = \frac{\dot{s}}{\dot{v}}$ , thus we have obtained a formula ex-

pressing the radius of curvature, by means of the fluxion of the arch of the curve, and the fluxion of the angle which a normal to the curve makes with the line of the abscissas. We proceed to deduce from this formula

Direct Method. other expressions which may involve the fluxions of $x$ and $y$ only.

96. Because PE is a normal to the curve at E, the tangent of the angle PEA or $v$ is equal to $\frac{\dot{x}}{y}$ (§ 75.),

put $\frac{\dot{x}}{y} = t$ , then because $\tan. v = t$ , we have by taking the fluxions (§ 60.), $\dot{v} \sec^2 v = \dot{t}$ , but $\sec^2 v = 1 + \tan^2 v$
$= 1 + \frac{\dot{x}^2}{y^2} = \frac{\dot{x}^2 + y^2}{y^2} = \frac{\dot{z}^2}{y^2}$ (§ 63.), therefore $\dot{v} \frac{\dot{z}^2}{y^2} = \dot{t}$
and $\dot{v} = \frac{\dot{t} y^2}{\dot{z}^2}$ .

Substituting now this value of $\dot{v}$ in the formula $r = \frac{\dot{z}^2}{\dot{v}}$

\text{it becomes } r = \frac{\dot{z}^2}{\dot{t} y^2}

If we now recollect that $t = \frac{\dot{x}}{y}$ , and that $\dot{z}^2 = \dot{x}^2 + \dot{y}^2$ , it will appear that this other expression which we have found for $r$ involves in effect the fluxions of $x$ and $y$ only.

97. In computing the values of $t = \frac{\dot{x}}{y}$ , and $\frac{\dot{z}^2}{t y^2}$ we may consider any two of the three quantities $x$ , $y$ , $z$ , as a function of the remaining quantity; and upon that hypothesis compute their fluxions.

Thus if we suppose that $y$ and $z$ are functions of $x$ , then, as in taking the fluxions of $y$ , $t$ , and $z$ , we must consider $\dot{x}$ as a given or constant quantity, from the equation $t = \frac{\dot{x}}{y}$ we have $\dot{t} = -\frac{\dot{x} \dot{y}}{y^2}$ (§ 39.), and sub-

stituting this value of $\dot{t}$ in $\frac{\dot{z}^2}{t y^2}$ , the value last found for $r$ , it becomes

r = \frac{\dot{y}}{-\dot{x} \dot{y}} = \frac{(\dot{x}^2 + \dot{y}^2)^{\frac{3}{2}}}{-\dot{x} \dot{y}}

If again, instead of considering $y$ and $z$ as functions of $x$ , we consider $x$ and $z$ as functions of $y$ , then from the equation $t = \frac{\dot{x}}{y}$ (as $\dot{y}$ must now be reckoned constant), and

we get $\dot{t} = \frac{\dot{x}}{y}$ , thus the formula $r = \frac{\dot{z}^2}{t y^2}$ becomes

r = \frac{\dot{z}^2}{\dot{x} \dot{y}} = \frac{(\dot{x}^2 + \dot{y}^2)^{\frac{3}{2}}}{\dot{x} \dot{y}}

We shall now apply these formulae to some examples.

98. Example 1.—It is required to find the general expression for the radius of curvature of a parabola.

The equation of the parabola is $y = a^{\frac{1}{2}} x^{\frac{1}{2}}$ , there-

fore, $\dot{y} = \frac{1}{2} a^{\frac{1}{2}} x^{-\frac{1}{2}} = \frac{a^{\frac{1}{2}} \dot{x}}{2x^{\frac{1}{2}}}$ , and, making $\dot{x}$ constant,

$\ddot{y} = -\frac{1}{4} a^{\frac{1}{2}} x^{-\frac{3}{2}} = \frac{-a^{\frac{1}{2}} \dot{x}^2}{4x^{\frac{3}{2}}}$ , therefore, $\dot{z} = \sqrt{(\dot{x}^2 + \dot{y}^2)}$

$= \frac{\dot{x}}{2} \sqrt{\left(\frac{4x+a}{x}\right)}$ , and, putting $r$ for the radius of curvature,

r = \frac{\dot{z}^2}{-\dot{x} \dot{y}} = \frac{(a+4x)^{\frac{1}{2}}}{2\sqrt{a}}.

If in this general expression, we put $x=0$ , we find $\frac{a^{\frac{1}{2}}}{2\sqrt{a}} = \frac{1}{2} a$ for the radius of curvature at the vertex of the curve.

Ex. 2. Suppose the curve to be an ellipse, required as in the last example.

Putting $a$ and $c$ to denote the two axes, the equation of the ellipse is $a^2 y^2 = c^2 (a x - x^2)$ . Hence taking the first and second fluxions, we have $2 a^2 y \dot{y} = c^2 x (a - 2x)$ , and $2 a^2 \dot{y}^2 + 2 a^2 y \ddot{y} = -2 c^2 \dot{x}^2$ ; whence $\dot{y} = \frac{c^2 x (a - 2x)}{2 a^2 y}$ , and $\ddot{y} = \frac{a^2 y^2 + c^2 x^2}{a^2 y}$ , which expressions, by substituting the values of $y$ and $\dot{y}$ , become

\begin{aligned} \dot{y} &= \frac{c^2 x (a - 2x)}{2 a \sqrt{(a x - x^2)}} \\ \ddot{y} &= \left\{ \frac{a^2 c^2 x^2 (a - 2x)^2}{4 a^2 c (a x - x^2) \sqrt{(a x - x^2)}} \right. \\ &\quad \left. + \frac{c^2 x^2}{a \sqrt{(a x - x^2)}} \right\} \\ &= \frac{c^2 x^2}{a} \times \frac{(a - 2x)^2 + 4(a x - x^2)}{4(a x - x^2) \sqrt{(a x - x^2)}} \\ &= \frac{c a \dot{x}^2}{4(a x - x^2)^{\frac{3}{2}}}; \end{aligned}

therefore,

\begin{aligned} \dot{z} &= \sqrt{(\dot{x}^2 + \dot{y}^2)} \\ &= \sqrt{\left( \frac{c^2 x^2 (a - 2x)^2}{4 a^2 (a x - x^2)} + \dot{x}^2 \right)} \\ &= \frac{\dot{x}}{2 a} \sqrt{\left( \frac{c^2 a^2 + (a^2 - c^2)(4 a x - 4 x^2)}{a x - x^2} \right)}, \end{aligned}

and

\begin{aligned} r &= \frac{\dot{z}^2}{-\dot{x} \dot{y}} \\ &= \frac{(a^2 c^2 + 4(a^2 - c^2)(a x - x^2))^{\frac{3}{2}}}{2 a^2 c}; \end{aligned}

which expression, when $x=0$ , becomes simply $\frac{c^2}{2 a}$ , the radius of curvature at the vertices of the transverse axis; but when $x = \frac{1}{2} a$ , it becomes $\frac{a^2}{2 c}$ , the radius of curvature at the vertices of the conjugate axis.

PART II. THE INVERSE METHOD OF FLUXIONS.

99. AS the DIRECT METHOD of fluxions treats of finding the relation between the fluxions of variable quantities, having given the relation subsisting between the quantities themselves; so the INVERSE METHOD treats of finding the relation subsisting between the variable quantities, having given the relation of their fluxions.

Whatever be the relation between variable quantities, we can in every case assign the relation of their fluxions; therefore the direct method of fluxions may in this respect be considered as perfect. But it is not the same with the inverse method, for there are no direct and general rules, by which we can in every case determine from the relation of the fluxions, that of their flowing quantities or fluents. All we can do is to compare any proposed fluxion with such fluxions as are derived from known fluents by the rules of the direct method, and if we find it to have the same form as one of these, we may conclude that the fluents of both, or at least the variable parts of these fluents, are identical.

100. In the direct method we have shewn, that by proper transformations, the finding of the fluxion of any proposed function is reducible to the finding of the fluxions of a few simple functions, and of the sums, or products, or quotients of such functions. In like manner, in the inverse method we must endeavour to transform complex fluxionary expressions into others more simple, so as to reduce them, if possible, to some fluxion, the fluent of which we already know.

SECT. I. Of the Fluents of Fluxions involving one variable quantity.

101. As when $y$ is such a function of a variable quantity $x$ , that $y = Ax^m + C$ , where $A$ , $m$ and $C$ denote constant quantities, we find by the direct method (§ 36. and § 26.) that $\dot{y} = mAx^{m-1}\dot{x}$ , or (putting $a$ instead of $mA$ , and $n$ instead of $m-1$ ), $\dot{y} = ax^n\dot{x}$ ; so, on the contrary, as often as we have the fluxional equation

\dot{y} = ax^n\dot{x},

we may conclude that the relation of the fluents is expressed by the equation.

y = \frac{ax^{n+1}}{n+1} + C;

for by substituting $mA$ instead of $a$ , and $m-1$ instead of $n$ in this equation, it becomes $y = Ax^m + C$ , the same equation as that from which the fluxional equation was derived.

102. The value of the constant quantity $C$ , which is generally called by writers on fluxions, the correction of the fluent, is to be determined from the particular

inquiry in which the fluxional equation $\dot{y} = ax^n\dot{x}$ occurs. If it be known that $y = 0$ , when $x$ acquires some known magnitude, which may be denoted by $b$ , then the general equation $y = \frac{ax^{n+1}}{n+1} + C$ , becomes in that particular case

0 = \frac{ab^{n+1}}{n+1} + C;

Hence, by subtracting each side of this last equation from the corresponding side of the former, we get

y = \frac{a(x^{n+1} - b^{n+1})}{n+1}

an equation that is independent of the constant and arbitrary quantity $C$ .

103. By giving particular values to $n$ in the fluxional equation $\dot{y} = ax^n\dot{x}$ , and in that of the fluents $y = \frac{a(x^{n+1} - b^{n+1})}{n+1}$ , we may obtain particular fluxional equations, and corresponding equations of the fluents. There is however one case which requires to be noticed; it is when $n = -1$ ; then the equation of the fluxions is $\dot{y} = ax^{-1}\dot{x} = \frac{a\dot{x}}{x}$ , and that of the fluents, according to the general formula $y = \frac{a(x^{-1+1} - b^{-1+1})}{-1+1}$

= \frac{a(x^0 - b^0)}{0} = \frac{a(1-1)}{0} = \frac{0}{0},

but from this expression it is manifest, that nothing can be concluded. The value of the function $\frac{a(x^{n+1} - b^{n+1})}{n+1}$ , in the particular case of $n+1=0$ may be found by the rule given in § 90 for determining the value of a function when it assumes the form $\frac{0}{0}$ ; but it may be otherwise found by proceeding thus.

Put $n+1=m$ , and let $p = \frac{\log. x}{\log. e}$ , and $q = \frac{\log. b}{\log. e}$ ; then, by the formula of § 54,

x^m = 1 + pm + \frac{p^2 m^2}{2} + \&c.

b^m = 1 + qm + \frac{q^2 m^2}{2} + \&c

and therefore

x^m - b^m = (p-q)m + \frac{(p^2 - q^2)m^2}{2} + \&c.

\text{and } \frac{x^m - b^m}{m} = (p-q) + \frac{(p^2 - q^2)m}{2} + \&c.

Thus.

Thus we have $\frac{x^n - b^n}{m}$ ; or $\frac{x^{n+1} - b^{n+1}}{n+1}$ expressed generally by a series, all the terms of which, except the first, being multiplied by $m$ or $n+1$ , will vanish when $n+1=0$ , or when $n=-1$ ; hence it appears that the general equation $y = \frac{a(x^{n+1} - b^{n+1})}{n+1}$ , becomes in the particular case of $n=-1$ , $y = a(p-q)$ , which, substituting for $p$ and $q$ their values, and observing that

\frac{\log. x}{\log. e} - \frac{\log. b}{\log. e} = \frac{1}{\log. e} \times \log. \frac{x}{b}, \text{ becomes}

= y \log. e \times \log. \frac{x}{b}

where $\log. e$ and $\log. \frac{x}{b}$ are to be taken according to the same system, which may be any system of logarithms whatever. So that if we take the Napierian system, in which $\log. e = 1$ , then

y = a \log. \frac{x}{b} = a \log. x - a \log. b = a \log. x + C,

where $C$ denotes a constant quantity, and where the letter $l$ , in this formula, and in others in which it may occur, is put as an abbreviation of the words Napierian logarithm, so that by $a \log. x$ is meant $a$ multiplied by the Napierian logarithm of $x$ , &c.

This expression which we have found for the value of $y$ in the particular case of $y = a x^{-1} \dot{x}$ , or $\frac{ax}{x}$ , coincides with what we might have found by considering that when $y = l. x$ , it has been shewn (§ 57.), that $y = \frac{x}{x}$ , so that conversely, when $y = \frac{ax}{x}$ , we may conclude that $y = a \log. x + C$ , where $C$ denotes a constant quantity, to be determined from the particular question in which the fluxional equation may occur.

104. It must now be evident that if

\dot{y} = a x^m \dot{x} + b x^n \dot{x} + c x^p \dot{x} + \&c.

where $m, n, p, \&c.$ are constant numbers, then

y = \frac{a x^{m+1}}{m+1} + \frac{b x^{n+1}}{n+1} + \frac{c x^{p+1}}{p+1} + \&c. + C;

here $C$ denotes a constant arbitrary quantity that may be considered as the sum of the constant quantities which ought to be added to the terms $\frac{a x^{m+1}}{m+1}, \frac{b x^{n+1}}{n+1}, \&c.$ each being regarded as a distinct fluent.

105. In general, since that when

y = a t + b v + c u + \&c. + C,

where $t, v, u, \&c.$ denote any functions of a variable quantity, and $C$ a constant quantity, we have (§ 35. and § 36.)

\dot{y} = a \dot{t} + b \dot{v} + c \dot{u} + \&c.

So, on the contrary, if we have any fluxional equation of this last form, we may conclude that

y = a t + b v + c u + \&c. + C.

And since that when $u = v t + C$ , where $u, v$ and $t$ denote any function of a variable quantity, and $C$ a constant quantity, we have, § 37, $u = v t + t \dot{v}$ , so, on the contrary, if

\dot{u} = v \dot{t} + t \dot{v},

we may conclude that

u = v t + C,

and in like manner if we have

\dot{u} = \frac{t \dot{v} - v \dot{t}}{t^2} = \frac{\dot{v}}{t} - \frac{v \dot{t}}{t^2},

we may infer from § 39. that

u = \frac{v}{t} + C.

106. It is often convenient to denote the fluent of a fluxional expression without actually exhibiting that fluent. For this purpose we shall employ the sign $\int$ , putting it before the fluxion whose fluent we mean to denote. Thus, by the expression $\int a x^n \dot{x}$ , is to be understood the fluent of $a x^n \dot{x}$ ; and as this fluent has been found to be $\frac{a x^{n+1}}{n+1} + C$ , we may express this conclusion in symbols shortly thus,

\int a x^n \dot{x} = \frac{a x^{n+1}}{n+1} + C.

107. Suppose we have $\dot{y} = (ax+b)^m \dot{x}$ , we may expand $(ax+b)^m$ into a series, and multiply the series by $\dot{x}$ , and find the fluent of each term of the result. But we may also find the fluent of this expression without employing the development of $(ax+b)^m$ , by proceeding thus. Put $ax+b = z$ , then $x = \frac{z-b}{a}$ , and $\dot{x} = \frac{\dot{z}}{a}$ .

Substitute now these values of $ax+b$ , and $\dot{x}$ , in the expression for $\dot{y}$ , and it becomes $\dot{y} = \frac{z^m \dot{z}}{a}$ ; hence we have

(§ 101.) $y = \frac{z^{m+1}}{a(m+1)} + C$ , and consequently, by substituting $(ax+b)$ for $z$ ,

y = \frac{(ax+b)^{m+1}}{a(m+1)} + C.

108. Suppose that $\dot{y} = (ax^n+b)^m x^{n-1} \dot{x}$ . By putting as before $ax^n+b = z$ , we have $n a x^{n-1} \dot{x} = \dot{z}$ , and $x^{n-1} \dot{x} = \frac{\dot{z}}{n a}$ ; hence $\dot{y} = \frac{z^m \dot{z}}{n a}$ , and $y = \frac{z^{m+1}}{n a(m+1)} + C$ , and substituting for $z$ its value $ax^n+b$ ,

y =

y = \frac{(ax^n + b)^{m+1}}{na(m+1)} + C.

109. Let us now consider fractional functions, and to begin with a simple case, let us suppose that

y = \frac{Ax^n}{(ax+b)^n}. \text{ Put } ax+b=z, \text{ then } x = \frac{z-b}{a}, \dot{x} = \frac{\dot{z}}{a},

and consequently,

\dot{y} = \frac{A(z-b)^{n-1}\dot{z}}{a^{n+1}z^n}.

We have now only to find the development of $(z-b)^n$ , to multiply each of its terms by $\dot{z}$ and divide it by $z^n$ , and take the fluent of the result.

Let us take for example the case of $m=3$ , and $n=2$ , then

\dot{y} = \frac{A(z-b)^2\dot{z}}{a^5z^2} \\ = \frac{A}{a^4} \left\{ z\dot{z} - 3b\dot{z} + 3b^2z^{-1}\dot{z} - b^3z^{-3}\dot{z} \right\}

Hence, taking the fluents of the several terms, as in § 105, we have

y = \frac{A}{a^4} \left\{ \frac{z^2}{2} - 3bz + 3b^2 \ln z + b^3 z^{-1} \right\} + C.

Let us now restore the value of $z$ , and then it appears, that when

\dot{y} = \frac{Ax^n}{(ax+b)^n},

y = \frac{A}{a^4} \left\{ \frac{1}{2}(ax+b)^2 - 3b(ax+b) + 3b^2 \ln(ax+b) + b^3(ax+b)^{-1} \right\} + C.

110. If we suppose that

\dot{y} = \frac{Ax^n + Bx^{n-1} + Cx^{n-2} + \&c.}{(ax+b)^n},

then, we may write the equation thus,

\dot{y} = \frac{Ax^n}{(ax+b)^n} + \frac{Bx^{n-1}}{(ax+b)^n} + \frac{Cx^{n-2}}{(ax+b)^n} + \&c.

and take the fluent of each term, in the same manner

as we have found the fluent of $\frac{Ax^n}{(ax+b)^n}$ .

Of the Fluents of rational Fractions.

111. Every fluxion that is a rational fraction is comprehended under this general formula,

\frac{(Ax^n + Bx^{n-1} + Cx^{n-2} + \&c.)\dot{x}}{A'x^{n'} + B'x^{n'-1} + C'x^{n'-2} + \&c.},

which, by putting $U$ to denote the expression between the parentheses in the numerator, and $V$ the denominator,

may be represented by $\frac{U\dot{x}}{V}$ . Now in the first

place, we remark that the greatest exponent of the powers of $x$ in the numerator may be supposed to be less than that of its powers in the denominator. For if it were not so, by dividing $U$ by $V$ , and calling $Q$ the quotient, and $R$ the remainder, we should have

\frac{U\dot{x}}{V} = Q\dot{x} + \frac{R\dot{x}}{V}, \text{ and}

\int \frac{U\dot{x}}{V} = \int Q\dot{x} + \int \frac{R\dot{x}}{V}.

Now, $Q$ being a rational and integer function $\int Q\dot{x}$ may be found, as in § 104, and it only remains to find

$\int \frac{R\dot{x}}{V}$ , an expression in which the highest exponent of the powers of $x$ in $R$ is less by unity than in $V$ ; so that the fraction $\frac{R\dot{x}}{V}$ may be generally expressed thus,

\frac{(Ax^{n-1} + Bx^{n-2} + Cx^{n-3} \dots + T)\dot{x}}{x^n + A'x^{n-1} + B'x^{n-2} + C'x^{n-3} \dots + T'}.

The general method of finding the fluent of a fractional expression of this form consists in decomposing it into a series of other fractions, the denominators of which are more simple. These fractions may be found by proceeding as follows: By putting the denominator of the proposed fraction equal to 0, we get this equation,

x^n + A'x^{n-1} + B'x^{n-2} \dots + T' = 0.

Suppose now that the roots of this equation are found, and that they are denoted by

-a, -a', -a'', -a''', \&c.

which quantities we shall suppose, in the first place, are all unequal. Then the expression which has been assumed as equal to 0, may (ALGEBRA, Sect. X.) be considered as the product of $n$ factors

x+a, x+a', x+a'', x+a''', \&c.

Let the proposed fraction $\frac{R}{V}$ be now assumed as equal to the sum of the simple fractions

\frac{N}{x+a}, \frac{N'}{x+a'}, \frac{N''}{x+a''}, \&c.

having for their denominators the simple factors of the denominator of the proposed fraction, and for their numerators quantities which are constant, but as yet are indetermined.

That we may avoid complicated calculations, and present a determinate object to the mind, let us suppose that the fluxion of which we are to find the fluent is,

\frac{(Ax^3 + Bx + C)\dot{x}}{x^3 + A'x^2 + B'x + C'}.

and that we have by the resolution of the cubic equation $x^3 + A'x^2 + B'x + C' = 0$ found

x^3 + A'x^2 + B'x + C' = (x+a)(x+a')(x+a'').

The

The fractions

\frac{N \dot{x}}{x+a}, \frac{N' \dot{x}}{x+a'}, \frac{N'' \dot{x}}{x+a''},

when reduced to a common denominator are

\frac{N(x+a')(x+a'')\dot{x}}{(x+a)(x+a')(x+a'')}, \frac{N'(x+a)(x+a'')\dot{x}}{(x+a)(x+a')(x+a'')},

\frac{N''(x+a)(x+a')\dot{x}}{(x+a)(x+a')(x+a'')}.

The common denominator of these fractions is the same as that of the proposed fraction, and each of the numerators, as well as their sum, is a function of $x$ of a degree lower than the denominator, that is, in the present case, it is a function of the second degree. By taking the actual products of the factors in the numerators, and adding the results, we find the sum of the fractions equal to

\frac{\dot{x}}{V} \left\{ \begin{aligned} & (N+N'+N'')x^2 \\ & + \{ N(a'+a'') + N'(a+a'') + N''(a+a') \} x \\ & + N'a'a'' + N'a'a' + N''a'a' \end{aligned} \right\} x

where $V$ denotes the common denominator $(x+a)(x+a')(x+a'') = x^3 + A'x^2 + B'x + C'$ . Setting aside

the factor $\frac{\dot{x}}{V}$ of the above expression, we are now to compare that part of it which involves the three indeterminate quantities $N, N', N''$ , with $A'x^2 + B'x + C'$ , the numerator of the proposed fraction, thus we obtain these three equations

\begin{aligned} N+N'+N'' &= A, \\ N(a'+a'') + N'(a+a'') + N''(a+a') &= B, \\ N'a'a'' + N'a'a' + N''a'a' &= C. \end{aligned}

By these equations, which are all of the first degree, we may determine the values of $N, N'$ and $N''$ , and thus

we have the proposed fraction $\frac{(A'x^2 + B'x + C')\dot{x}}{x^3 + A'x^2 + B'x + C'}$ equal to

\frac{N \dot{x}}{x+a} + \frac{N' \dot{x}}{x+a'} + \frac{N'' \dot{x}}{x+a''},

where $N, N', N''$ , and $a, a', a''$ , are constant and known quantities.

Put $x+a = x$ , then $\dot{x} = \dot{x}$ , and the fraction $\frac{N \dot{x}}{x+a}$ is transformed to $\frac{N \dot{x}}{x}$ , of which the fluent is $N \ln x = N \ln 1$ .

$(x+a)$ (§ 103). In like manner we find $\int \frac{N' \dot{x}}{x+a'} = N' \ln(x+a')$ , and $\int \frac{N'' \dot{x}}{x+a''} = N'' \ln(x+a'')$ , and consequently

\int \frac{(A'x^2 + B'x + C')\dot{x}}{x^3 + A'x^2 + B'x + C'} = N \ln(x+a) + N' \ln(x+a') + N'' \ln(x+a'') + \text{const.}

= \ln \left\{ (x+a)^N (x+a')^{N'} (x+a'')^{N''} \right\} + \text{const.}

where by const. is meant a constant quantity.

It is easy to extend this mode of proceeding to the general formula given at the beginning of this §; and it is obvious, that as often as the denominator of a rational fraction can be decomposed into real and unequal factors, the determination of the fluent of that fraction is attended with no other difficulty than this decomposition, which requires the numerical resolution of equations.

112. We have supposed that the factors of the denominator of the proposed fraction are unequal among themselves, and it is only when this is the case that the fraction can be decomposed into others, having all this form $\frac{N}{x+a}$ . If we suppose that the denominator $x^n + A'x^{n-1} + B'x^{n-2} + \dots + T'$ has a factor of the form $(x+a)^p$ , then the proposed fraction

\frac{(Ax^{n-1} + Bx^{n-2} + Cx^{n-3} + \dots) \dot{x}}{x^n + A'x^{n-1} + B'x^{n-2} + C'x^{n-3} + \dots + T'}

must be assumed equal to

\frac{(Px^{p-1} + Qx^{p-2} + Rx^{p-3} + \dots + Y) \dot{x}}{(x+a)^p} + \frac{N' \dot{x}}{x+a'} = \frac{N'' \dot{x}}{x+a''} + \dots

where $P, Q, R, \dots, Y, N', N''$ , &c. denote indeterminate but constant quantities, and $x+a', x+a'',$ &c. are the remaining factors of the denominator of the proposed fraction. To determine the quantities $P, Q, R, \dots, Y, N', N''$ , &c. we must now proceed in all respects as in last §, that is, we must reduce the fractions involving these quantities to a common denominator, which will be the same as the denominator of the proposed fraction; then we must add the numerators, and put the coefficient of each power of $x$ in the sum equal to the coefficient of the same power in the numerator of the proposed fraction. Thus we shall have as many equations as indeterminate quantities, and by resolving these equations, the value of these quantities will be found.

Having thus determined all the quantities $P, Q, R, \dots, Y$ , which enter into the fraction

\frac{(Px^{p-1} + Qx^{p-2} + \dots + Y) \dot{x}}{(x+a)^p}

its fluent may be found as shown in § 109. But we may also assume it equal to

\frac{M \dot{x}}{(x+a)^p} + \frac{M' \dot{x}}{(x+a)^{p-1}} + \frac{M'' \dot{x}}{(x+a)^{p-2}} + \dots + \frac{M''' \dot{x}}{x+a}

and, it is easy to see, that by reducing these fractions to

Inverse Method. a common denominator and adding them, the numerator of their sum will have the same form as that of the fraction whose fluent we are seeking; so that the values of the indeterminate quantities $M, M', \&c.$ will be found by putting the coefficients of the same power of $x$ in both numerators equal to each other. To find the fluent of

$\frac{M \dot{x}}{(x+a)^r}$ we may assume $x+a=z$ , then $\dot{x}=\dot{z}$ , and

\int \frac{M \dot{x}}{(x+a)^r} = \int \frac{M \dot{z}}{z^r} = \frac{M z^{-r+1}}{1-r} = \frac{M}{(1-r)(x+a)^{r-1}}.

In like manner

\int \frac{M' \dot{x}}{(x+a)^{r-1}} = \frac{M'}{(2-r)(x+a)^{r-2}}

and so on, all the fluents being algebraic, except the last $\int \frac{M' \dots \dot{x}}{x+a}$ which is $M' \dots 1. (x+a)$ , a logarithmic function.

113. In resolving the equation

x^n + A'x^{n-1} + B'x^{n-2} \dots + T = 0,

it may happen that some of its roots $a, a', a'', \&c.$ are imaginary quantities, and then some of the simple factors $x+a, x+a', x+a'', \&c.$ will be imaginary. These factors always occur in pairs (ALGEBRA, § 179.) and have this form

x+a+\beta\sqrt{-1}, \quad x+a-\beta\sqrt{-1},

so that their product

x^2 + 2ax + a^2 + \beta^2

is a real factor of the second degree. As every corresponding pair of imaginary simple factors may be united in this manner into a real factor of the second degree, if these factors are all unequal, we may avoid introducing imaginary quantities into the fluent of the proposed fraction by proceeding thus. Let $x+a+\beta\sqrt{-1}$ and $x+a-\beta\sqrt{-1}$ , denote two corresponding imaginary simple factors of the denominator. Instead of the two simple fractions

\frac{N \dot{x}}{x+a+\beta\sqrt{-1}}, \quad \frac{N' \dot{x}}{x+a-\beta\sqrt{-1}},

which would have been assumed if the factors had been real, assume a single fraction

\frac{(Kx+L) \dot{x}}{x^2+2ax+a^2+\beta^2}

the denominator of which is a real function of $x$ of the second degree, viz. that which is the product of the two imaginary factors. Here $K$ and $L$ denote two constant but indeterminate coefficients, the values of which, as

VOL. VIII. Part II.

also those of the other indeterminate coefficients are to be found as before.

If the denominator of the proposed fraction have several equal factors of the second degree resulting from its imaginary simple factors, so that the product of those equal factors is

(x^2+2ax+a^2+\beta^2)^s;

then, corresponding to this product, we must, among the fractions having indeterminate coefficients, assume one of this form

\frac{(Q'x^{2s-1}+R'x^{2s-3}+\dots+Y') \dot{x}}{(x^2+2ax+a^2+\beta^2)^s},

where $Q', R' \dots Y'$ denote constant and indeterminate coefficients, the values of which will be found in all respects as those of the others.

We are now to find the fluents of these two fluxional expressions, beginning with the first, viz.

\frac{(Kx+L) \dot{x}}{x^2+2ax+a^2+\beta^2}, \quad \text{or} \quad \frac{(Kx+L) \dot{x}}{(x+a)^2+\beta^2}.

Put $x+a=z$ , then it becomes

\frac{(Kz+L-Ka) \dot{z}}{z^2+\beta^2};

and this again, by putting $L-Ka=M$ , is resolved into these two fluxions

\frac{Kz \dot{z}}{z^2+\beta^2} + \frac{M \dot{z}}{z^2+\beta^2}.

We can immediately find the fluent of the first of these, by putting $z^2+\beta^2=v$ , for then $z \dot{z} = \frac{v}{2}$ , and

\int \frac{Kz \dot{z}}{z^2+\beta^2} = \frac{K}{2} \int \frac{v}{v} = K \frac{1}{2} \ln v, \quad (\text{§ 103.}), \\ = K \ln \sqrt{z^2+\beta^2}.

With respect to the other fluxion, if we put $z=\beta y$ , we have

\frac{M \dot{z}}{z^2+\beta^2} = \frac{M}{\beta} \frac{\dot{y}}{1+y^2},

but we have seen (§ 60.) that $\frac{\dot{y}}{1+y^2}$ is the fluxion of an arch of which the tangent is $y$ , therefore

\int \frac{M}{\beta} \frac{\dot{y}}{1+y^2} = \frac{M}{\beta} \text{arc}(\tan. = y) + \text{const.} \\ = \frac{M}{\beta} \text{arc}(\tan. = \frac{z}{\beta}) + \text{const.}

It is proper to remark that if $\frac{z}{\beta}$ be the tangent of an arch, then the sine of that arch is $\frac{z}{\sqrt{z^2+\beta^2}}$ , and its cosine is $\frac{\beta}{\sqrt{z^2+\beta^2}}$ , thus we may express the fluent

Inverse Method. under different forms, by introducing the sine or cosine of the arch instead of its tangent.

If instead of $x$ we substitute in these two fluents $x + \alpha$ we find that the fluent of $\frac{(Kx+L)\dot{x}}{x^2+2\alpha x+\alpha^2+\beta^2}$ is

K \cdot \sqrt{x^2+2\alpha x+\alpha^2+\beta^2} + \frac{L-K\alpha}{\beta} \arctan\left(\frac{x+\alpha}{\beta}\right) + \text{const.}

14. To find the fluent of the expression

\frac{(Q'x^{q-1} + R'x^{q-2} + \dots + Y')\dot{x}}{(x^2+2\alpha x+\alpha^2+\beta^2)^r}

we first transform it to

\frac{(Kx+L)\dot{x}}{(x^2+2\alpha x+\alpha^2+\beta^2)} + \frac{(K'x+L')\dot{x}}{(x^2+2\alpha x+\alpha^2+\beta^2)^{r-1}} + \dots + \frac{(K''x+L'')\dot{x}}{x^2+2\alpha x+\alpha^2+\beta^2}

where $K, L, K', L', \dots$ denote indeterminate but constant coefficients, which may be determined by reducing these fractions to a common denominator, and proceeding as in the two preceding $\S$ . Then the whole difficulty is reduced to the finding of the fluxion of the expression

\frac{(Kx+L)\dot{x}}{(x^2+2\alpha x+\alpha^2+\beta^2)^r} = \frac{(Kx+L)\dot{x}}{((x+\alpha)^2+\beta^2)^r}

where $q$ denotes some integer number. To simplify this expression put $x+\alpha = \infty$ , and $L-K\alpha = M$ , then it becomes $\frac{(K\infty \times M)\dot{\infty}}{(\infty^2+\beta^2)^r}$ , which we shall now show may be reduced to $\int \frac{H\dot{\infty}}{(\infty^2+\beta^2)^{r-1}}$ . To effect this reduction we decompose its fluent into two parts.

\int \frac{K\infty\dot{\infty}}{(\infty^2+\beta^2)^r} + \int \frac{M\dot{\infty}}{(\infty^2+\beta^2)^r}

The fluent of the first part may be immediately found by putting $\infty^2+\beta^2 = v$ ; for then $\infty\dot{\infty} = \frac{v}{2}$ and

\int \frac{K\infty\dot{\infty}}{(\infty^2+\beta^2)^r} = \int \frac{K\frac{v}{2}}{2v^r} = \frac{Kv^{-r+2}}{2(1-q)}

Let us now suppose that the fluent of the second part $\frac{M\dot{\infty}}{(\infty^2+\beta^2)^r}$ is equal to the sum of the algebraic function

$\frac{G\infty}{(\infty^2+\beta^2)^{r-1}}$ , and another function, which is the fluent of $\frac{H\dot{\infty}}{(\infty^2+\beta^2)^{-1}}$ , that is, let us assume

\int \frac{M\dot{\infty}}{(\infty^2+\beta^2)^r} = \frac{G\infty}{(\infty^2+\beta^2)^{r-1}} + \int \frac{H\dot{\infty}}{(\infty^2+\beta^2)^{-1}}

where $G$ and $H$ are constant but indeterminate coefficients. To determine these, let the fluxion of each side of this equation be taken (observing that the fluxion of a quantity having the sign $\int$ prefixed to it is the same quantity only without that sign); thus we have

\frac{M\dot{\infty}}{(\infty^2+\beta^2)^r} = \frac{G\dot{\infty}}{(\infty^2+\beta^2)^{r-1}} - \frac{2(q-1)G\infty^2\dot{\infty}}{(\infty^2+\beta^2)^r} + \frac{H\dot{\infty}}{(\infty^2+\beta^2)^{-1}}

and from this equation, by rejecting what is common to each term, we find

M = G(\infty^2+\beta^2) - 2(q-1)G\infty^2 + H(\infty^2+\beta^2),

and hence

M = G\beta^2 + H\beta^2 + (G-2(q-1)G+H)\infty^2;

Therefore by comparing together like terms, we find

M = G\beta^2 + H\beta^2, \quad G-2(q-1)G+H=0;

and from these equations we get

G = \frac{M}{(2q-2)\beta^2}, \quad H = \frac{(2q-3)M}{(2q-2)\beta^2}.

Let these values of $G$ and $H$ be now substituted in our assumed equation, and it becomes

\int \frac{M\dot{\infty}}{(\infty^2+\beta^2)^r} = \frac{M}{(2q-2)\beta^2} \frac{\infty}{(\infty^2+\beta^2)^{r-1}} + \frac{M(2q-3)}{(2q-2)\beta^2} \int \frac{\dot{\infty}}{(\infty^2+\beta^2)^{-1}}

Thus we have reduced the determination of the fluent of $\frac{M\dot{\infty}}{(\infty^2+\beta^2)^r}$ to that of $\frac{\dot{\infty}}{(\infty^2+\beta^2)^{r-1}}$ , and by proceeding in the same manner with this last fluxion, its fluent may be made to depend on that of $\frac{\dot{\infty}}{(\infty^2+\beta^2)^{-1}}$ ; but this will be more readily effected by simply substituting $q-1$ instead of $q$ , and supposing $M=1$ in the preceding equation.

Thus

Thus we shall obtain

\int \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}} = \frac{1}{(2q-4)\beta^2} \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}} + \frac{(2q-5)}{(2q-4)\beta^2} \int \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}}

Substituting now this value of

\int \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}}

in the

former equation, we have

\int \frac{M \dot{x}}{(x^2 + \beta^2)^q}

equal to

\frac{M}{(2q-2)\beta^2} \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}} = \frac{(2q-3)M}{(2q-2)(2q-4)\beta^2} \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}} + \frac{(2q-3)(2q-5)M}{(2q-2)(2q-4)\beta^2} \int \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}}

It is easy to see, that like as we obtained an expression for the fluent of $\frac{\dot{x}}{(x^2 + \beta^2)^{q-1}}$ by substituting $q-1$ for $q$ , and supposing $M=1$ , in the equation preceding the last but one; so by substituting $q-2$ for $q$ , we shall obtain an expression for the fluent of $\frac{\dot{x}}{(x^2 + \beta^2)^{q-2}}$ , which expression will consist of two terms; one an algebraic function of $x$ , and the other $\int \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}}$ multiplied by a constant and given coefficient. This value of $\int \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}}$ when substituted in the last equation will produce an expression for $\int \frac{M \dot{x}}{(x^2 + \beta^2)^q}$ consisting of algebraic quantities and $\int \frac{\dot{x}}{(x^2 + \beta^2)^{q-1}}$ . By continuing this process it is evident that we shall at last have $\int \frac{M \dot{x}}{(x^2 + \beta^2)^q}$ expressed by a series of algebraic quantities, and $\int \frac{\dot{x}}{(x^2 + \beta^2)^q}$ , and here we must stop, for if we repeat the process with a view to make the fluent depend on $\int \frac{\dot{x}}{(x^2 + \beta^2)^q}$ that is on $\int \dot{x}$ , or $x$ , we shall find that the coefficient of this quantity becomes infinite. As to the fluent of $\frac{\dot{x}}{x^2 + \beta^2}$ we have exhibited the expression for it in last §.

In comparing together the results which have been obtained in the preceding articles, it must appear that when a fluxion is expressed by a rational fraction, if we grant the resolution of equations, the fluent may always

be assigned either algebraically, or by means of arches of a circle or logarithms; and that to prepare it for a solution, we must decompose the fraction into others, whose denominators may be either binomial or trinomial quantities. This decomposition may always be effected by the method of indeterminate coefficients. There are, however, several analytical artifices by which the labour of calculation may be greatly shortened. These we now proceed to explain.

115. Let us recur to the fraction $\frac{U}{V}$ and suppose that $x+a$ is one of the unequal factors of the denominator $V$ , so that we have $V = (x+a)Q$ . Let us now put $\frac{U}{V} = \frac{A}{x+a} + \frac{P}{Q}$ , $A$ being supposed a constant quantity, and $P$ an indeterminate function of $x$ , but such as not to be divisible by $x+a$ . Then we have $U = AQ + P(x+a)$ , and hence $P = \frac{U - AQ}{x+a}$ . As $P$ is an integer function with respect to $x$ , it follows from this equation that $U - AQ$ , which is also a rational and integer function of $x$ , is divisible by $x+a$ , and consequently has $x+a$ for a factor; therefore, the function $U - AQ$ will vanish when we substitute $-a$ in it instead of $x$ , seeing that $-a$ is the value of $x$ that makes the factor $x+a=0$ . Let us denote by $u$ and $q$ , what $U$ and $Q$ become by this substitution, which however will not affect the indeterminate quantity $A$ , because it is independent of $x$ . We have therefore $u - Aq = 0$ , and consequently $A = \frac{u}{q}$ .

This value of $A$ requires that we should know the function $Q$ given by the equation $V = (x+a)Q$ , and we may always find it by dividing $V$ by $x+a$ . The direct method of fluxions affords also a very simple method of determining it. For by taking the fluxion of the above equation we have

\frac{\dot{V}}{x} = Q + (x+a) \frac{\dot{Q}}{x};

if in this result we make $x+a=0$ , or $x=-a$ , and denote by $v$ what $\frac{\dot{V}}{x}$ becomes by that substitution, we shall have $v=q$ , and consequently $A = \frac{u}{v}$ .

The expression $A = \frac{u}{v}$ has always a finite value, for the numerator and denominator can never become $=0$ , because we suppose the fraction $\frac{U}{V}$ reduced to its lowest terms, and consequently, that the numerator $U$ has not for a factor $x+a$ , which is a factor of the denominator, but which being contained in it only once does not enter into $Q$ .

116. Let us now consider how the numerators of the fractions, into which the proposed fraction $\frac{U}{V}$ is to be decomposed, are to be found in the case of the denominator

Invertor V having equal factors of the first degree. In this case we have $V = Q(x+a)^n$ , and we assume

\frac{U}{V} = \frac{A}{(x+a)^n} + \frac{B}{(x+a)^{n-1}} + \frac{C}{(x+a)^{n-2}} + \dots + \frac{N}{x+a} + \frac{P}{Q}

By reducing to a common denominator, we find U equal to

Q \left\{ A + B(x+a) + C(x+a)^2 + \dots + N(x+a)^{n-1} \right\} + P(x+a)^n

and P equal to

\frac{U - Q(A + B(x+a) + C(x+a)^2 + \dots + N(x+a)^{n-1})}{(x+a)^n}

and as P ought to be an integer function of $x$ , the numerator of its value is necessarily divisible $n$ times successively by $x+a$ ; therefore, that numerator ought to be equal to 0, when $-a$ is substituted in it instead of $x$ . Now this substitution being made, each of the terms of the numerator which is multiplied by $x+a$ vanishes, so that there remains only $U - AQ$ , but that this quantity may be divisible by $x+a$ , it is necessary that $U - qA = 0$ , where $u$ and $q$ denote the same as in last §, hence $A = \frac{u}{q}$ .

This value of A changes $U - QA$ into $U - \frac{u}{q}Q$ , which must be divisible by $x+a$ . Let us, with a view to abridge, put $U - \frac{u}{q}Q = U'(x+a)$ , then substituting this quantity in the value of P, and dividing both numerator and denominator by $x+a$ , we have P equal to

\frac{U' - Q(B + C(x+a) + \dots + N(x+a)^{n-2})}{(x+a)^{n-1}}

Now to obtain B we make $x+a=0$ , then, putting $u'$ to denote what $U'$ becomes by substituting $-a$ in it in place of $x$ , we have $u' - qB = 0$ , and $B = \frac{u'}{q}$ .

Instead of B let its value be substituted in $U' - QB$ , and this quantity becomes $U' - \frac{u'}{q}Q$ , which vanishing when $x+a=0$ , will have $x+a$ for a divisor; therefore, we may put $U' - \frac{u'}{q}Q = U''(x+a)$ , then, substituting this last quantity instead of the former in the value of P, and dividing the numerator and denominator by $x+a$ , we have P equal to

\frac{U'' - Q(C + D(x+a) + \dots + N(x+a)^{n-3})}{(x+a)^{n-2}}

By continuing the same mode of reasoning, and the same notation, we find $u'' - qC = 0$ , and $C = \frac{u''}{q}$ . And so on with the remaining quantities.

The direct method of fluxions facilitates greatly the preceding operations. For the numerator of P being divisible by $(x+a)^n$ is necessarily of this form $X(a+x)^n$ , X being an integer function of $x$ , but which does not contain the factor $x+a$ . Now agreeably to what has been shown in § 88, the successive fluxions of this numerator, as far as the $n-1$ order inclusive, vanish when $x+a$ is supposed $=0$ . By giving to the numerator the following form

Q \left( \frac{U}{Q} - A - B(x+a) - C(x+a)^2 - \dots \right)

and observing that the function Q does not contain the factor $x+a$ , it is manifest that it is only the part of this expression between the parentheses which ought to be divisible by $(x+a)^n$ . Let us put $\frac{U}{Q} = Z$ , then the successive fluxions of that part are

\dot{Z} - Bx - 2C(x+a)x - 3D(x+a)^2x \dots

\ddot{Z} - 2Cx^2 - 2 \cdot 3D(x+a)x^2 \dots

\ddot{\dot{Z}} - 2 \cdot 3Dx^3 \dots

and these results ought all to vanish when we put $x+a=0$ . Thus we have

Z - A = 0, \text{ and } A = \frac{u}{q},

\dot{Z} - Bx = 0, B = \frac{\dot{Z}}{x},

\ddot{Z} - 2Cx^2 = 0, C = \frac{\ddot{Z}}{2x^2},

\ddot{\dot{Z}} - 2 \cdot 3Dx^3 = 0, D = \frac{\ddot{\dot{Z}}}{2 \cdot 3x^3},

&c. &c.

observing that in each of these functions $\frac{\dot{Z}}{x}, \frac{\ddot{Z}}{2x^2}$ , we must substitute $-a$ instead of $x$ .

The most simple way to find the value of Q in this case is to divide V by $(x+a)^n$ , but we may also find it by the direct method of fluxions, as in the preceding §; for, since $V = Q(x+a)^n$ , if we take the fluxion of each side of this equation $n$ times, and then make $x+a=0$ , we shall find, § 88, the $n$ th fluxion of V equal to $1 \cdot 2 \cdot 3 \dots n Q x^n$ ; and consequently

Q = \frac{n \text{th flux. of } V}{1 \cdot 2 \dots n x^n}.

117. Let us now consider how we are to find the numerator of the fraction which forms a part of $\frac{U}{V}$ when it has this form

\frac{Ax+B}{x^3+2ax+a^2+\beta^2}

Assume

Inverse Method. Assume.

\frac{U}{V} = \frac{Ax+B}{x^3+2\alpha x+\alpha^2+\beta^2} + \frac{P}{Q};

then, reducing the latter part of this equation to a common denominator, we find

U = Q(Ax+B) + P(x^3+2\alpha x+\alpha^2+\beta^2).

Hence we deduce

P = \frac{U - Q(Ax+B)}{x^3+2\alpha x+\alpha^2+\beta^2}.

As $P$ is supposed to be an integer function with respect to $x$ , it follows that $U - Q(Ax+B)$ is divisible by $x^3+2\alpha x+\alpha^2+\beta^2$ ; therefore, the former of these two quantities must contain among its factors those of the latter, and the quantities, which, being substituted for $x$ , cause the latter to vanish, must also make the former vanish. But the factors of $x^3+2\alpha x+\alpha^2+\beta^2$ are $x+\alpha+\beta\sqrt{-1}$ , and $x+\alpha-\beta\sqrt{-1}$ , and these, being put each $=0$ , gives us $x = -(\alpha+\beta\sqrt{-1})$ , and $x = -(\alpha-\beta\sqrt{-1})$ , therefore, each of these values of $x$ being substituted in $U - Q(Ax+B)$ ought to make that quantity vanish. Let us denote by $u \pm i\sqrt{-1}$ , and by $q \pm i\sqrt{-1}$ what $U$ and $Q$ respectively become when $-(u \pm i\sqrt{-1})$ is substituted in each instead of $x$ , then, after this transformation, we have

-(q \pm i\sqrt{-1}) \left\{ -A(u \pm i\sqrt{-1}) + B \right\} = 0.

This equation is twofold, because of the sign $\pm$ with which several of its terms are affected, and it is equivalent to those which would be formed by putting the real part equal to 0, and the imaginary part $=0$ ; from this consideration we have

\begin{aligned} u + q\alpha A - q\beta A - qB &= 0, \\ u' + q\beta A + q\alpha A - q'B &= 0, \end{aligned}

two equations which give us the values of $A$ and $B$ .

The function $Q$ may be found as in § 115. For, if we take the fluxions of each side of the equation

Q(x^3+2\alpha x+\alpha^2+\beta^2) = V,

and afterwards make

x^3+2\alpha x+\alpha^2+\beta^2=0,

we find $Q(2\alpha x+2\alpha^2) = \dot{V}$ and hence

Q = \frac{\dot{V}}{2\alpha x+2\alpha^2};

Let the two values of $x$ , to wit $-(u \pm i\sqrt{-1})$ , be substituted instead of it in this equation, then, putting $v \pm i\sqrt{-1}$ to denote what the expression $\frac{\dot{V}}{x}$ becomes by that substitution, and writing $q \pm i\sqrt{-1}$ instead of $Q$ , we have

q \pm i\sqrt{-1} = \frac{v \pm i\sqrt{-1}}{\pm 2\beta\sqrt{-1}},

which, by multiplying the terms of the fraction on the latter side of the equation by $\sqrt{-1}$ , becomes

q \pm i\sqrt{-1} = \frac{v \pm i\sqrt{-1}}{\pm 2\beta}.

Hence by putting the real parts of each side of this equation equal to each other, and also the imaginary parts equal to each other, we find

q = -\frac{v}{2\beta}, \quad q' = \frac{v}{2\beta}.

118. If the factor $x^3+2\alpha x+\alpha^2+\beta^2$ is found several times in the denominator of $V$ , so that

$V = Q(x^3+2\alpha x+\alpha^2+\beta^2)^n$ , then, § 113, we assume in this case $\frac{U}{V}$ equal to

\frac{Ax+B}{(x^3+2\alpha x+\alpha^2+\beta^2)^n} + \frac{A'x+B'}{(x^3+2\alpha x+\alpha^2+\beta^2)^{n-1}} + \frac{A''x+B''}{(x^3+2\alpha x+\alpha^2+\beta^2)^{n-2}} + \dots + \frac{P}{Q}

reducing this expression to a common denominator, and so ordering the equation as to bring $P$ to stand alone on one side, we find $P$ equal to

U = Q \left\{ \frac{Ax+B + (A'x+B')(x^3+2\alpha x+\alpha^2+\beta^2)}{(x^3+2\alpha x+\alpha^2+\beta^2)^n} + \frac{(A''x+B'')(x^3+2\alpha x+\alpha^2+\beta^2)^2}{(x^3+2\alpha x+\alpha^2+\beta^2)^n} + \dots \right\}

By reasoning in this as in the preceding case, it may be concluded that the numerator of this expression ought to vanish when $-(u \pm i\sqrt{-1})$ is substituted in it instead of $x$ ; therefore putting $u \pm i\sqrt{-1}$ , and $q \pm i\sqrt{-1}$ to denote the same things as before, we deduce from that substitution

-(q \pm i\sqrt{-1}) \left\{ -A(u \pm i\sqrt{-1}) + B \right\} = 0

the very same equation for the determination of $A$ and $B$ , as we have already found in last §.

Having found the values of these quantities, they may be substituted in the numerator of $P$ , and the terms $U - Q(Ax+B)$ becoming divisible by $x^3+2\alpha x+\alpha^2+\beta^2$ , the whole expression becomes divisible by the same quantity. Calling therefore $U'$ the quotient arising from the division of $U - Q(Ax+B)$ by $x^3+2\alpha x+\alpha^2+\beta^2$ , we have $P$ equal to

U' = Q \left[ A'x+B' + (A''x+B'')(x^3+2\alpha x+\alpha^2+\beta^2) + \dots \right]

If in this numerator we substitute instead of $x$ its values deduced from the equation $x^3+2\alpha x+\alpha^2+\beta^2=0$ , and put the result $=0$ , we may determine $A'$ and $B'$ in the very same way that we have already determined $A$ and $B$ , and by proceeding in this manner we shall find the remaining coefficients $A''$ , $B''$ , &c.

This case is quite analogous to that which has been already treated in § 116, and the direct method of fluxions applies to it in the same manner as to the other. For since $Q$ does not contain the factor $x^3+2\alpha x+\alpha^2+\beta^2$ , if the numerator of $P$ be divided by the function $Q$ , the result, which may be denoted by $r$ , ought to be of this form

Inverse Method. form $r = X(x^2 + 2ax + a^2 + \beta^2)^n$ and consequently ought to vanish, as well as all its fluxions, from the first order to the $n-1$ order, inclusively, when $x^2 + 2ax + a^2 + \beta^2 = 0$ ; this being the case, we have these equations,

r = 0, \dot{r} = 0, \ddot{r} = 0, \dots

and so on to the $n-1$ fluxion of $r$ , which ought also to be $= 0$ ; each of these equations becomes twofold when we substitute, instead of $x$ , the values of which it is susceptible in consequence of the equation $x^2 + 2ax + a^2 + \beta^2 = 0$ . By putting the real and the imaginary parts separately $= 0$ , we shall obtain as many equations as are sufficient to determine $A, B, A', B', \dots$

It may also be remarked, that from the equation

V = Q(x^2 + 2ax + a^2 + \beta^2)^n

we find $Q$ equal to the quotient arising from the division of the $n$ th fluxion of $V$ by the $n$ th fluxion of $x^2 + 2ax + a^2 + \beta^2$ , observing to assume

x^2 + 2ax + a^2 + \beta^2 = 0.

119. We shall now give some applications of what has been said relative to the fluents of rational fractions. Suppose the fraction to be

\frac{x}{x^2 + x^2 - x^4 - x^3}

The factors of its denominator are easily found, for it may be put under this form

x^3(x^2 + x^4 - x - 1) = x^3(x+1)(x^4-1),

the factor $x^4-1$ may be decomposed into $x^2-1$ and $x^2+1$ , or $x-1, x+1$ , and $x^2+1$ , thus we have the denominator equal to

x^3(x-1)(x+1)^2(x^2+1)

therefore (§ 111, § 112, and § 113.) the proposed fraction is to be decomposed as follows

\frac{Ax}{x-1} + \frac{Bx}{(x+1)^2} + \frac{Cx}{x+1} + \frac{Dx}{x^2} + \frac{Ex}{x^2} + \frac{Fx}{x} + \frac{(Gx+H)x}{1+x^2}

By reducing these fractions to a common denominator, and comparing the numerator of their sum with that of the proposed fraction, we might determine the unknown quantities $A, B, C, \dots$ we shall, however, rather employ the methods that have just been explained.

By comparing this particular example $\frac{x}{x^2+x^2-x^4-x^3}$

with the general expression $\frac{Ux}{V}$ , it appears that $U=1$ ,

and $V=x^2+x^2-x^4-x^3$ . First let us investigate the numerator of the fraction $\frac{A}{x-1}$ , and for this purpose

we employ the formula $A = \frac{u}{v}$ (§ 115.). As we have

$U=1$ , it is evident that $u=1$ ; and since $V=x^2+x^2$

$-x^4-x^3$ , therefore $\frac{\dot{V}}{x} = 8x^7 + 7x^6 - 4x^3 - 3x^2$ . If in this expression we substitute $+1$ instead of $x$ (viz. the value of $x$ deduced from the equation $x-1=0$ ) we find the result to be 8, therefore $v=8$ . So that

A = \frac{u}{v} = \frac{1}{8}, \text{ and } \frac{A}{x-1} = \frac{1}{8} \frac{1}{x-1}.

Let us next investigate the values of $B$ and $C$ in the fractions $\frac{B}{(x+1)^2}$ , and $\frac{C}{x+1}$ , by means of the rule of § 116, and that we may make the symbols expressing the quantity under consideration agree with those employed in that formula, let us exchange the letters $B$ and $C$ for $A$ and $B$ , so that we are to consider $\frac{A}{(x+1)^2}$

+ $\frac{B}{x+1}$ . In the first place we have

Q = \frac{x^2 + x^2 - x^4 - x^3}{(x+1)^2} = x^6 - x^5 + x^4 - x^3;

Put $x+1=0$ , then $x=-1$ ; substituting now this value of $x$ in the value of $Q$ , the result is $4=q$ , therefore

$A = \frac{u}{q} = \frac{1}{4}$ . Let this value of $A$ be substituted for $A$ in the expression for $U'$ in the § above cited, and we have

U' = \frac{U - AQ}{x+1} = \frac{4 - x^6 + x^5 - x^4 + x^3}{4(x+1)} = \frac{-x^5 + 2x^4 - 3x^3 + 4x^2 - 4x + 4}{4}

Hence putting $-1$ instead of $x$ in the expression for $U'$ , we have $u' = \frac{9}{2}$ and $B = \frac{u'}{q} = \frac{9}{8}$ . Thus the two fractions under consideration are found to be $\frac{1}{4} \cdot \frac{1}{(x+1)^2}$ and $\frac{9}{8} \cdot \frac{1}{x+1}$ .

We might have deduced the value of $B$ from the formula $B = \frac{Z}{x}$ , § 116, where $Z$ is put for $\frac{U}{Q}$ for we have

Z = \frac{U}{Q} = \frac{1}{(x^6 - x^5 + x^4 - x^3)},

\text{and } \frac{\dot{Z}}{x} = -\frac{6x^5 - 5x^4 + 4x^3 - 3x^2}{(x^6 - x^5 + x^4 - x^3)^2}.

If in this expression we substitute $-1$ instead of $x$ , it becomes $\frac{18}{16} = \frac{9}{8}$ , the same value for $B$ as before.

Let us now consider the fractions $\frac{D}{x^2} + \frac{E}{x^2} + \frac{F}{x}$ , or exchanging the symbols $D, E, F$ for $A, B, C$ ,

\frac{A}{x^2} + \frac{B}{x^2} + \frac{C}{x}.

The numerators $A, B, C$ may all be found from these formulas of § 116.

$A =$

A = \frac{U}{Q} = Z, \quad B = \frac{\dot{Z}}{x}, \quad C = \frac{\ddot{Z}}{x^2},

observing that in this case $Q = x^5 + x^4 - x - 1$ ; and that we must substitute 0 instead of $x$ in each formula, after taking the fluxions. Now we have

Z = \frac{U}{Q} = \frac{x}{x^5 + x^4 - x - 1},

\dot{Z} = -\frac{5x^4 + 4x^3 - 1}{(x^5 + x^4 - x - 1)^2},

\ddot{Z} = -\frac{20x^3 + 12x^2}{(x^5 + x^4 - x - 1)^3},

+\frac{2(5x^4 + 4x^3 - 1)^2}{(x^5 + x^4 - x - 1)^4};

Hence putting $x=0$ , we find

A = -1, \quad B = +1, \quad C = -1,

\frac{A}{x^3} + \frac{B}{x^2} + \frac{C}{x} = -\frac{1}{x^3} + \frac{1}{x^2} - \frac{1}{x}.

There yet remains the fraction $\frac{Gx+H}{x^4+1}$ , or $\frac{Ax+B}{x^4+1}$ to be considered. It may be found by subtracting the sum of all the others from the proposed fraction; we proceed, however, to find it directly by the formulas of § 117. In the first place we have $Q = x^5 + x^4 - x^4 - x^3$ ; next, the factor $x^4+1$ being put $=0$ gives $x = \pm\sqrt{-1}$ , $\alpha=0$ , $\beta=0$ . Hence we find

q \pm q' \sqrt{-1} = -2 \pm 2\sqrt{-1}, \quad u = 1, \quad \text{and} \quad u' = 0.

The equations which determine $A$ and $B$ thus become

1 + 2A + 2B = 0, \quad 2A - 2B = 0,

therefore $A = B = -\frac{1}{4}$ , and

\frac{Ax+B}{x^2+1} = -\frac{1}{4} \cdot \frac{x+1}{x^2+1}.

Thus, upon the whole, the proposed fraction

$\frac{x}{x^5+x^4-x^4-x^3}$ is decomposed into the following

\frac{1}{8} \cdot \frac{x}{x-1} + \frac{1}{4} \cdot \frac{x}{(x+1)^2} + \frac{9}{8} \cdot \frac{x}{x+1} - \frac{x}{x^3} + \frac{x}{x^2} - \frac{x}{x} - \frac{1}{4} \cdot \frac{(x+1)x}{x^2+1}.

The manner of finding the fluent of each of these has been already explained, and the result of taking all the fluents is

\left. \begin{aligned} & \frac{1}{8} \ln(x-1) - \frac{1}{4} \ln \frac{x+1}{x} \\ & + \frac{9}{8} \ln(x+1) + \frac{1}{2x^2} - \frac{1}{x} - \ln x \\ & - \frac{1}{8} \ln(x^2+1) - \frac{1}{4} \arctan(x) + \text{const.} \end{aligned} \right\}

The union of all the algebraic terms produces the fraction $\frac{2-2x-5x^3}{4x^2-(1+x)}$ , and that of the logarithmic quantities gives

\begin{aligned} & \frac{1}{8} \ln(x-1) + \frac{1}{8} \ln(x+1) + \ln(x+1) \\ & - \frac{1}{8} \ln(x^2+1) - \ln x \\ & = \frac{1}{8} \ln \left( \frac{x^2-1}{x^2+1} \right) + \ln \left( \frac{x+1}{x} \right). \end{aligned}

We have therefore upon the whole $\int \frac{x}{x^5+x^4-x^4-x^3}$ equal to

\begin{aligned} & \frac{2-2x-5x^3}{4x^2(1+x)} + \frac{1}{8} \ln \left( \frac{x^2-1}{x^2+1} \right) \\ & + \ln \left( \frac{x+1}{x} \right) - \frac{1}{4} \arctan(x) + \text{const.} \end{aligned}

120. When a fluxion is a rational fraction having either of these forms

\frac{x^m \dot{x}}{x^n \pm a^n}, \quad \frac{x^m \ddot{x}}{x^n \pm 2pa^{n-1}x^n + a^{2n}},

we can always, by the application of a particular theorem in analysis, resolve its denominator into real factors of the first and second degrees. The theorem to which we allude is this. Let $n$ be any positive integer, and let $x$ denote any arch of a circle, of which the radius is unity, then

(\cos x \pm \sqrt{-1} \sin x)^n = \cos nx \pm \sqrt{-1} \sin nx.

We proceed to prove this theorem. Because

\begin{aligned} & (\cos x + \sqrt{-1} \sin x)(\cos x - \sqrt{-1} \sin x) \\ & = \cos^2 x + \sin^2 x = 1. \end{aligned}

\text{If we put } \cos x \pm \sqrt{-1} \sin x = v,

\text{Then } \cos x \mp \sqrt{-1} \sin x = \frac{1}{v};

Therefore taking the sum of these two equations,

2 \cos x = v + \frac{1}{v}.

Now by the arithmetic of fines, (ALGEBRA, § 358.)

\begin{aligned} 2 \cos 2x &= 2(2 \cos x \times \cos x - 1), \\ 2 \cos 3x &= 2(2 \cos x \times \cos 2x - \cos x), \\ 2 \cos 4x &= 2(2 \cos x \times \cos 3x - \cos 2x), \\ 2 \cos 5x &= 2(2 \cos x \times \cos 4x - \cos 3x), \\ & \text{&c.} \end{aligned}

Therefore, substituting in the first of these equations

$v + \frac{1}{v}$ instead of $2 \cos x$ , we have

2 \operatorname{cos}. 2z = \left(v + \frac{1}{v}\right) \left(v + \frac{1}{v}\right) - 2

= v^2 + \frac{1}{v^2}.

In like manner, substituting in the second equation, $v + \frac{1}{v}$ instead of $2 \operatorname{cos}. z$ , and $v^2 + \frac{1}{v^2}$ instead of $2 \operatorname{cos}. 2z$ , we have

2 \operatorname{cos}. 3z = \left(v + \frac{1}{v}\right) \left(v^2 + \frac{1}{v^2}\right) - \left(v + \frac{1}{v}\right)

= v^3 + \frac{1}{v^3}.

Proceeding in the same way with the third and following equations, we find

2 \operatorname{cos}. 4z = v^4 + \frac{1}{v^4};

2 \operatorname{cos}. 5z = v^5 + \frac{1}{v^5};

To that we may conclude in general that

2 \operatorname{cos}. nz = v^n + \frac{1}{v^n};

Hence we have this quadratic equation

v^{2n} - 2 \operatorname{cos}. nz \times v^n + 1 = 0,

from which, by completing the square, we find

v^n = \operatorname{cos}. nz \pm \sqrt{(\operatorname{cos}.^2 nz - 1)};

therefore, by substituting for $v$ the quantity it was put to represent, and observing that $\sqrt{(\operatorname{cos}.^2 nz - 1)} = \sqrt{(-\sin^2 nz)} = \sqrt{-1} \sin. nz$ , we have $(\operatorname{cos}. nz \pm \sqrt{-1} \sin. nz)^n = \operatorname{cos}. nz \pm \sqrt{-1} \sin. nz$ , as was to be proved.

121. The function $x^n \pm a^n$ is transformed to $a^n(y^n \pm 1)$ by putting $x = ay$ , and to discover its factors, we must resolve the equation

y^n \pm 1 = 0.

The expression $y = \operatorname{cos}. z + \sqrt{-1} \sin. z$ satisfies this equation, by a very simple determination of the arch $z$ ; for we have $y^n = (\operatorname{cos}. z + \sqrt{-1} \sin. z)^n = \operatorname{cos}. nz + \sqrt{-1} \sin. nz$ , and as by putting $\pi$ to denote half the circumference, and $m$ any whole number, we have (ALGEBRA, § 352)

\sin. m\pi = 0, \operatorname{cos}. m\pi = \pm 1,

where the sign $+$ is to be taken, if $m$ be an even number, but $-$ if it be odd, we have only to suppose $nz = m\pi$ , in order to obtain $y^n = \pm 1$ .

That we may distinguish the case in which $m$ is even,

from that in which it is odd, we shall write for the first $2m$ , and for the second $2m+1$ ; we therefore make

nz = 2m\pi, \text{ and } nz = (2m+1)\pi.

By the first hypothesis, we find

y^n = +1, y = \operatorname{cos}. \frac{2m\pi}{n} + \sqrt{-1} \sin. \frac{2m\pi}{n},

and by the second

y^n = -1,

y = \operatorname{cos}. \frac{(2m+1)\pi}{n} + \sqrt{-1} \sin. \frac{(2m+1)\pi}{n}.

122. By means of the indeterminate number $m$ , each of these expressions for $y$ furnishes all the values of which this quantity is susceptible, for we may take successively

m=0, m=1, m=2, m=3, \&c.

The first formula gives

y = \operatorname{cos}. 0 \cdot \pi = 1

y = \operatorname{cos}. \frac{2\pi}{n} + \sqrt{-1} \sin. \frac{2\pi}{n},

y = \operatorname{cos}. \frac{4\pi}{n} + \sqrt{-1} \sin. \frac{4\pi}{n},

&c.

It is evident that we shall always have different results as far as $m=n-1$ . If, however, we suppose $m=n$ ; then we have $y = \operatorname{cos}. 2\pi = 1$ , which is the same as the first of the values already obtained, and if we suppose $m=n+1$ , then (ALGEBRA, § 25.)

\operatorname{Cos}. \frac{(2n+2)\pi}{n} = \operatorname{cos}. (2\pi + \frac{2\pi}{n}) = \operatorname{cos}. \frac{2\pi}{n},

\operatorname{Sin}. \frac{(2n+2)\pi}{n} = \sin. (2\pi + \frac{2\pi}{n}) = \sin. \frac{2\pi}{n},

which is the same as the second value, and so on with respect to the others.

By this mode of proceeding we shall not only obtain the $n$ roots of the equation $y^n = 1$ , or $y^n - 1 = 0$ , but, with a little attention, we shall discover that these roots may be arranged in pairs, by bringing together those that only differ in the sign of the radical $\sqrt{-1}$ ; for since

$\operatorname{Cos}. (2\pi - \rho) = \operatorname{cos}. \rho$ , and $\sin. (2\pi - \rho) = -\sin. \rho$ , it follows that

y = \operatorname{cos}. \frac{(2n-2m)\pi}{n} + \sqrt{-1} \sin. \frac{(2n-2m)\pi}{n}

= \operatorname{cos}. \frac{2m\pi}{n} - \sqrt{-1} \sin. \frac{2m\pi}{n}.

Hence it appears that we may comprehend all the roots of the equation $y^n = 1$ in the single expressions

$y =$

y = \text{cos.} \frac{2m\pi}{n} \pm \sqrt{-1} \text{ sin.} \frac{2m\pi}{n},

by giving to

m

only these values

0, 1, 2, \dots, \frac{n}{2},

n

is even, and these values

0, 1, 2, \dots, \frac{n-1}{2},

n

is odd; and it may be observed that in the former case the last value of

y

y = \text{cos.} \pi = -1,

because that then the equation

y^2 - 1 = 0

has two real roots.

The two values comprehended in the formula,

y = \text{cos.} \frac{2m\pi}{n} + \sqrt{-1} \text{ sin.} \frac{2m\pi}{n},

give for factors of the first degree of the quantity

y^2 - 1

, the two imaginary expressions

y = \left( \text{cos.} \frac{2m\pi}{n} + \sqrt{-1} \text{ sin.} \frac{2m\pi}{n} \right),

y = \left( \text{cos.} \frac{2m\pi}{n} - \sqrt{-1} \text{ sin.} \frac{2m\pi}{n} \right),

and the product of these is the expression

y^2 - 2y \text{ cos.} \frac{2m\pi}{n} + 1,

which comprehends all the real factors of the second degree.

As an example of the formula

y = \text{cos.} \frac{2m\pi}{n} \pm \sqrt{-1} \text{ sin.} \frac{2m\pi}{n},

the simple factors, or those of the first degree, contained in the function

y^6 - 1

will be

y - 1,

y = \left( \text{cos.} \frac{2\pi}{6} \pm \sqrt{-1} \text{ sin.} \frac{2\pi}{6} \right),

y = \left( \text{cos.} \frac{4\pi}{6} \pm \sqrt{-1} \text{ sin.} \frac{4\pi}{6} \right),

y + 1.

y^2 - 2y \text{ cos.} \frac{2m\pi}{n} + 1

gives as factors of the second degree

y^2 - 2y + 1,

y^2 - 2y \text{ cos.} \frac{2\pi}{6} + 1,

y^2 - 2y \text{ cos.} \frac{4\pi}{6} + 1,

y^2 + 2y + 1.

The first and the last of the factors of the second degree are the squares of $y - 1$ , and $y + 1$ , factors of the first degree, each of which only enters once into the proposed function; it will therefore be necessary, when we employ the factors of the second degree, to reject the first and last, and take instead of them

(y - 1)(y + 1) = y^2 - 1.

The factors of the first degree of the function

y^5 - 1

are

y - 1

y = \left( \text{cos.} \frac{2\pi}{5} \pm \sqrt{-1} \text{ sin.} \frac{2\pi}{5} \right)

y = \left( \text{cos.} \frac{4\pi}{5} \pm \sqrt{-1} \text{ sin.} \frac{4\pi}{5} \right)

Those of the second degree are

y^2 - 2y + 1,

y^2 - 2y \text{ cos.} \frac{2\pi}{5} + 1,

y^2 - 2y \text{ cos.} \frac{4\pi}{5} + 1;

but it is to be observed that the first factor of the second degree is the square of

y - 1

, which enters only once into the proposed function.

123. When the function to be decomposed into factors is

y^n + 1

, the formula

y = \text{cos.} \frac{(2m+1)\pi}{n} + \sqrt{-1} \text{ sin.} \frac{(2m+1)\pi}{n},

which corresponds to that case (§ 121.) is also susceptible of the double sign

\pm

, provided we stop at the value of

m

which gives

2m + 1 = n, \text{ or } 2m + 1 = n - 1,

according as

n

is odd or even; hence it follows that

m = \frac{n-1}{2}, \quad m = \frac{n-2}{2};

the factors of the first degree are

y = \left( \text{cos.} \frac{(2m+1)\pi}{n} \pm \sqrt{-1} \text{ sin.} \frac{(2m+1)\pi}{n} \right),

and those of the second

y^2 - 2y \text{ cos.} \frac{(2m+1)\pi}{n} + 1.

When among these last there is found some which are squares, we must take only one of their simple factors, in the same way as in the two preceding examples.

When the function is

y^n - 1

, the factors of the first degree are

y = \left( \text{cos.} \frac{\pi}{5} \pm \sqrt{-1} \text{ sin.} \frac{\pi}{5} \right),

y = \left( \text{cos.} \frac{3\pi}{5} \pm \sqrt{-1} \text{ sin.} \frac{3\pi}{5} \right)

y + 1;

and those of the second,

y^2 - 2y \cos. \frac{\pi}{5} + 1,

y^2 - 2y \cos. \frac{3\pi}{5} + 1,

y^2 + 2y + 1.

The function

y^2 + 1

has for factors of the first degree

y - (\cos. \frac{\pi}{6} \pm \sqrt{-1} \sin. \frac{\pi}{6}),

y - (\cos. \frac{3\pi}{6} \pm \sqrt{-1} \sin. \frac{3\pi}{6}),

y - (\cos. \frac{5\pi}{6} \pm \sqrt{-1} \sin. \frac{5\pi}{6}),

and those of the second,

y^2 - 2y \cos. \frac{\pi}{6} + 1,

y^2 - 2y \cos. \frac{3\pi}{6} + 1, \text{ or } y^2 + 1,

y^2 - 2y \cos. \frac{5\pi}{6} + 1.

124. Such functions as are of this form $x^2 + 2px + q$ may be treated in the same manner as those which consist of only two terms. By putting the function $= 0$ , and resolving the equation which is thus produced, in the same manner as if it were of the second degree, we find the factors to be

x + (p \pm \sqrt{p^2 - q});

p^2

exceed

q

, the second term of these factors is real, and by making

\pm a^n = p \pm \sqrt{p^2 - q},

we have the functions of the form

x \pm a^n

to decompose into factors.

When

p^2 < q

, then we put

p = a^n

q = b^n

x = by

, and the function becomes

b^n y^2 + 2a^n b^n y + b^{2n}

= b^n (y^2 + \frac{2a^n}{b^n} y + 1);

but the condition

p^2 < q

, or

a^{2n} < b^{2n}

, makes

a^n < b^n

, and

\frac{a^n}{b^n} < 1

, therefore

\frac{a^n}{b^n}

may be represented by the cosine of a given arch

\delta

, and the proposed function will be reduced to

b^n (y^2 + 2y \cos. \delta + 1),

we have then only to resolve the equation

y^2 + 2y \cos. \delta + 1 = 0,

and we immediately find

y^2 = -\cos. \delta \pm \sqrt{-1} \sin. \delta;

we now assume, as in § 121,

y = \cos. z \pm \sqrt{-1} \sin. z;

then we find (§ 120),

y^2 = \cos. nz \pm \sqrt{-1} \sin. nz,

which expression for

y^2

, being compared with its other value, gives

\cos. nz = \cos. \delta, \sin. nz = \sin. \delta.

These relations will be satisfied if we suppose

nz = 2m\pi + \delta

m

being any whole number whatever, for

\cos. (2m\pi + \delta) = \cos. \delta, \sin. (2m\pi + \delta) = \sin. \delta;

we have therefore

z = \frac{2m\pi + \delta}{n},

y = \cos. \frac{2m\pi + \delta}{n} \pm \sqrt{-1} \sin. \frac{2m\pi + \delta}{n}.

The factors of the first degree of the function

y^2 + 2y \cos. \delta + 1

will consequently be comprehended in this formula,

y - \left\{ \cos. \frac{2m\pi + \delta}{n} \pm \sqrt{-1} \sin. \frac{2m\pi + \delta}{n} \right\}.

p

the coefficient of the second term of the proposed function be negative, the only change necessary is to make

p = -a^n

, and to take the arch

\delta

greater than a quadrant.

Of the Fluents of Irrational Functions.

125. When a fluxionary expression involves irrational functions, we must endeavour either to transform it into another that is rational, or to reduce it to a series of irrational terms of this form $Ax^{\frac{1}{n}}$ , and then, in either case its fluent may be found by the rules already delivered.

Let us take for example the fluxion

\frac{(1 + \sqrt{x-1} \sqrt{x^2}) \dot{x}}{1 + \sqrt{x}}

It is evident that by putting

x = z^6

, all the extractions indicated by the radical signs may be effected, and the fluxion may be transformed to

\frac{6z^5(1 + z^3 - z^4) \dot{z}}{1 + z^3}

, which, by dividing the numerator by

1 + z^3

, may be otherwise expressed thus,

-6(z^7 \dot{z} - z^6 \dot{z} - z^5 \dot{z} + z^4 \dot{z} - z^3 \dot{z} + \dot{z} - \frac{\dot{z}}{1+z^3}).

The fluent of which is

-6 \left\{ \frac{z^8}{8} - \frac{z^7}{7} - \frac{z^6}{6} + \frac{z^5}{5} - \frac{z^4}{4} + \frac{z^3}{3} \right\} + \text{const.}

Inverse Method. 126. We shall first consider such fluxions as contain the irrational function $\sqrt{A+Bx+Cx^2}$ , and which have necessarily one or other of these forms.

X \dot{\times} \sqrt{A+Bx+Cx^2}, \frac{X \dot{\times}}{\sqrt{A+Bx+Cx^2}},

$X$ being put for any rational function of $x$ ; and it may be remarked, that the latter form comprehends the former, which may be written thus:

\frac{X \dot{\times} \sqrt{A+Bx+Cx^2} \times \sqrt{A+Bx+Cx^2}}{\sqrt{A+Bx+Cx^2}} \\ = \frac{X(A+Bx+Cx^2) \dot{\times}}{\sqrt{A+Bx+Cx^2}};

and here the numerator of the fluxion is a rational function of $x$ .

Before we transform the expression $\sqrt{A+Bx+Cx^2}$ into another that is rational with respect to the variable quantity it contains, we shall put the quantity $A+Bx+Cx^2$ under this form,

C \left( \frac{A}{C} + \frac{B}{C} x + x^2 \right);

and, in order to abridge, we shall put

C=c^2, \frac{A}{C}=a, \frac{B}{C}=b,

\sqrt{A+Bx+Cx^2} = c \sqrt{a+bx+x^2}.

Let us now assume $\sqrt{a+bx+x^2} = x+\dot{x}$ , then, squaring both sides of the equation, we find $a+bx = 2x\dot{x} + \dot{x}^2$ , hence we get $x = \frac{a-\dot{x}^2}{2\dot{x}-b}$ , and consequently

\sqrt{A+Bx+Cx^2} = c(x+\dot{x}) = c \left( \frac{a-b\dot{x}+\dot{x}^2}{2\dot{x}-b} \right)

\dot{x} = - \frac{2(a-b\dot{x}+\dot{x}^2)\dot{x}}{(2\dot{x}-b)^2}.

By means of these values the fluxion $\frac{X \dot{\times}}{\sqrt{A+Bx+Cx^2}}$ ,

is transformed into another fluxion $Z \dot{\times}$ , where $Z$ denotes a rational function of $x$ , which is real when $C$ or $c^2$ is positive; but as when $C$ is negative $c$ becomes imaginary, the fluxion $Z \dot{\times}$ which involves $c$ becomes also imaginary.

In this case we have to consider $\sqrt{A+Bx-Cx^2}$ , and making

C=c^2, \frac{A}{C}=a, \frac{B}{C}=b,

it becomes $c \sqrt{a+bx-x^2}$ . The quantity $x^2-bx+a$ may always be decomposed into real factors of the first degree; let us represent these factors by $x-a$ , and $x-a'$ , then it is evident that

a+bx-x^2 = -(x^2-bx+a) \\ = (x-a)(x-a').

Let us now assume

\sqrt{(x-a)(x-a')} = (x-a)z,

then squaring both sides of the equation it becomes divisible by $x-a$ , and we have $x-a = (x-a)z^2$ , from which we find

x = \frac{ax^2+a'}{x^2+1}, (x-a)z = \frac{(x-a)z}{x^2+1},

\dot{x} = \frac{2(a-a')z\dot{x}}{(x^2+1)^2}

values which render the proposed fluxion $\frac{X \dot{\times}}{\sqrt{A+Bx+Cx^2}}$ rational.

127. Let us now take for example the fluxion $\frac{\dot{x}}{\sqrt{A+Bx+Cx^2}}$ ; by applying to it the first of the

preceding transformations it becomes $\frac{-2\dot{x}}{c(2\dot{x}-b)}$ , the fluent of which is $-\frac{1}{c} \log(2\dot{x}-b) + \text{const.}$ Substituting now for $z$ its value $-x + \sqrt{a+bx+x^2}$ , and for $a, b$ , and $c$ , the quantities they severally represent, the fluent becomes

\frac{-1}{\sqrt{C}} \log \left\{ \frac{1}{\sqrt{C}} \left( \frac{B}{\sqrt{C}} - 2x\sqrt{C} \right) \right\} + \text{const.}

a result to which we may also give this form,

\frac{-1}{\sqrt{C}} \log \left\{ \frac{-B}{2\sqrt{C}} - x\sqrt{C} \right\} \\ + \sqrt{A+Bx+Cx^2} \Big\} \\ + 1. \frac{2}{\sqrt{C}} + \text{const.}

By uniting the constant quantities into one, and observing that the radical quantity $\sqrt{C}$ may have the sign $\pm$ prefixed to it, we have at last

\int \frac{\dot{x}}{\sqrt{A+Bx+Cx^2}} \text{ equal to} \\ \frac{1}{\sqrt{C}} \log \left\{ \frac{B}{2\sqrt{C}} + x\sqrt{C} + \sqrt{A+Bx+Cx^2} \right\} \\ + \text{const.}

128. Let us take for the second example $\frac{\dot{x}}{\sqrt{A+Bx-Cx^2}}$ . By employing the latter transformation of § 126, we have $\frac{-2\dot{x}}{c(x^2+1)}$ , of which the

fluent is

-\frac{2}{c} \text{ arc } (\tan. = x) + \text{const.}

Substituting now instead of $x$ its value

$\frac{\sqrt{(x-a)(x-a')}}{\sqrt{(x-a)}}$ deduced from the equation $x-a = (x-a)z^2$ ,

Invert Method. and putting $\sqrt{C}$ for $c$ , we get $\int \frac{x}{\sqrt{(A+Bx-Cx^2)}}$ equal to

-\frac{2}{\sqrt{C}} \operatorname{arc} \left( \tan. \frac{\sqrt{(x'-x)}}{\sqrt{(x-x')}} \right) + \text{const.}

$x$ and $x'$ being the roots of the equation

x^2 - \frac{B}{C}x - \frac{A}{C} = 0.

Let us suppose that $A=C=1$ and $B=0$ , then the proposed fluxion becomes in this particular case $\frac{x}{\sqrt{(1-x^2)}}$ and the preceding formula gives for its fluent $-2 \operatorname{arc} \left( \tan. \frac{\sqrt{(1-x)}}{\sqrt{(1+x)}} \right) + \text{const.}$ for $x$ and $x'$ being the roots of the equation $x^2 - 1 = 0$ , we must take $x=-1$ , and $x'=1$ .

We may, however, give this fluent another form by proceeding thus: Let $v$ be the arch whose $\tan.$ $= \frac{\sqrt{(1-x)}}{\sqrt{(1+x)}}$ , then $\tan^2 v = \frac{1-x}{1+x}$ , and $x = \frac{1-\tan^2 v}{1+\tan^2 v} = \frac{2}{1+\tan^2 v} - 1 = \frac{2}{\sec^2 v} - 1 = 2 \cos^2 v - 1$ , but $2 \cos^2 v - 1 = \cos 2v$ (ALGEBRA, § 358.), therefore, $x = \cos 2v$ . Put $s$ for the arch whose sine is $x$ , and $\pi$ for half the circumference, then $x = \cos (\frac{1}{2}\pi - s)$ , therefore $2v = \frac{1}{2}\pi - s$ , and since it has been shewn that $\int \frac{x}{\sqrt{(1-x^2)}} = -2v + \text{const.}$ therefore $\int \frac{x}{\sqrt{(1-x^2)}} = s - \frac{1}{2}\pi + \text{const.}$ or, by including the arch $\frac{1}{2}\pi$ in the constant quantity, $\int \frac{x}{\sqrt{(1-x^2)}} = s + \text{const.}$ This conclusion agrees with what has been shewn in § 59.

Instead of finding the fluent of

\frac{x}{\sqrt{(A+Bx-Cx^2)}} = \frac{x}{c\sqrt{(a+bx-x^2)}}

by first transforming it to a rational expression, we may reduce it directly to an arch of a circle by proceeding as follows. Put $x = \frac{b}{2} = z$ , then $x = z$ , and the fluxion is transformed to $\frac{x}{c\sqrt{(a+\frac{1}{2}b^2-x^2)}}$ ; again, put $a+\frac{1}{2}b^2 = g^2$ , and $z = gu$ , then $x = gu$ , and this last fluxion is transformed to $\frac{u}{c\sqrt{(1-u^2)}}$ , the fluent of which is $\frac{1}{c} \operatorname{arc} (\sin. u) + \text{const.}$

Of the Fluents of Binomial Fluxions.

129. Let us now consider such fluxions as have this form

x^{m-1} \cdot x(a+bx^2)^q,

and which are sometimes called binomial fluxions. We may here suppose $m$ and $n$ to be whole numbers,

without affecting the generality of the expression.

For if we had $x^{\frac{p}{q}} \cdot x(a+bx^2)^q$ , we may assume $x=x^{\frac{p}{q}}$ , then $x = x^{\frac{p}{q}}$ , and the fluxion becomes $6x^{\frac{p}{q}} \cdot x^{\frac{p}{q}}(a+bx^2)^q$ .

We may also suppose $n$ to be positive, for if it were negative, so that the fluxion were $x^{m-1} \cdot x^{\frac{p}{q}}(a+bx^2)^q$ we have only to assume $x = \frac{1}{x}$ , and the

fluxion is transformed to $-x^{-m-1} \cdot x^{\frac{p}{q}}(a+x^{-2})^q$ .

Let us inquire in what case the fluxion $x^{m-1} \cdot x^{\frac{p}{q}}(a+bx^2)^q$ may become rational. Assume $a+bx^2 = x^r$ , then $(a+bx^2)^q = x^{rq}$ , $x^r = \frac{x^r-a}{b}$ , $x^m = \left(\frac{x^r-a}{b}\right)^{\frac{m}{r}}$ ,

and $x^{m-1} \cdot x^{\frac{p}{q}} = \frac{q}{nb} x^{r-1} \left(\frac{x^r-a}{b}\right)^{\frac{m}{r}-1} \cdot x^{\frac{p}{q}}$ ; hence the proposed fluxion is transformed to

\frac{q}{nb} x^{p+q-1} \cdot \left(\frac{x^r-a}{b}\right)^{\frac{m}{r}-1};

this expression is evidently rational as often as $\frac{m}{r}$ is a whole number.

There are yet other cases in which the fluxion may become rational, and which may be determined by assuming $a+bx^2 = x^r u^s$ , thus we have

x^m = \frac{a}{u^s-b}, x^r = \frac{a}{(u^s-b)^{\frac{r}{s}}}, \dots

$x^{m-1} \cdot x^{\frac{p}{q}} = \frac{-q a^{\frac{m}{r}} u^{\frac{r}{s}-1} \cdot u^{\frac{p}{q}}}{n(u^s-b)^{\frac{m}{r}+\frac{1}{q}}}$ , and because that $(a+bx^2)^q$

$= \frac{a^q u^{sq}}{(u^s-b)^q}$ the fluxion $x^{m-1} \cdot x^{\frac{p}{q}}(a+bx^2)^q$ is transformed to

formed to

\frac{-q a^{\frac{m}{r}+\frac{p}{q}} u^{\frac{r}{s}+\frac{p}{q}-1} \cdot u^{\frac{r}{s}+\frac{p}{q}}}{n(u^s-b)^{\frac{m}{r}+\frac{1}{q}}},

an expression which is rational if $\frac{m}{r} + \frac{p}{q}$ is a whole number.

130. As it is not possible, in every case, to express in finite terms the formula $\int x^{m-1} \cdot x^{\frac{p}{q}}(a+bx^2)^q$ , we may try to reduce it to its most simple case, as we have done

Inverse Method. done with respect to $\int \frac{z}{(x^2 + \beta^2)^n}$ in § 114, which we

have succeeded in reducing to $\int \frac{z}{x^2 + \beta^2}$ . To effect this reduction, we remark, that, since when $u$ and $v$ denote any functions of a variable quantity, the fluxion of $uv$ is $u\dot{v} + v\dot{u}$ , (§ 37.) therefore $\int u \dot{v} = uv - \int v \dot{u}$ . Now

if we can decompose the expression $x^{m-1} \dot{x} (a + bx^n)^p$ into two factors, such that we can find the fluent of one of them, then denoting that factor by $\dot{v}$ , and the other by $u$ , the fluent of the proposed fluxion will be made to depend on that of $v\dot{u}$ , which in some cases will be more simple than the proposed fluxion. That we may abridge a little the results, we shall write $p$ instead of $\frac{p}{q}$ , so that $p$ will represent any fraction; the proposed fluxion thus simplified in its form is

x^{m-1} \dot{x} (a + bx^n)^p.

Among the different ways of resolving this fluxion into two factors, we shall choose that which diminishes the exponent of $x$ without the parentheses, we therefore write the fluxion thus

x^{m-n} \times x^{n-1} \dot{x} (a + bx^n)^p,

now the fluent of the factor $x^{m-n} \dot{x} (a + bx^n)^p$ may always be determined, whatever be the value of $p$ , by § 108; let us denote this factor by $\dot{v}$ , then

v = \frac{(a + bx^n)^{p+1}}{(p+1)nb}, \text{ and } u = x^{m-n},

thus the formula $\int u \dot{v} = uv - \int v \dot{u}$ give us $\int x^{m-n} \dot{x} (a + bx^n)^p$ equal to

\frac{x^{m-n} (a + bx^n)^{p+1}}{(p+1)nb} - \frac{m-n}{(p+1)nb} \int x^{m-n-1} \dot{x} (a + bx^n)^{p+1}.

\text{But } \int x^{m-n-1} \dot{x} (a + bx^n)^{p+1} =

\int x^{m-n-1} \dot{x} (a + bx^n)^p (a + bx^n) =

a \int x^{m-n-1} \dot{x} (a + bx^n)^p + b \int x^{m-n-1} \dot{x} (a + bx^n)^{p+1};

Substituting now this last value in the preceding equation, and collecting into one the terms involving the fluent $\int x^{m-n-1} \dot{x} (a + bx^n)^p$ , we find

\left(1 + \frac{m-n}{(p+1)n}\right) \int x^{m-n-1} \dot{x} (a + bx^n)^p =

\frac{1}{(p+1)nb} \left\{ \begin{array}{l} x^{m-n} (a + bx^n)^{p+1} \\ - a(m-n) \int x^{m-n-1} \dot{x} (a + bx^n)^p \end{array} \right\}

hence at last we get

\int x^{m-n-1} \dot{x} (a + bx^n)^p = \frac{1}{b(pn+m)} \left\{ \begin{array}{l} x^{m-n} (a + bx^n)^{p+1} \\ - a(m-n) \int x^{m-n-1} \dot{x} (a + bx^n)^p \end{array} \right\} \quad (A)

It is easy to see that, as we have, by this formula, reduced the determination of the fluent of $x^{m-1} \dot{x} (a + bx^n)^p$ to that of $x^{m-n-1} \dot{x} (a + bx^n)^p$ , we may reduce this last to that of $x^{m-2n-1} \dot{x} (a + bx^n)^p$ by writing $m-n$ in place of $m$ in equation (A), then by changing $m$ into $m-2n$ we may reduce the fluent of $x^{m-2n-1} \dot{x} (a + bx^n)^p$ to that of $x^{m-3n-1} \dot{x} (a + bx^n)^p$ , and so on.

In general, if $r$ denote the number of reductions, we shall at last come to

$\int x^{m-rn-1} \dot{x} (a + bx^n)^p$ , and the last formula will be

\int x^{m-(r-1)n-1} \dot{x} (a + bx^n)^p = \frac{x^{m-rn} (a + bx^n)^{p+1}}{b(pn+m-(r-1)n)} - \frac{a(m-rn) \int x^{m-rn-1} \dot{x} (a + bx^n)^p}{b(pn+m-(r-1)n)}.

It appears by the last formula, that if $m$ is a multiple of $n$ , then $\int x^{m-1} \dot{x} (a + bx^n)^p$ will be an algebraic quantity, for in that case the coefficient $m-rn$ will be $=0$ , and therefore the term containing $\int x^{m-rn-1} \dot{x} (a + bx^n)^p$ will vanish. This result coincides with what we have already found, § 129.

131. We may also obtain a reduction, by which the exponent $p$ will be diminished by unity. For this purpose it is sufficient to observe that $\int x^{m-1} \dot{x} (a + bx^n)^p$ is equal to

\int x^{m-1} \dot{x} (a + bx^n)^{p-1} (a + bx^n) = a \int x^{m-1} \dot{x} (a + bx^n)^{p-1} + b \int x^{m+n-1} \dot{x} (a + bx^n)^{p-1},

and that the formula (A) by changing $m$ into $m+n$ , and $p$ into $p-1$ gives

\frac{\int x^{m+n-1} \dot{x}(a+bx^n)^{p-1}}{x^m(a+bx^n)^p = am \int x^{n-1} \dot{x}(a+bx^n)^{p-1}} = \frac{b(pn+m)}{pn+m}.

Substitute now this value in the preceding equation, we have

(B)

\int x^{m-1} \dot{x}(a+bx^n)^p = \frac{x^m(a+bx^n)^p + pna \int x^{m-1} \dot{x}(a+bx^n)^{p-1}}{pn+m}.

By means of this general formula we may take away successively from $p$ as many units as it contains, and by the application of this formula, and formula (A), we may cause the fluent $\int x^{m-1} \dot{x}(a+bx^n)^p$ to depend on

\int x^{m-rn-1} \dot{x}(a+bx^n)^{p-s}, \quad rn \text{ being the greatest multiple of } r \text{ contained in } m-1, \text{ and } s \text{ the greatest whole number contained in } p.

The fluent $\int x^2 \dot{x}(a+bx^3)^{\frac{5}{3}}$ , for example, may, by the application of formula (A), be reduced successively to

\int x^4 \dot{x}(a+bx^3)^{\frac{2}{3}}, \text{ and } \int x \dot{x}(a+bx^3)^{\frac{1}{3}},

and by formula (B) $\int x \dot{x}(a+bx^3)^{\frac{1}{3}}$ is reduced successively to

\int x \dot{x}(a+bx^3)^{\frac{1}{3}}, \text{ and } \int x \dot{x}(a+bx^3)^{-\frac{2}{3}}.

132. It is evident, that if $m$ and $n$ were negative, the formulas (A) and (B) would not answer the purpose for which they have been investigated, because, in that case, they would increase the exponents instead of diminishing them. If, however, we reverse them, we shall find that they then apply to the case under consideration.

From formula (A) we get

\int x^{m-n-1} \dot{x}(a+bx^n)^p = \frac{x^{m-n}(a+bx^n)^{p+1}}{a(m-n)} - \frac{b(m+n)p \int x^{m-1} \dot{x}(a+bx^n)^p}{a(m-n)};

Substitute now $m+n$ in place of $m$ , and it becomes

(C)

\int x^{m-1} \dot{x}(a+bx^n)^p =

\frac{x^m(a+bx^n)^{p+1}}{am} - \frac{b(m+n+np) \int x^{m+n-1} \dot{x}(a+bx^n)^p}{am}.

This formula diminishes the exponents without the parentheses, because $m+n-1$ becomes $-m+n-1$ , when $-m$ is substituted instead of $m$ .

To reverse formula (B) we first take

\int x^{m-1} \dot{x}(a+bx^n)^{p-1} = \frac{-x^m(a+bx^n)^p}{pna} + \frac{(m+np) \int x^{m-1} \dot{x}(a+bx^n)^p}{pna};

Then, writing $p+1$ instead of $p$ , we find

(D)

\int x^{m-1} \dot{x}(a+bx^n)^p = \frac{-x^m(a+bx^n)^{p+1}}{(p+1)na} + \frac{(m+n+np) \int x^{m-1} \dot{x}(a+bx^n)^{p+1}}{(p+1)na}.

This formula answers the purpose we have in view, because $p+1$ becomes $-p+1$ when $p$ is negative.

These formulas (A), (B), (C), (D), are inapplicable when their denominators vanish. This is the case with formula (A); for example, when $m=-np$ ; but, in every such case, the proposed fluxion may have its fluent determined either algebraically or by logarithms.

133. Let the fluent be $\int \frac{x^{m-1} \dot{x}}{\sqrt{1-x^2}}$ , $m$ being a whole

positive number. Formula (A) immediately applies to this case, so that by putting $a=1, b=-1, n=2, p=-\frac{1}{2}$ , we have

\int \frac{x^{m-1} \dot{x}}{\sqrt{1-x^2}} = \left\{ \begin{array}{l} -\frac{x^{m-2} \sqrt{1-x^2}}{m-1} \\ + \frac{m-2}{m-1} \int \frac{x^{m-3} \dot{x}}{\sqrt{1-x^2}} \end{array} \right.

or, substituting $m$ in place of $m-1$ ,

\int \frac{x^m \dot{x}}{\sqrt{1-x^2}} = \left\{ \begin{array}{l} -\frac{x^{m-1} \sqrt{1-x^2}}{m} \\ + \frac{m-1}{m} \int \frac{x^{m-2} \dot{x}}{\sqrt{1-x^2}} \end{array} \right.

Let

Let us suppose, for example, that

m=1

, then

\int \frac{x \dot{x}}{\sqrt{(1-x^2)}} = -\sqrt{(1-x^2)} + \text{const.}

Let us now suppose that

m=3

, then

\int \frac{x^3 \dot{x}}{\sqrt{(1-x^2)}} = \left\{ -\frac{1}{2} x^2 \sqrt{(1-x^2)} + \frac{1}{2} \int \frac{x \dot{x}}{\sqrt{(1-x^2)}} \right\}

or, substituting for

\int \frac{x \dot{x}}{\sqrt{(1-x^2)}}

its value,

\int \frac{x^3 \dot{x}}{\sqrt{(1-x^2)}} = -\left( \frac{1}{3} x^2 + \frac{2}{3} \right) \sqrt{(1-x^2)} + \text{const.}

If we suppose

m=2

, then

\int \frac{x^2 \dot{x}}{\sqrt{(1-x^2)}} = \left\{ -\frac{1}{2} x \sqrt{(1-x^2)} + \frac{1}{2} \int \frac{\dot{x}}{\sqrt{(1-x^2)}} \right\}

But we have already found, § 128, that

\int \frac{\dot{x}}{\sqrt{(1-x^2)}} = \text{arc}(\text{fin.} = x),

therefore, putting

A

for

\text{arc}(\text{fin.} = x)

\int \frac{x^2 \dot{x}}{\sqrt{(1-x^2)}} = -\frac{1}{2} x \sqrt{(1-x^2)} + \frac{1}{2} A + \text{const.}

In the very same way we find that

\int \frac{x^4 \dot{x}}{\sqrt{(1-x^2)}} \text{ is equal to } -\left( \frac{1}{4} x^2 + \frac{3}{8} x \right) \sqrt{(1-x^2)} + \frac{3}{8} A + \text{const.}

134. In the case of

m

, a negative number, we must have recourse to formula (C), from which we find

\int \frac{x^{-m-1} \dot{x}}{\sqrt{(1-x^2)}} = \left\{ -\frac{x^{-m} \sqrt{(1-x^2)}}{m} + \frac{m-1}{m} \int \frac{x^{-m+1} \dot{x}}{\sqrt{(1-x^2)}} \right\}

which formula, by writing

-m

instead of

-m-1

, becomes

\int \frac{\dot{x}}{x^m \sqrt{(1-x^2)}} = \left\{ -\frac{\sqrt{(1-x^2)}}{(m-1)x^{m-1}} + \frac{m-2}{m-1} \int \frac{\dot{x}}{x^{m-2} \sqrt{(1-x^2)}} \right\}

We cannot here suppose

m=1

, for that value would render the denominator

=0

; therefore, before we can apply this formula, it is necessary to investigate the

fluent of

\frac{\dot{x}}{x \sqrt{(1-x^2)}}

. We may easily find it from § 126,

or otherwise thus, put

1-x^2=z^2

, then

x = \sqrt{(1-z^2)}, \quad \dot{x} = \frac{-z \dot{z}}{\sqrt{(1-z^2)}}.

Therefore

\frac{\dot{x}}{x \sqrt{(1-x^2)}} = \frac{-z \dot{z}}{1-z^2} = \frac{-\frac{1}{2} \dot{z}}{1+\frac{1}{2}z} = \frac{-\frac{1}{2} \dot{z}}{1-z}.

The fluent of the right hand side of this equation is evidently (§ 103.)

-\frac{1}{2} \ln(1+z) + \frac{1}{2} \ln(1-z) = -\frac{1}{2} \ln \left( \frac{1+z}{1-z} \right);

or, since

\frac{1+z}{1-z} = \frac{(1+z)^2}{1-z^2}

, the same fluent may be expressed thus,

-\frac{1}{2} \ln \left( \frac{(1+z)^2}{1-z^2} \right) = -\ln \frac{1+z}{\sqrt{(1-z^2)}};

therefore, by substituting

\sqrt{(1-x^2)}

for

z

, and

x

for

\sqrt{(1-z^2)}

, we have

\int \frac{\dot{x}}{x \sqrt{(1-x^2)}} = -\ln \left( \frac{1+\sqrt{(1-x^2)}}{x} \right) + \text{const.}

If we suppose

m=2

, the formula becomes

\int \frac{\dot{x}}{x^2 \sqrt{(1-x^2)}} = -\frac{\sqrt{(1-x^2)}}{x} + \text{const.}

If we suppose

m=3

, then

\int \frac{\dot{x}}{x^3 \sqrt{(1-x^2)}} = \left\{ -\frac{\sqrt{(1-x^2)}}{2x^2} + \frac{1}{2} \int \frac{\dot{x}}{x \sqrt{(1-x^2)}} \right\},

which expression, by substituting for

\int \frac{\dot{x}}{x \sqrt{(1-x^2)}}

its value, becomes

\int \frac{\dot{x}}{x^3 \sqrt{(1-x^2)}} = \left\{ -\frac{\sqrt{(1-x^2)}}{2x^2} - \frac{1}{2} \ln \left( \frac{1+\sqrt{(1-x^2)}}{x} \right) + \text{const.} \right\}

Of Finding Fluents by Series.

135. We can always easily find an expression for the fluent

\int X \dot{x}

, where

X

denotes any function of

x

; when that function is expanded into a series, each term of which is some power of

x

multiplied by a constant quantity; thus suppose

X = A x^m + B x^{m+1} + C x^{m+2} + \dots

then

X \dot{x}

is equal to

A x^m

Ax^m \dot{x} + Bx^{m+n} \dot{x} + Cx^{m+2n} \dot{x} + \&c.

and taking the fluent of each term by § 101,

\int X \dot{x} = \frac{Ax^{m+1}}{m+1} + \frac{Bx^{m+n+1}}{m+n+1} + \frac{Cx^{m+2n+1}}{m+2n+1} + \&c. + \text{const.}

If in the development of $x$ there be any term of this form $\frac{A}{x}$ , the fluent corresponding to that term will be $A \log x$ (§ 103).

136. The most simple function of $x$ that can be expanded into a series is $\frac{1}{a+x}$ , which becomes

\frac{1}{a} - \frac{x}{a^2} + \frac{x^2}{a^3} - \frac{x^3}{a^4} + \&c.

Hence we find

\frac{\dot{x}}{a+x} = \frac{\dot{x}}{a} - \frac{x \dot{x}}{a^2} + \frac{x^2 \dot{x}}{a^3} - \frac{x^3 \dot{x}}{a^4} + \&c.

and taking the fluents

\int \frac{\dot{x}}{a+x} = \frac{x}{a} - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} - \frac{x^4}{4a^4} + \&c. + \text{const.}

Now we know that $\int \frac{\dot{x}}{a+x} = \log(a+x)$ (§ 57.) therefore

\log(a+x) = \frac{x}{a} - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} - \frac{x^4}{4a^4} + \&c. + \text{const.}

To find the value of the constant quantity we have only to make $x=0$ , for then the equation becomes $\log a = \text{const.}$ therefore

\log(a+x) = \log a + \frac{x}{a} - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} - \&c.

hence, if we subtract $\log a$ from each side, and observe that $\log(x+a) - \log a = \log\left(\frac{x+a}{a}\right) = \log\left(1 + \frac{x}{a}\right)$ we get

\log\left(1 + \frac{x}{a}\right) = \frac{x}{a} - \frac{x^2}{2a^2} + \frac{x^3}{3a^3} - \frac{x^4}{4a^4} + \&c.

From this conclusion we may deduce rules for computing the logarithms of numbers.

137. Let the fluxion be $\frac{ax^m \dot{x}}{a^2+x^2}$ , which may be put under this form $\frac{\dot{x}}{1+\frac{x^2}{a^2}}$ , and which consequently belongs

to the arch of which the tangent $= \frac{x}{a}$ (§ 60.). By reducing $\frac{a}{a^2+x^2}$ into a series, we find

\frac{a}{a^2+x^2} = \frac{1}{a} - \frac{x^2}{a^3} + \frac{x^4}{a^5} - \frac{x^6}{a^7} + \&c.

Hence, multiply both sides by $\dot{x}$ , and taking the fluent of each term, we get

\int \frac{ax \dot{x}}{a^2+x^2} = \arctan\left(\frac{x}{a}\right) + \text{const.} = \frac{x}{a} - \frac{x^3}{3a^2} + \frac{x^5}{5a^4} - \frac{x^7}{7a^6} + \&c. + \text{const.}

If we wish to deduce from this equation, the value of the least arch whose tangent is $\frac{x}{a}$ , it is necessary to suppress the arbitrary constant quantity, for when that arch $= 0$ , then $x=0$ ; thus we have the arch whose tangent is $\frac{x}{a}$ expressed by the infinite series

\frac{x}{a} - \frac{x^3}{3a^2} + \frac{x^5}{5a^4} - \frac{x^7}{7a^6} + \&c.

Let $\pi$ denote the circumference of a circle whose diameter is unity, or half the circumference of a circle whose radius is unity, then, as the sine of $30^\circ$ degrees, or $\frac{1}{2}\pi$ , is $\frac{1}{2}$ , and its cosine $\sqrt{1-\frac{1}{4}} = \frac{1}{2}\sqrt{3}$ , we have $\tan \frac{\pi}{6} = \frac{\sin \frac{1}{2}\pi}{\cos \frac{1}{2}\pi} = \sqrt{\frac{1}{3}}$ . Let $\sqrt{\frac{1}{3}}$ be substituted instead of $x$ in the above series, and $a$ be supposed $= 1$ , thus we get

\frac{\pi}{6} = \sqrt{\frac{1}{3}} \times \left(1 - \frac{1}{3 \cdot 3} + \frac{1}{5 \cdot 3^3} - \frac{1}{7 \cdot 3^5} + \&c.\right)

and therefore

\pi = \sqrt{12} \times \left(1 - \frac{1}{3 \cdot 3} + \frac{1}{5 \cdot 3^3} - \frac{1}{7 \cdot 3^5} + \&c.\right);

by taking the sum of about fifteen terms of this series, we shall find $\pi = 3.1415927$ . The determination of this number is of great importance in every branch of mathematics.

138. By proceeding with the fluxion $\frac{x^m \dot{x}}{a^2+x^2}$ in the same manner as we have done with $\frac{ax \dot{x}}{a^2+x^2}$ , we get

\int \frac{x^m \dot{x}}{a^2+x^2} = \frac{x^{m+1}}{(m+1)a^2} - \frac{x^{m+n+1}}{(m+n+1)a^4} + \frac{x^{m+2n+1}}{(m+2n+1)a^6} \&c.

This series proceeds by the positive powers of $x$ , or is an ascending series, but we may also expand $\frac{1}{a^2+x^2}$ into a series proceeding by the negative powers of $x$ , and which will therefore be called a descending series. Thus because

\frac{1}{a^2+x^2}

\text{Inverse Method.} \quad \frac{1}{x^n + a^n} = \frac{1}{x^n} - \frac{a^n}{x^{2n}} + \frac{a^{2n}}{x^{3n}} - \frac{a^{3n}}{x^{4n}} + \&c.

we have, after multiplying both sides by $x^n$ and taking the fluents

\int \frac{x^n}{x^n + a^n} = - \frac{1}{(n-m-1)x^{n-m-1}} + \frac{a^n}{(2n-m-1)x^{3n-m-1}} - \frac{a^{2n}}{(3n-m-1)x^{5n-m-1}} + \&c. + \text{const.}

This last series will be convergent when $x$ is greater than $a$ , and at the same time $m \leq n$ , and $n > 1$ . But besides, that it may contain algebraic terms only, it is necessary that none of the divisors $n-m-1$ , $2n-m-1$ , $3n-m-1$ , &c. become $\equiv 0$ ; this circumstance will take place as often as $m+1$ is a multiple of $n$ , and in which case

the series $\frac{x^n}{x^{2n-m}} - \frac{a^n x^n}{x^{3n-m}} + \&c.$ which is the development of the fluxion, will contain a term of this form, $\frac{a^n x^n}{x}$ , the fluent of which is $a^n 1.x$ .

If in this result we put $m=2$ , $n=0$ , and $a=1$ , we get

\int \frac{x}{1+x^2} = - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \&c. + \text{const.}

But although the expression $\frac{x}{1+x^2}$ is the fluxion of the arch having $x$ for its tangent, we must not conclude that this series is the development of that arch, for $x$ being supposed $\equiv 0$ , each of the terms of the series becomes infinite.

The consideration of the constant quantity added to the fluent will remove this apparent difficulty, if we remark, that to know the true value of a series, it is always necessary to begin with the case in which it is convergent. Now the series

- \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \&c.

converges so much the faster as $x$ is greater, and it vanishes when $x$ is infinite; but in this extreme case the equation

\text{arc}(\tan. = x) = - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \&c. + \text{const.}

becomes simply $\text{arc} \frac{\pi}{2} = \text{const.}$ where $\pi$ denotes half the circumference of the circle; therefore, substituting this value of the constant quantity, we have

\text{arc}(\tan. = x) = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \&c.

We may easily find an expression for the fluent of the rational fraction $\frac{U}{V}$ (§ 111.) by expanding the

quantity $\frac{U}{V}$ into a series, but the result thus obtained is in general very complicated, and seldom convergent; besides, this manner of finding the fluent is hardly of any use, since it may be expressed by means of arches of a circle and logarithms, both of which are readily obtained from the common trigonometrical tables.

139. The fluent of $x^{m-1} \cdot (a+bx^n)^{\frac{p}{q}}$ is easily obtained by first expanding the quantity $(a+bx^n)^{\frac{p}{q}}$ into a series by the binomial theorem, then multiplying each term of that series by $x^{m-1}$ , and taking the fluents of the results by § 101. Thus we have $\int x^{m-1} \cdot (a+bx^n)^{\frac{p}{q}} =$

(a+bx^n)^{\frac{p}{q}} = a^{\frac{p}{q}} \left\{ \frac{x^m}{m} + \frac{p b x^{m+n}}{q a m+n} + \frac{p(p-q)b^2 x^{m+2n}}{1.2 q^2 a^2 m+2n} + \frac{p(p-q)(p-2q)b^3 x^{m+3n}}{1.2.3 q^3 a^3 m+3n} + \&c. \right\} + \text{const.}

This is an ascending series, but to get a descending series we must divide $(a+bx^n)^{\frac{p}{q}}$ by $x^{\frac{p}{q}}$ and multiply $x^{m-1}$ , the remaining part of the fluxional expression, by the same quantity, thus the fluxion is transformed to

x^{m+\frac{p}{q}-1} \cdot (b+a x^{-n})^{\frac{p}{q}}

the fluent of which, by proceeding as in the former case, is

b^{\frac{p}{q}} \left\{ \frac{q x^{m+\frac{p}{q}}}{m q + n p} - \frac{p a q x^{m+\frac{p}{q}}}{q b m q + (p-q)n} + \frac{p(p-q) a^2 q x^{m+\frac{p}{q}}}{1.2 q^2 b^2 m q + (p-2q)n} + \&c. \right\} + \text{const.}

either of these series may be employed if $a$ and $b$ are both positive, or $q$ an odd number, but if $q$ be an even number, the first formula becomes imaginary on account of the factor $a^{\frac{p}{q}}$ if $a^{\frac{p}{q}}$ be negative, and the same thing happens to the second formula if $a^{\frac{p}{q}}$ be negative.

140. Let it be required to express by a series the fluent of $\frac{x}{\sqrt{1-x^2}}$ . That we may develop the radical quantity $\frac{1}{\sqrt{1-x^2}}$ we put it under this form

\frac{1}{\sqrt{1-x^2}} = \frac{1}{5 C} (1-x^2)

$(1-x^2)^{-\frac{1}{2}}$ , an expression which when expanded by the binomial theorem, is

1 + \frac{1}{2}x^2 + \frac{1 \cdot 3}{2 \cdot 4}x^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}x^6 + \dots

therefore, multiplying each term of this series by $x$ , and taking the fluent, we get

\int \frac{x}{\sqrt{1-x^2}} = x + \frac{1 \cdot x^3}{2 \cdot 3} + \frac{1 \cdot 3 \cdot x^5}{2 \cdot 4 \cdot 5} + \frac{1 \cdot 3 \cdot 5 \cdot x^7}{2 \cdot 4 \cdot 6 \cdot 7} + \text{const.}

If we suppose $x$ to denote the fine of an arch, then $\sqrt{1-x^2}$ is its cofine, and $\frac{x}{\sqrt{1-x^2}}$ is the fluxion of the arch itself (§ 59.); therefore the series which we have just found, expresses the length of the arch of a circle, radius being unity, and the fine of the arch $x$ . If we suppose the series to express the smallest arch that corresponds to the fine $x$ , then, as when the fine of that arch $= 0$ , the arch itself $= 0$ , the series expressing the arch must vanish when $x = 0$ , therefore we must suppress the constant quantity added to complete the fluent; or suppose it $= 0$ . The same series has already been found by the direct method of fluxions in § 72.

Let $\pi$ denote the same as in § 137, then, as the fine of 30 degrees, or of $\frac{1}{6}\pi$ , is $\frac{1}{2}$ , we have, by substituting $\frac{1}{6}$ instead of $x$ in the preceding series, and multiplying both sides by 6,

\pi = 3 \left( 1 + \frac{1}{2 \cdot 3 \cdot 2^2} + \frac{1 \cdot 3}{2 \cdot 4 \cdot 5 \cdot 2^4} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7 \cdot 2^6} + \dots \right)

by means of this series, which involves only rational numbers, we may compute (but with more labour), the value of $\pi$ as before.

Suppose the fluxion to be $\frac{x}{\sqrt{ax-x^2}}$ , which may be otherwise expressed thus, $\frac{x}{\sqrt{a} \sqrt{1 - \frac{x^2}{a}}}$ . By the binomial theorem $(1 - \frac{x^2}{a})^{-\frac{1}{2}} =$

1 + \frac{x^2}{2a} + \frac{1 \cdot 1 \cdot x^4}{2 \cdot 4 \cdot a^2} + \frac{1 \cdot 1 \cdot 3 \cdot x^6}{2 \cdot 4 \cdot 6 \cdot a^3} + \dots

Let each term of this series be multiplied by $a^{\frac{1}{2}} x^{\frac{1}{2}} \frac{x}{\sqrt{a}}$ , and the fluent taken by § 101, thus we get $\int \frac{x}{\sqrt{ax-x^2}} =$

a^{\frac{1}{2}} \left( \frac{2x^{\frac{1}{2}}}{3} + \frac{1}{2} \frac{2x^{\frac{5}{2}}}{5a} + \frac{1 \cdot 1}{2 \cdot 4} \frac{2x^{\frac{9}{2}}}{7a^2} - \frac{1 \cdot 1 \cdot 3}{2 \cdot 4 \cdot 6} \frac{2x^{\frac{13}{2}}}{9a^3} + \dots \right) + \text{const.}

141. By resolving a fluxion into an infinite series, the object in view is to transform it into a series of other fluxions, each of which may have its fluent determined by known methods; but it is not always neces-

sary that the terms of the series should be each simply a power of $x$ multiplied by $x$ and constant quantities.

If for example we have this fluxion

\frac{x \sqrt{1-e^2 x^2}}{\sqrt{1-x^2}},

in which $e$ is supposed to denote a small constant quantity, we may expand $\sqrt{1-e^2 x^2}$ or $(1-e^2 x^2)^{\frac{1}{2}}$ into a series, which will thus become

1 - \frac{1}{2} e^2 x^2 - \frac{1 \cdot 1}{2 \cdot 4} e^4 x^4 - \frac{1 \cdot 1 \cdot 3}{2 \cdot 4 \cdot 6} e^6 x^6 - \dots

and the fluxion $\frac{x \sqrt{1-e^2 x^2}}{\sqrt{1-x^2}}$ will be transformed to

\frac{x}{\sqrt{1-x^2}} \left\{ 1 - \frac{1}{2} e^2 x^2 - \frac{1 \cdot 1}{2 \cdot 4} e^4 x^4 - \frac{1 \cdot 1 \cdot 3}{2 \cdot 4 \cdot 6} e^6 x^6 - \dots \right\}

the series will converge very fast when $e$ is small, for that $\sqrt{1-x^2}$ may be a real quantity, $x^2$ must be less than 1. We must now multiply each term of the series

by the common factor $\frac{x}{\sqrt{1-x^2}}$ and take the fluents, which being all contained in the general expression

$\int \frac{x^m x}{\sqrt{1-x^2}}$ , will be found by § 133. Thus, putting $A$ to denote an arch of which $x$ is the fine, we have

\begin{aligned} & \int \frac{x \sqrt{1-e^2 x^2}}{\sqrt{1-x^2}} = \\ & A + \frac{1}{2} e^2 \left\{ \frac{1}{2} x \sqrt{1-x^2} - \frac{1}{2} A \right\} \\ & + \frac{1 \cdot 1}{2 \cdot 4} e^4 \left\{ \left( \frac{1}{4} x^3 + \frac{1 \cdot 3}{2 \cdot 4} x \right) \sqrt{1-x^2} - \frac{1 \cdot 3}{2 \cdot 4} A \right\} \\ & + \frac{1 \cdot 1 \cdot 3}{2 \cdot 4 \cdot 6} e^6 \left\{ \left( \frac{1}{6} x^5 + \frac{1 \cdot 3}{4 \cdot 6} x^3 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x \right) \sqrt{1-x^2} - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} A \right\} \\ & + \dots + \text{const.} \end{aligned}

Of the Fluents of such Fluxions as involve Logarithmic and Exponential functions.

142. Let it be required to find the fluent of $x^m x(1.x)$ , where $1.x$ denotes the Napierian logarithm of $x$ . In this case, as well as in some following examples, we shall have recourse to the principle already employed in § 130, namely, that if $v$ and $x$ denote any functions of a variable quantity $x$ , then

\int x v = v x - \int v x.

Let us therefore assume $x^m x = v$ , and $1.x = x$ , then,

(§ 101.) $\frac{x^{m+1}}{m+1} = v$ , and (§ 57.) $\frac{x}{x} = 1$ , therefore, substituting

Inverse Method. substituting these values of $v, x, v', x'$ in the formula becomes

\int x^m \dot{x} (1.x) = \frac{x^{m+1} \cdot 1.x}{m+1} - \int \frac{x^m \dot{x}}{m+1},

or, since $\int \frac{x^m \dot{x}}{m+1} = \frac{x^{m+1}}{(m+1)^2} + \text{const.}$

\int x^m \dot{x} (1.x) = x^{m+1} \left\{ \frac{1.x}{m+1} - \frac{1}{(m+1)^2} \right\} + \text{const.}

Let us next suppose that the proposed fluxion is $x^m \dot{x} (1.x)^n$ . Put $x^m \dot{x} = v$ , and $(1.x)^n = x$ , then (§ 101.) $\frac{x^{m+1}}{m+1} = v$ , and (§ 57.) $\frac{n x (1.x)^{n-1}}{x} = \dot{x}$ , therefore, substituting as before these values in the formula $\int x \dot{v} = v x - \int v \dot{x}$ , we get

\int x^m \dot{x} (1.x)^n = \frac{x^{m+1} (1.x)^n}{m+1} - \int \frac{n x^m \dot{x} (1.x)^{n-1}}{m+1}.

It is evident, that by this formula the determination of the fluent of $x^m \dot{x} (1.x)^n$ is reduced to that of $x^m \dot{x} (1.x)$ which we have already found, and in like manner that the determination of the fluent of $x^m \dot{x} (1.x)^3$ is reduced to that of $x^m \dot{x} (1.x)^2$ , and so on, from which it appears that the fluent of $x^m \dot{x} (1.x)^n$ is expressible in finite terms when $n$ is a whole positive number. The formula, however, will not apply when $m = -1$ , because of the denominator $m+1 = -1+1=0$ . But in this case we have

\int \frac{\dot{x}}{x} (1.x)^n = \frac{1}{n+1} (1.x)^{n+1} + \text{const.}

If $n$ be negative or fractional, the fluent of $x^m \dot{x} (1.x)^n$ can only be expressed by an infinite series.

143. As an example of an exponential function, let it be required to find the fluent of $a^x \dot{x}$ . Here we may put $\dot{v}$ for $\dot{x}$ , and $x$ for $a^x$ , then we have $v = \frac{x^{\log a}}{\log a}$ , § 101, and $\dot{x} = (1.a) a^x \dot{x}$ , (§ 56.) therefore, substituting these values of $v, x, v', x'$ in the formula $\int x \dot{v} = v x - \int v \dot{x}$ , we get

\int a^x \dot{x} = \frac{a^x x^{\log a}}{\log a} - \int \frac{(1.a) a^x x^{\log a} \dot{x}}{\log a};

therefore, substituting $n = -1$ everywhere instead of $n$ ,

\int a^x \dot{x} = \frac{a^x x^{\log a}}{\log a} - \int \frac{(1.a) a^x x^{\log a} \dot{x}}{\log a};

hence, bringing $\int a^x \dot{x}$ to stand alone one side of the equation

\int a^x \dot{x} = \frac{a^x x^{\log a}}{1.a} - \frac{n}{1.a} \int a^x x^{\log a} \dot{x}.

If we suppose $n=1$ , then, observing that $\int a^x \dot{x} = \frac{a^x}{1.a}$ Inverse Method.

(§ 56.),

\int a^x \dot{x} = \frac{a^x x}{1.a} - \frac{a^x}{(1.a)^2} + \text{const.}

If $n=2$ , then

\int a^x \dot{x} = \frac{a^x x^2}{1.a} - \frac{2}{1.a} \int a^x \dot{x}.

In this expression we substitute the value of $\int a^x \dot{x}$ just found, thus it becomes

\int a^x \dot{x} = a^x \left\{ \frac{x^2}{1.a} - \frac{2x}{(1.a)^2} + \frac{2}{(1.a)^3} \right\} + \text{const.}

Proceeding in this way we may find the fluent when $n=3$ , or when $n=4$ , or in general, when $n$ is any integer number whatever, the number of terms in the fluent being in this case always finite; it is not so however when $n$ is either negative or fractional.

Of the Fluents of such Fluxions as contain functions related to a circle.

144. Let us suppose that $x$ is an arch whose fine is $x$ , and that it is required to find the fluent of $x \dot{x}$ . Put $x^2 \dot{x} = v$ , then

\int x \dot{x} = \int x \dot{v} = v x - \int v \dot{x};

but since $\dot{v} = x \dot{x}$ , we have $v = \frac{x^{n+1}}{n+1}$ , § 101, and since $x = \text{fin. } x$ , we have $\dot{x} = \dot{v} \cos. x = \dot{v} \sqrt{1-x^2}$ (§ 59.), and therefore $\dot{x} = \frac{\dot{v}}{\sqrt{1-x^2}}$ ; thus we have

\int x \dot{x} = \frac{x x^{n+1}}{n+1} - \frac{1}{n+1} \int \frac{x^{n+1} \dot{v}}{\sqrt{1-x^2}};

hence the determination of the proposed fluent is reduced to $\int \frac{x^{n+1} \dot{v}}{\sqrt{1-x^2}}$ which we have already considered in § 133. By the same mode of reasoning we may determine the fluent when $x$ denotes the cosine of the arch $x$ .

145. It appears from § 59. that $n x$ being put to denote any arch of a circle to radius unity, the fluxion of the fine of that arch is $n \dot{x} \cos. n x$ ; therefore, on the contrary,

\int \dot{x} \cos. n x = \frac{1}{n} \text{fin. } n x + \text{const.}

In like manner, from the formulas of § 59. and § 60. we find

\int \dot{x} \sin. nx = \frac{-1}{n} \cos. nx + \text{const.}

\int \frac{\dot{x}}{\cos. nx} = \frac{1}{n} \tan. nx + \text{const.}

\int \frac{\dot{x}}{\sin. nx} = \frac{-1}{n} \cot. nx + \text{const.}

\int \frac{\dot{x} \sin. nx}{\cos. nx} = \frac{1}{n} \sec. nx + \text{const.}

\int \frac{\dot{x} \cos. nx}{\sin. nx} = \frac{-1}{n} \operatorname{cosec}. nx + \text{const.}

146. By the second of these expressions we find the fluent of

\dot{x} (A + B \sin. x + C \sin. 2x + \&c.)

to be

Ax - B \cos. x - \frac{1}{2} C \cos. 2x - \&c. + \text{const.}

and from the first expression we find the fluent of

\dot{x} (A + B \cos. x + C \cos. 2x + \&c.)

to be

Ax + B \sin. x + \frac{1}{2} C \sin. 2x + \&c. + \text{const.}

147. It has been shewn in the Arithmetic of fines, (see ALGEBRA, § 356.) that

\sin. x = \frac{1}{2} (-\cos. 2x + 1),

therefore, by what has been shewn in § 145.

\begin{aligned} \int \dot{x} \sin. x &= \frac{1}{2} \int (-\dot{x} \cos. 2x + \dot{x}) \\ &= \frac{1}{2} (-\frac{1}{2} \sin. 2x + x) + \text{const.} \end{aligned}

It has also been shewn that

\sin. 3x = \frac{1}{2} (-\sin. 3x + 3 \sin. x),

therefore, multiplying each term of this expression by $\dot{x}$ , and taking the fluents,

\int \dot{x} \sin. 3x = \frac{1}{2} (\frac{1}{2} \cos. 3x - 3 \cos. x) + \text{const.}

In the same manner may the fluent of $\dot{x} \sin. nx$ be found, $n$ being any positive integer number whatever.

Again, it has been shewn (ALGEBRA, § 356.), that

\cos. x = \frac{1}{2} (\cos. 2x + 1);

therefore

\begin{aligned} \int \dot{x} \cos. x &= \frac{1}{2} \int (\dot{x} \cos. 2x + \dot{x}) \\ &= \frac{1}{2} (\frac{1}{2} \sin. 2x + x) + \text{const.} \end{aligned}

and because

\cos. 3x = \frac{1}{2} (\cos. 3x + 3 \cos. x)

therefore, multiplying by $\dot{x}$ , and taking the fluents,

\int \dot{x} \cos. 3x = \frac{1}{2} (\frac{1}{2} \sin. 3x + 3 \sin. x) + \text{const.}

and proceeding in this way we may find the fluents of $\dot{x} \cos. nx$ , $n$ being any positive integer number.

148. The fluents of $\dot{x} \sin. nx$ , and $\dot{x} \cos. nx$ may be expressed under another form, by proceeding as in § 142. Thus, beginning with $\dot{x} \sin. nx$ , and resolving it into $\dot{x} \sin. x \times \sin. (n-1)x$ , if we put $\dot{x} \sin. x = v$ , and $\sin. (n-1)x = t$ , we have by § 145. $v = -\cos. x$ , and (by § 26 and § 59) $t = (n-1)x \cos. x \sin. (n-2)x$ , therefore, substituting in the formula $\int t v = v t - \int v t$ , we have

\begin{aligned} \int \dot{x} \sin. nx &= -\cos. x \sin. (n-1)x \\ &+ (n-1) \int \dot{x} \cos. x \sin. (n-2)x; \end{aligned}

but $\cos. x = 1 - \sin. x$ , therefore $\int \dot{x} \sin. nx$ is equal to

\begin{aligned} &-\cos. x \sin. (n-1)x + (n-1) \int \dot{x} \sin. (n-2)x \\ &-(n-1) \int \dot{x} \sin. nx; \end{aligned}

which expression, by bringing together the terms containing $\int \dot{x} \sin. nx$ becomes

\begin{aligned} \int \dot{x} \sin. nx &= -\frac{1}{n} \cos. x \sin. (n-1)x \\ &+ \frac{n-1}{n} \int \dot{x} \sin. (n-2)x. \end{aligned}

By giving particular values to $n$ we have

\begin{aligned} \int \dot{x} \sin. x &= -\frac{1}{2} \cos. x \sin. x + \frac{1}{2} \int \dot{x} \\ &= -\frac{1}{2} \cos. x \sin. x + \frac{1}{2} x + \text{const.} \end{aligned}

\begin{aligned} \int \dot{x} \sin. 3x &= -\frac{1}{3} \cos. x \sin. 3x + \frac{2}{3} \int \dot{x} \sin. x \\ &= -\frac{1}{3} \cos. x \sin. 3x - \frac{2}{3} \cos. x + \text{const.} \end{aligned}

We may proceed in this way as far as we please, deducing the fluent of $\dot{x} \sin. nx$ from that of $\dot{x} \sin. (n-1)x$ , and the fluent of $\dot{x} \sin. (n-1)x$ from that of $\dot{x} \sin. (n-2)x$ , and so on.

If in the general formula we substitute every where $2-n$ instead of $n$ , it becomes

\begin{aligned} \int \dot{x} \sin. (2-n)x &= \frac{1}{n-2} \cos. x \sin. (1-n)x \\ &+ \frac{n-1}{n-2} \int \dot{x} \sin. nx. \end{aligned}

Inverse Method. an expression which, by bringing $\int \ddot{x} \text{fin.}^{-n} \ddot{x}$ or $\int \frac{\ddot{x}}{\text{fin.}^n \ddot{x}}$ to stand on one side of the equation, becomes

\int \frac{\ddot{x}}{\text{fin.}^n \ddot{x}} = -\frac{1}{(n-1)} \frac{\text{cos.} \ddot{x}}{\text{fin.}^{n-1} \ddot{x}} + \frac{n-2}{n-1} \int \frac{\ddot{x}}{\text{fin.}^{n-1} \ddot{x}}.

This formula is not applicable to the case of $n=1$ , because then each of the terms of the fluent is divided by $n-1=0$ , and therefore becomes infinite. In order to obtain the expression for the fluent in this particular case, we proceed thus. It is evident that $\frac{1}{\text{fin.}^2 \ddot{x}} = \frac{1}{1-\text{cos.}^2 \ddot{x}}$ , but $\frac{1}{1-\text{cos.}^2 \ddot{x}} = \frac{1}{2(1-\text{cos.} \ddot{x})} + \frac{1}{2(1+\text{cos.} \ddot{x})}$ as will be found by reducing the fractions to a common denominator, therefore $\frac{1}{\text{fin.}^2 \ddot{x}} = \frac{\text{fin.} \ddot{x}}{2(1-\text{cos.} \ddot{x})} + \frac{\text{fin.} \ddot{x}}{2(1+\text{cos.} \ddot{x})}$ , and consequently,

\int \frac{\ddot{x}}{\text{fin.}^2 \ddot{x}} = \frac{1}{2} \int \frac{\ddot{x} \text{fin.} \ddot{x}}{1-\text{cos.} \ddot{x}} + \frac{1}{2} \int \frac{\ddot{x} \text{fin.} \ddot{x}}{1+\text{cos.} \ddot{x}};

but if it be considered that the fluxion of $\text{cos.} \ddot{x}$ is $-\ddot{x} \text{fin.} \ddot{x}$ (§ 59.) it will appear by § 103 that $\int \frac{\ddot{x} \text{fin.} \ddot{x}}{1-\text{cos.} \ddot{x}} = 1 \cdot (1-\text{cos.} \ddot{x})$ , and that $\int \frac{\ddot{x} \text{fin.} \ddot{x}}{1+\text{cos.} \ddot{x}} = -1 \cdot (1+\text{cos.} \ddot{x})$ , therefore

\begin{aligned} \int \frac{\ddot{x}}{\text{fin.}^2 \ddot{x}} &= \frac{1}{2} \cdot 1 \cdot (1-\text{cos.} \ddot{x}) - \frac{1}{2} \cdot 1 \cdot (1+\text{cos.} \ddot{x}) + \text{const.} \\ &= \frac{1}{2} \cdot 1 \cdot \left( \frac{1-\text{cos.} \ddot{x}}{1+\text{cos.} \ddot{x}} \right) + \text{const.} \\ &= 1 \cdot \frac{\sqrt{1-\text{cos.}^2 \ddot{x}}}{\sqrt{1+\text{cos.}^2 \ddot{x}}} + \text{const.} \end{aligned}

If in the general formula for $\int \frac{\ddot{x}}{\text{fin.}^n \ddot{x}}$ we suppose $n=2$ we have

$\int \frac{\ddot{x}}{\text{fin.}^2 \ddot{x}} = -\frac{\text{cos.} \ddot{x}}{\text{fin.} \ddot{x}} = -\text{cotan.} \ddot{x} + \text{const.}$ which agrees with what has been already observed (§ 145.), and if we suppose $n=3$ , then

\begin{aligned} \int \frac{\ddot{x}}{\text{fin.}^3 \ddot{x}} &= -\frac{\text{cos.} \ddot{x}}{2 \text{fin.}^2 \ddot{x}} + \frac{1}{2} \int \frac{\ddot{x}}{\text{fin.} \ddot{x}} \\ &= -\frac{\text{cos.} \ddot{x}}{2 \text{fin.}^2 \ddot{x}} + \frac{1}{4} \cdot 1 \cdot \left( \frac{1-\text{cos.} \ddot{x}}{1+\text{cos.} \ddot{x}} \right) + \text{const.} \end{aligned}

In this way we proceed as far as we please, deducing

$\int \frac{\ddot{x}}{\text{fin.}^4 \ddot{x}}$ from $\int \frac{\ddot{x}}{\text{fin.}^3 \ddot{x}}$ , and so on.

149. If in the general formulas for $\int \ddot{x} \text{fin.}^n \ddot{x}$ and

$\int \frac{\ddot{x}}{\text{fin.}^n \ddot{x}}$ we substitute $\frac{1}{4} \pi - \ddot{x}$ instead of $\ddot{x}$ , (where $\frac{1}{4} \pi$ denotes a quadrant), and observe that

\text{fin.} \left( \frac{1}{4} \pi - \ddot{x} \right) = \text{cos.} \ddot{x}, \quad \text{cos.} \left( \frac{1}{4} \pi - \ddot{x} \right) = \text{fin.} \ddot{x},

and that the fluxion of $\left( \frac{1}{4} \pi - \ddot{x} \right)$ is $-\ddot{x}$ , we shall immediately obtain

\begin{aligned} \int \ddot{x} \text{cos.}^n \ddot{x} &= \frac{1}{n} \text{fin.} \ddot{x} \text{cos.}^{n-1} \ddot{x} \\ &+ \frac{n-1}{n} \int \ddot{x} \text{cos.}^{n-1} \ddot{x} \end{aligned}

\begin{aligned} \int \frac{\ddot{x}}{\text{cos.}^n \ddot{x}} &= \frac{1}{n-1} \frac{\text{fin.} \ddot{x}}{\text{cos.}^{n-1} \ddot{x}} \\ &+ \frac{n-2}{n-1} \int \frac{\ddot{x}}{\text{cos.}^{n-1} \ddot{x}}; \end{aligned}

and in like manner from the formula expressing the fluent of $\int \frac{\ddot{x}}{\text{fin.} \ddot{x}}$ we deduce

\begin{aligned} \int \frac{\ddot{x}}{\text{cos.} \ddot{x}} &= \frac{1}{2} \cdot 1 \cdot \left( \frac{1+\text{fin.} \ddot{x}}{1-\text{fin.} \ddot{x}} \right) + \text{const.} \\ &= 1 \cdot \frac{\sqrt{1+\text{fin.}^2 \ddot{x}}}{\sqrt{1-\text{fin.}^2 \ddot{x}}} + \text{const.} \end{aligned}

150. It has been shewn in ALGEBRA, § 357, that $16 \text{cos.}^3 \ddot{x} \text{fin.}^3 \ddot{x} = -\text{fin.} 5 \ddot{x} + \text{fin.} 3 \ddot{x} + 2 \text{fin.} \ddot{x}$ , therefore

\int \ddot{x} \text{cos.}^3 \ddot{x} \text{fin.}^3 \ddot{x} = \frac{1}{16} \left( \frac{1}{5} \text{cos.} 5 \ddot{x} - \frac{1}{3} \text{cos.} 3 \ddot{x} - 2 \text{cos.} \ddot{x} \right) + \text{const.}

The same mode of finding the fluent will apply to any fluxion of this form $\ddot{x} \text{fin.}^m \ddot{x} \text{cos.}^n \ddot{x}$ ; or by resolving the fluxion into two parts, the determination of its fluent may be reduced to that of a fluxion in which the exponents $m$ and $n$ are less than in the proposed fluxion, by the method of proceeding already employed in § 148.

151. Let us now denote $\text{fin.} \ddot{x}$ by $x$ , then $\text{cos.} \ddot{x} = \sqrt{1-x^2}$ and since $\ddot{x} \text{cos.} \ddot{x} = \ddot{x}$ (§ 59.) therefore $\ddot{x} = \frac{\ddot{x}}{\sqrt{1-x^2}}$ ; these values being substituted in any function involving $\ddot{x}$ , $\text{fin.} \ddot{x}$ and $\text{cos.} \ddot{x}$ will immediately reduce it to an algebraic form. Thus for example, we shall have $\ddot{x} \text{fin.}^m \ddot{x} \text{cos.}^n \ddot{x}$ transformed to

x^m (1-x^2)^{\frac{n-1}{2}}

an expression which may have its fluent determined by the formulas of § 133 and § 134.

SECT. II. Application of the Inverse Method of Fluxions to the Resolution of Problems.

To find the Areas of Curves.

152. It has been shewn in § 61, that if the abscissa of a curve be denoted by $x$ , the ordinate by $y$ , and the

Inverse Method. area by $s$ , then $s = yx$ . Therefore the general formula expressing the area of any curve will be

s = \int yx.

Hence to find the area of any curve, we must either find from the equation of the curve the value of $y$ in terms of $x$ , or else the value of $x$ in terms of $y$ , and $y$ , and either the one or the other of these being substituted in the above formula, and the fluent found by the methods already explained, we shall have a general expression for the curvilinear area as required.

Ex. 1. Let it be required to find the area of any curve of the parabolic kind, of which, putting the abscissa $AB = x$ , and the ordinate $BP = y$ , the equation is $y^2 = ax^n$ .

Then we have $y = a^{\frac{1}{n}} x^{\frac{m}{n}}$ , and

\begin{aligned} s &= \int yx = a^{\frac{1}{n}} \int x^{\frac{m+n}{n}} \\ &= \frac{na^{\frac{1}{n}}}{m+n} x^{\frac{m+n}{n}} + C, \end{aligned}

where $C$ denotes the constant quantity that may be required to complete the fluent. As in the present case, it is required to find the area of the portion of the curve next its vertex, so that when $s=0$ , then $x=0$ , there-

fore, also $C=0$ , and the area is simply $\frac{na^{\frac{1}{n}}}{m+n} x^{\frac{m+n}{n}}$ .

If it be required to find the area comprehended between two ordinates $PB, pb$ , put $AB=d$ , then when $s=0$ , we have $x=d$ , therefore the general expression

s = \frac{na^{\frac{1}{n}}}{m+n} x^{\frac{m+n}{n}} + C \text{ becomes in this case } 0 = \frac{na^{\frac{1}{n}}}{m+n} d^{\frac{m+n}{n}} + C,

hence $C = -\frac{na^{\frac{1}{n}}}{m+n} d^{\frac{m+n}{n}}$ , and consequently the area $BPpb$ , or $s$ , is equal to

\frac{na^{\frac{1}{n}}}{m+n} \left\{ x^{\frac{m+n}{n}} - d^{\frac{m+n}{n}} \right\}.

When $n$ is an even number, the expression for the

area, viz. $\frac{na^{\frac{1}{n}}}{m+n} x^{\frac{m+n}{n}}$ may be considered as negative as well as positive, on account of the radical quantity $\frac{m+n}{n}$ , or $\sqrt[n]{x^{m+n}}$ , which has then a twofold value, it may therefore have the sign $\pm$ prefixed to it; but in this case the same abscissa $AB$ belongs to two branches of the curve $APp$ and $AP'p'$ , as in fig. 19. N^o 1, there-

fore the two values of the expression $\pm \frac{na^{\frac{1}{n}}}{m+n} x^{\frac{m+n}{n}}$ may be considered as indicating the two areas $APB, AP'B$ ,

on each side of the axis, corresponding the one to the positive, and the other to the negative ordinates.

When the exponents $m$ and $n$ are both odd numbers,

the quantity $x^{\frac{m+n}{n}}$ has only one sign, and remains always positive whatever be the sign of $x$ , but in this case one of the two branches of the curve has its abscissas and its ordinates negative at the same time (as in fig. 19. N^o 2.): it follows therefore that the areas corresponding to the negative abscissas and ordinates ought to be regarded as positive.

If $n$ alone is odd, then the quantity $x^{\frac{m+n}{n}}$ becomes negative at the same time as $x$ , but in this case the two branches of the curve are on the same side of the line in which the abscissas are taken (as in fig. 19. N^o 3.): and the ordinates remain always positive.

Upon the whole it may be concluded, that the area of a curve is positive when the abscissas and the ordinates have the same sign, and negative when they have contrary signs.

If we suppose $m=1$ , and $n=2$ , then the curve is the common parabola, the area of which from the general

formula is found to be $\frac{2}{3} a^{\frac{1}{2}} x^{\frac{3}{2}} = \frac{2}{3} xy$ ; hence it appears that the parabola is $\frac{2}{3}$ of its circumscribing parallelogram.

Ex. 2. Suppose the curve to be a circle. Put Fig. 20. $AB=x, BP=y$ , the diameter $AD=a$ , the area $ABP=s$ . From the nature of the circle $y^2 = a^2 - x^2$ , therefore $y = \sqrt{a^2 - x^2}$ , and

s = \int yx = \int x \sqrt{a^2 - x^2}.

In this case the fluxion is not of such a form as to admit of an algebraic fluent in finite terms, we must therefore have recourse to the method of series, but we have already found the fluent in this way in § 140, therefore, from the series there brought out we have

\begin{aligned} s &= \sqrt{ax} \left( \frac{2x}{3} - \frac{1}{2} \frac{2x^3}{5a} - \frac{1.1}{2.4} \frac{2x^5}{7a^3} \right. \\ &\quad \left. - \frac{1.1.3}{2.4.6} \frac{2x^7}{9a^5} - \dots \right) \end{aligned}

this expression does not require a constant quantity to be added to it, because when $x=0$ we must also have $s=0$ .

If we suppose the arch $AP$ to be $\frac{1}{4}$ of the quadrant $AE$ , then it is known that $PB = \frac{1}{4}$ the rad. $AC = \frac{1}{4} a$ , therefore, if we suppose the radius $= 1$ , we have in this case $BC = \frac{1}{4} \sqrt{3}$ , and $AB = 1 - \frac{1}{4} \sqrt{3} = 0.1339746$ nearly. If this number be substituted instead of $x$ , and a few terms of the series computed, we shall find the area $ABP = 0.0452931$ ; to this add the triangle $CBP = \frac{1}{4} \times \sqrt{3} = 0.2165063$ , and we have the sector $ACP = 0.2617994$ , which number when multiplied by 3 gives 7853982 for the area of the quadrant. This number also expresses the area of a circle of which the diameter is 1.

Ex. 3. Suppose the curve to be an ellipse. Put Fig. 21. the

Inverse Method. the transverse axis $AD = a$ , the conjugate axis $2CE = b$ , also $AB = x$ , $BP = y$ ; then by the nature of the curve $y = \frac{b}{a} \sqrt{a^2 - x^2}$ , and $\dot{s} = y \dot{x} = \frac{b}{a} \dot{x} \sqrt{a^2 - x^2}$ ; but if a circle be described on $AD$ as a diameter, and $BP$ the ordinate of the ellipse be produced to meet the circle, it appears from last example that $\dot{x} \sqrt{a^2 - x^2}$ is the fluxion of $AQB$ the segment of the circle corresponding to the elliptic area $APB$ or $s$ ; therefore, putting $v$ for the segment $AQB$ , we have $\dot{s} = \frac{b}{a} \dot{v}$ and $s = \frac{b}{a} v$ , here the constant quantity $c$ must be suppressed, because $s$ and $v$ must vanish together. Hence it appears that the area of any segment of an ellipse is to the area of the corresponding segment of its circumscribing circle as the lesser axis of the ellipse is to the greater; therefore the whole ellipse must be to the whole circle in the same ratio.

Fig. 22. Ex. 5. Let the curve be a hyperbola, of which $C$ is the centre. Put the semi-transverse axis $CA = a$ , the semiconjugate axis $= b$ , $CB = x$ , $BP = y$ , the area $APB = s$ . From the nature of the curve $y = \frac{b}{a} \sqrt{x^2 - a^2}$ , therefore

s = \int y \dot{x} = \frac{b}{a} \int \dot{x} \sqrt{x^2 - a^2}.

But it appears from formula B (§ 131.) that

\int \dot{x} \sqrt{x^2 - a^2} = \frac{1}{2} x \sqrt{x^2 - a^2} - \frac{1}{2} a^2 \int \frac{\dot{x}}{\sqrt{x^2 - a^2}}

and again by § 127,

\int \frac{\dot{x}}{\sqrt{x^2 - a^2}} = \frac{1}{a} \left\{ x + \sqrt{x^2 - a^2} \right\} + c,

s = \frac{b}{2a} x \sqrt{x^2 - a^2} - \frac{ab}{2} \frac{1}{a} \left\{ x + \sqrt{x^2 - a^2} \right\} + c.

To discover the value of the constant quantity $c$ , we must consider that when $x = a$ , then $s = 0$ , and in this extreme case the general equation just found becomes

0 = -\frac{ab}{2} \frac{1}{a} a + c,

hence $c = \frac{ab}{2} \frac{1}{a} a$ , and consequently, observing that

\begin{aligned} & -\frac{ab}{2} \frac{1}{a} \left\{ x + \sqrt{x^2 - a^2} \right\} + \frac{ab}{2} \frac{1}{a} a \\ & = -\frac{ab}{2} \frac{1}{a} \left\{ \frac{x + \sqrt{x^2 - a^2}}{a} \right\} \end{aligned}

s = \frac{b}{2a} x \sqrt{x^2 - a^2} - \frac{ab}{2} \frac{1}{a} \left\{ \frac{x + \sqrt{x^2 - a^2}}{a} \right\}.

It immediately follows from this formula that

\frac{b}{2a} x \sqrt{x^2 - a^2} - s = \frac{ab}{2} \frac{1}{a} \left\{ \frac{x + \sqrt{x^2 - a^2}}{a} \right\}

but if a straight line be drawn from $C$ to $P$ so as to form the triangle $CBP$ , it is manifest that $\frac{b}{2a} x \sqrt{x^2 - a^2}$ is equal to $\frac{1}{2} CB \times BP$ , that is, to the triangle $CBP$ , therefore the excess of the triangle $CBP$ above the area $s$ , that is the hyperbolic sector $CAP$ , is equal to the logarithmic function

\frac{ab}{2} \frac{1}{a} \left\{ \frac{x + \sqrt{x^2 - a^2}}{a} \right\}.

Ex. 5. Suppose the curve to be an equilateral hyperbola, that is a hyperbola whose axes are equal, and that it is required to find the curvilinear area $DCBP$ comprehended between $DC$ , a perpendicular from $D$ (a given point in the curve) to one of the asymptotes, and $PB$ , a perpendicular from any other point in the curve to the same asymptote.

Let $A$ be the centre, put $AC = a$ , $CD = b$ , $AB = x$ , $BP = y$ , the area $DCBP = s$ . From the property of the asymptotes we have $xy = ab$ , and therefore $y = \frac{ab}{x}$ , hence (§ 103.)

s = \int y \dot{x} = \int \frac{ab}{x} \dot{x} = ab \frac{1}{x} \int \dot{x} = ab \ln x + c.

To find the value of $c$ , let us suppose $x = a$ , then $s = 0$ , and the general formula becomes in this case

0 = ab \ln a + c, \text{ and hence } c = -ab \ln a,

therefore

s = ab \ln x - ab \ln a = ab \ln \frac{x}{a}.

If we suppose $a = b = 1$ , then $s = \ln x$ , from which it appears that in this case the hyperbolic area $DCBP$ represents the Napierian logarithm of the number $x$ ; it was from the consideration of this property that the logarithms originally invented by Napier were called hyperbolic logarithms.

But the logarithms of any other system may also be represented by areas of the same hyperbola; for this purpose it is only necessary to determine the magnitudes of $a$ , and $b$ , so that $\frac{b}{a} = M$ , where $M$ denotes the modulus of the system, thus we shall have $ab = a^2 M$ , and $s = a^2 M \ln \frac{x}{a}$ , or, putting $a = 1$ , $s = M \ln x$ , an expression for the logarithm of $x$ according to any system whatever of which the modulus is $M$ (ALGEBRA, § 287.)

Ex. Let the curve be the cycloid of which $AE$ is the axis and $A$ the vertex, let a semicircle be described on $AE$ as a diameter, draw $AG$ perpendicular to the axis, and from any point in the curve draw $PB$ perpendicular to $AG$ and $PD$ perpendicular to $AE$ , meeting the circle in $Q$ , and draw $QC$ to $C$ the centre of the circle. Put $AC = a$ , $AB = x$ , $BP = y$ , the area $ABP = s$ , and put $\psi$ for the angle $ACQ$ , that is for the arc of a circle which measures $ACQ$ , the radius of

Inverse Method. of that circle being unity, then $AD=a(1-\cos. v)$ , $DQ=a \sin. v$ , and $\text{arch } AQ=a v$ , and since from the nature of the curve, $PD=\text{arch } BQ+DQ$ , therefore $PD=a v+a \sin. v=a(v+\sin. v)$ ; hence

x=a(v+\sin. v), \dot{x}=a\dot{v}(1+\cos. v), (\S 59.)

y=a(1-\cos. v)

s=\int \dot{x} y = \int a^2 \dot{v} (1-\cos. v)

=a^2 \int v \sin. v = \frac{1}{2} a^2 v - \frac{1}{2} a^2 \sin. v \cos. v (\S 148.)

= \frac{1}{2} AC \times \text{arch } AQ - \frac{1}{2} CD \times DQ;

and here no constant quantity is wanted to complete the fluent; because upon the supposition that $AQ=0$ both sides of the equation vanish as they ought to do; now it is obvious that $\frac{1}{2} AC \times AQ = \text{area of sector } ACQ$ , and $\frac{1}{2} CD \times DQ = \text{area of triangle } DCQ$ , therefore

s = \text{area of circ. seg. } ADQ.

Let $AG$ be the greatest value of $x$ ; complete the parallelogram $AGFE$ , then from the general expression for the cycloidal area, it follows that the whole cycloidal space $APFG$ is equal to the semicircle $AQE$ ; but from the nature of the curve, $EF$ is equal to $AQE$ , half the circumference, therefore the rectangle $EG$ is equal to four times the semicircle $AQE$ ; from these equals take away the external cycloidal space $AGE$ , and the semicircle $AQE$ , which have been shewn to be equal, and the remainders, viz. the internal cycloidal space $APFE$ , and three times the semicircle $AQE$ , are equal to each other.

153. In some cases it is more convenient to refer a curve line to a fixed point than to an axis. Thus instead of expressing the nature of the circle by the equation $y^2=a^2-x^2$ , where $y$ denotes a perpendicular from any point of the curve upon $a$ the diameter, and $x$ the distance of that perpendicular from one end of the diameter, we may otherwise express it by the equation $z=r v$ , where $z$ denotes a variable arch of the circle reckoned from one end of its diameter, $r$ its radius, and $v$ the angle contained by a line drawn from the centre of the circle through the extremity of $z$ , which angle is measured by an arch of a circle having its radius unity.

The nature of the different conic sections may be defined in the same manner. Let $P$ be any point in a conic section, of which $F$ is one focus, and $FA$ a part of the axis; let $DC$ the directrix of the section meet $FA$ in $C$ , join $PF$ , and draw $PB$ perpendicular to the axis, and $PD$ to the directrix; then from the nature of the curve (CONIC SECTIONS) $PF$ has a given ratio to $PD$ , that is to $FC=FB$ ; put $FC=a$ , $FP=r$ , the angle $PFC=v$ , and suppose $PF:PD::n:1$ , then $PF=n.PD=n.FC=n.FB$ , hence observing that $FB=FP \times \cos. v$ , we get $r=a n-n r \cos. v$ , and $r = \frac{an}{1+n \cos. v}$ which equation expresses generally the nature of a conic section.

154. The formula which we have found for the flux-

ion of a curvilinear area, in § 61, is not immediately applicable when the nature of a curve is expressed in this way, we shall therefore investigate another formula suited to this particular manner of considering curves.

Let us suppose that $APR$ is a curve the position of any point $P$ of which is determined by $PF$ , its distance from a given point $F$ , and the angle which $PF$ makes with $AF$ a line given by position. Let a circle be described on $F$ as a centre with a rad. $=1$ , then $FP$ , as also the area $FAP$ , may be considered as functions of $BD$ the arch of that circle which measures the angle $PFA$ . From $F$ draw $FP'$ to any other point $P'$ of the curve meeting the circle $BD$ in $D'$ . Put $FP=r$ , the area $FAP=s$ , the angle $AFP$ , or the arch $BD=v$ , then the area $FPP'$ , and the arch $DD'$ , will be the corresponding increments of $s$ and $v$ , therefore, § 21,

\frac{\dot{s}}{\dot{v}} = \text{limit of } \frac{\text{area } FPP'}{DD' \times FD}.

Here $DD'$ the increment of $v$ is multiplied by $FD=1$ , to render the terms of the ratio homogeneous. On $F$ as a centre, with $FP$ as a radius, describe an arch of a circle meeting $FP'$ in $Q$ , then, as the sectors $FDD'$ , $FPQ$ are similar, we have

FE^2 : FP^2 :: FD \times DD' : FP \times PQ = 2 \text{ sect. } FPQ,

\text{hence } DD' \times FD = \frac{2 FD^2}{FP^2} \text{ sect. } FPQ = \frac{2}{r^2} \text{ sect. } FPQ

\text{and } \frac{\dot{s}}{\dot{v}} = \frac{r^2}{2} \times \lim. \frac{\text{area } FPP'}{\text{sect. } FPQ}

but the point $P'$ being supposed to approach continually to $P$ , it is manifest that the limit of $\frac{\text{area } FPP'}{\text{sect. } FPQ}$ is unity, or 1, therefore

\frac{\dot{s}}{\dot{v}} = \frac{r^2}{2}, \text{ and } \dot{s} = \frac{1}{2} r^2 \dot{v}.

154. By means of this formula we may find the areas of that class of curves called spirals. Let us take for example the spiral of ARCHIMEDES, which may be defined thus. Conceive a straight line $FR$ to revolve about $F$ the centre of a given circle, departing from a given position $FB$ ; conceive also a point $P$ to move in the revolving line, so that $PF$ its distance from the centre may be to $BD$ the arch of the circle passed over by the revolving line, as $m$ to $n$ , then the point $P$ will generate the spiral.

Put $BF=a$ , the angle $BFR=v$ , the line $FP=r$ , and the area generated by the line $FP=s$ , then the arch $BD=av$ , and since from the nature of the curve

r : av :: m : n, \text{ therefore } v = \frac{nr}{am}, \text{ and } \dot{v} = \frac{n \dot{r}}{am}, \text{ hence}

the general formula $\dot{s} = \frac{1}{2} r^2 \dot{v}$ becomes $\dot{s} = \frac{nr^2 \dot{r}}{2am}$ , therefore,

s = \int \frac{nr^2 \dot{r}}{2am} = \frac{nr^3}{6am}

this fluent does not require a constant quantity to be added,

added, as both $s$ and $r$ evidently vanish at the same time.

Fig. 28. 156. As the general expression for a curvilinear area $BCPD$ is $\int y x$ , where $x=AB$ the abscissa reckoned from a given point $A$ in the axis, and $y=BC$ the ordinate, it follows that $X$ being put to denote any function of a variable quantity $x$ , the fluent of $X x$ may always be exhibited by means of a curvilinear area. Thus let $CP\rho$ be a curve of such a nature that $AD$ and $DP$ the co-ordinates being denoted by $x$ and $y$ , the equation of the curve is $y=X$ , then, assuming any ordinate $BC$ as given by position, we have evidently

\int X x = \text{area } CBPD.

As the ordinate $BC$ (which is assumed as given by position) may be taken any where, the fluxion of the area being the same wherever it is taken, it appears, as has been already observed (§ 101) that the function $\int X x$ may be considered as indeterminate, for it admits of innumerable values corresponding to any particular value of $x$ , and in this respect it differs from an algebraic function, which for a given value of $x$ has always a determinate number of values. If however $x$ be supposed to increase from any determinate magnitude $a$ , to any other determinate magnitude $a'$ , then, taking the abscissa $AD=a$ , and $Ad=a'$ , and drawing the ordinates $DP, dp$ , we have

\text{when } x=a, \int X x = \text{area } CBPD,

\text{and when } x=a', \int X x = \text{area } CBdp,

Fig. 29. therefore, while $x$ increases from $a$ to $a'$ , or receives the increment $a'-a$ , the function $\int X x$ increases from area $CBPD$ to area $CBdp$ , and thus receives the increment area $PDdp$ , which is of a determinate magnitude as the ordinates $PD, dp$ have both a determinate position. Hence it appears that in assigning the fluent of $X x$ , we only determine the change that takes place in the value of the function $\int X x$ while $x$ passes from one particular value to another particular value.

157. As there are general and known methods by which an approximate value of any curvilinear area may be found, when a fluent is expressed by such an area, those methods may be applied to find an approximate value of the fluent. Let $PDdp$ be a curvilinear area, supposed to represent the fluent $\int X x$ between the limits of $x=AD$ and $x=Ad$ . Conceive $Dd$ to be divided into a number of equal parts $DD', D'D'', Dd$ , and the ordinates $P'D', P''D''$ drawn, and the two sets of parallelograms $DE, D'E', D''E''$ and $D'e, D''e, d''e''$ to be completed, the former constituting a rectilinear figure circumscribed about the curvilinear space $DD'P'D''pd$ , and the latter a rectilinear figure inscribed in that space; then as the circumscribed figure must necessarily be greater than the curvilinear space, that is, greater than

Inverse Method. $\int X x$ taken between the limits of $x=AD$ and $x=Ad$ , and the inscribed figure must be less, it follows that if we compute the areas of the circumscribed and inscribed figures we shall obtain two limits, the one greater, and the other less than $\int X x$ . And as by increasing the number of equal parts into which $Dd$ is divided we may bring the circumscribed and inscribed rectilinear figures as near to a ratio of equality as we please, it is always possible to find two limits which shall differ from each other, and consequently from $\int X x$ (which lies between them) by less than any assignable quantity.

158. If we join $P, P', P'', p$ , the tops of the ordinates, the rectilinear space formed by the trapeziums $DPP'D', D'P''P''D'', D''P''P''p$ will be more nearly equal to the curvilinear area, than the circumscribed rectilinear figure formed by the parallelograms $DE, D'E', D''E''$ ; therefore the sum of those trapeziums being found, it will be equal to the fluent $\int X x$ nearly.

Suppose, for example, that it is required to find the value of $\int \frac{x}{1+x^2}$ between the limits of $x=0$ , and $x=1$ . In this case $X = \frac{1}{1+x^2}$ , so that the equation of the curve $Pp$ is $y = \frac{1}{1+x^2}$ ; let us suppose $Dd$ the distance between the extreme ordinates to be divided into ten equal parts, then putting $0, 1, 2, \dots$ to $1$ instead of $x$ in the formula $y = \frac{1}{1+x^2}$ , we obtain eleven successive values of $y$ , or eleven equidistant ordinates, the numeral values of which will be as follows,

The first = 1.00000	the 7th = .73529
the 2d = .99010	the 8th = .67114
the 3d = .96154	the 9th = .60975
the 4th = .91743	the 10th = .55249
the 5th = .86207	the 11th = .50000
the 6th = .80000

By the elements of geometry the area of the rectilinear figure formed by the trapeziums is found by adding together all the ordinates except the first and last, and half the sum of the first and last, and multiplying that sum by the breadth of one of the trapeziums; now the sum of the ordinates, with the exception of the first and last, together with half the sum of the first and last, is $7.84981$ , and the common breadth of the trapeziums is $.1$ , therefore $\int \frac{x}{1+x^2} = 7.84981 \times .1 = .785$ nearly, as required. It is evident from § 137, that $\int \frac{x}{1+x^2}$ taken between the limits of $x=0$ , and $x=1$ , is accurately equal to an arch of $45^\circ$ , radius being unity, which arch being $\frac{1}{2}\pi$ will be found $= .7854$ nearly. If we recur to the series which has been found to express the above fluent in § 137, and put $a=1$ , and $x=1$ , we shall have $\int \frac{x}{1+x^2}$ taken between the prescribed limits equal to

x = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots

It is impossible, however, to compute the fluent simply by the addition of the terms of this series, on account of the slowness of its convergency.

159. As the curvilinear area $DP\rho d$ is the limit of the sum of the parallelograms $DP', D'P'', \dots$ which constitute the rectilinear figure inscribed in that area; as also the limit of the sum of the parallelograms $DE, D'E, \dots$ which constitute the circumscribed figure, the number of parts into which $Dd$ is divided being in either case conceived to be increased indefinitely, so that each may be continually diminished (and this being the case it is of no consequence whether those parts be conceived as all equal or as unequal) so from analogy we may conclude that if $x', x'', x''', \dots x^{(n)}$ be put to denote successive values of a variable quantity $x$ , and $X', X'', X''', \dots X^{(n)}$ the corresponding values of $X$ any function of $x$ , the limit of the sum of the products $X(x' - x), X'(x'' - x'), X''(x''' - x''), \dots$ when the number of successive values of $x$ and $X$ is continually increased, so that the difference between any two of them immediately following each other may be continually diminished, is equal to $\int X x$ , the fluent to be taken between the two extreme values of $x$ .

160. It was in this manner that the first writers on the differential calculus conceived a fluent; as the difference between any two of its succeeding values is the product of the function $X$ by $x' - x$ the increment of $x$ , they called that product (when $x' - x$ was conceived to be infinitely diminished) the Differential of the fluent; and as the fluent is the sum of all the products, or differentials, instead of calling it a fluent they called it an Integral; the process by which an integral is found from its differential or fluxion they called Integration. The terms Integral and Integration are sometimes employed by writers in applying the method of fluxions to mathematical enquiries.

To find the Lengths of Curves.

161. It has been shewn in § 63, that if the abscissa of a curve be denoted by $x$ , the ordinate by $y$ , and the curve line by $z$ , then $z = \sqrt{(x^2 + y^2)}$ , hence the general formula for finding the length of a curve is

z = \int \sqrt{(x^2 + y^2)}.

Therefore, if from the equation expressing the nature of a curve we find the value of $y$ in terms of $x$ and $x$ , or else the value of $x$ in terms of $y$ and $y$ , and substitute the one or the other in the above general formula, we shall obtain a fluxion the fluent of which will be the length of the curve.

Ex. 1. Suppose the curve to be a parabola, and that $AB = x$ , $BP = y$ , the arch $AP = z$ , the parameter $= a$ , then the equation of the curve being $ax = y^2$ , we have

z = \frac{2y^2}{a}, \text{ and}

z = \int \sqrt{(x^2 + y^2)} = \int \sqrt{\left(\frac{4y^2}{a^2} + y^2\right)} \\ = \frac{1}{a} \int y \sqrt{(a^2 + 4y^2)}.

By comparing this expression with formula (B) in § 131, we find

\int y \sqrt{(a^2 + 4y^2)} = \frac{1}{2} y \sqrt{(a^2 + 4y^2)} \\ + \frac{1}{2} a^2 \int \frac{y}{\sqrt{(a^2 + 4y^2)}};

again by § 127,

\int \frac{y}{\sqrt{(a^2 + 4y^2)}} = \frac{1}{2} \left\{ 2y + \sqrt{(a^2 + 4y^2)} \right\}

therefore

z = \frac{y \sqrt{(a^2 + 4y^2)}}{2a} \\ + \frac{a}{4} \left\{ 2y + \sqrt{(a^2 + 4y^2)} \right\} + C.

To determine the value of $C$ , we must consider that when $y = 0$ , then $z = 0$ , so that then the general formula will become simply

0 = \frac{a}{4} \left\{ 2y + \sqrt{(a^2 + 4y^2)} \right\} + C, \text{ and hence } C = -\frac{a}{4} \left\{ 2y + \sqrt{(a^2 + 4y^2)} \right\},

therefore, substituting the value of $C$ , and bringing together the logarithmic quantities,

z = \frac{y \sqrt{(a^2 + 4y^2)}}{2a} \\ + \frac{a}{4} \left\{ 2y + \sqrt{(a^2 + 4y^2)} \right\}.

Ex. 2. Suppose the curve to be a circle, and that $C$ Fig. 20. is its centre, and $AE$ a quadrant of the circle. Put $CB = x$ , $BP = y$ , the arch $EP = z$ , the radius of the circle $= a$ , then $x^2 + y^2 = a^2$ , and $y = \sqrt{(a^2 - x^2)}$ , and $\dot{y} = \frac{-x \dot{x}}{\sqrt{(a^2 - x^2)}}$ , hence

z = \int \sqrt{(x^2 + y^2)} = \int \sqrt{\left(x^2 + \frac{x^2 \dot{x}^2}{a^2 - x^2}\right)} \\ = \int \frac{a \dot{x}}{\sqrt{(a^2 - x^2)}}.

This fluent can only be expressed by an infinite series, under which form it has been already exhibited in § 140, the radius being there supposed unity.

Ex. 3. Let the curve be an ellipse, and let it be required to find the length of the curve between $E$ the vertex of the lesser axis, and $P$ any point in the curve. To simplify the calculation, let us suppose that the semi-transverse axis $AC = 1$ ; put the semi-conjugate axis $CE = b$ , the eccentricity (that is $\sqrt{(1 - b^2)} = e$ ), the abscissa $CB = x$ , the ordinate $PB = y$ , the arch $EP = z$ ; then, the equation of the curve being $y^2 = b^2(1 - x^2)$ , we have

Inverse Method. have $y = b\sqrt{1-x^2}$ , and $\dot{y} = \frac{-bx\dot{x}}{\sqrt{1-x^2}}$ , and therefore

\begin{aligned} \dot{z} &= \int \sqrt{\dot{x}^2 + \dot{y}^2} = \int \sqrt{\dot{x}^2 + \frac{b^2 x^2 \dot{x}^2}{1-x^2}} \\ &= \int \frac{\dot{x} \sqrt{1-x^2}}{\sqrt{1-x^2}} \end{aligned}

This fluent can only be expressed by means of an infinite series, and it has been already given in this form in § 141.

If we take $x=1$ , then all the quantities in that series which are multiplied by $\sqrt{1-x^2}$ will vanish, but in this particular case $z$ is the elliptic quadrant $EA$ , and $A$ is a quadrant of the circumscribing circle, or $\frac{1}{4}\pi$ , therefore the elliptic quadrant is equal to

\frac{1}{4}\pi \left( 1 - \frac{1}{2.2} e^2 - \frac{1.1.3}{2.2.4.4} e^4 - \frac{1.1.3.3.5}{2.2.4.4.6.6} e^6 - \dots \right)

This series converges very fast if $e$ be a small fraction.

Fig. 11. Ex. 4. Suppose the curve to be a cycloid. Let a circle be described on its axis meeting the ordinate $PB$ in $Q$ , and draw $CQ$ to the centre of the circle. Put $AB=x$ , $BP=y$ , the cycloidal arch $AP=\dot{z}$ , the radius $AC=a$ , the angle $ACQ=v$ , then $AB=a(1-\cos v)$ , $BQ=a \sin v$ , the circ. arch $AQ=a v$ , so that $x=a(1-\cos v)$ , and from the nature of the curve $y=a(v+\sin v)$ , therefore (§ 59.)

\begin{aligned} \dot{x} &= a \dot{v} \sin v, \quad \dot{y} = a \dot{v} (1 + \cos v), \\ \dot{x}^2 + \dot{y}^2 &= a^2 \dot{v}^2 \{ \sin^2 v + (1 + \cos v)^2 \} \\ &= a^2 \dot{v}^2 (2 + 2 \cos v); \end{aligned}

but $2+2 \cos v = 4 \cos^2 \frac{1}{2} v$ (ALGEBRA, § 356.) therefore

\begin{aligned} \dot{z} &= \int \sqrt{\dot{x}^2 + \dot{y}^2} = 2a \int \dot{v} \cos \frac{1}{2} v \\ &= 4a \sin \frac{1}{2} v + C, \quad (\text{§ 145.}) \end{aligned}

but as when $v=0$ , then $z=0$ , therefore $C=0$ , and $z=4a \sin \frac{1}{2} v$ ; but if the chord $AQ$ be drawn, $2a \sin \frac{1}{2} v =$ chord $AQ$ , therefore $z=2$ chord $AQ$ .

Fig. 30. 162. The formula $\dot{z} = \int \sqrt{\dot{x}^2 + \dot{y}^2}$ not being applicable in its present form to curves of the spiral kind, we shall here investigate another suited to that particular class of curves.

Let $APR$ be a curve of such a nature that the position of any point $P$ in the curve is determined by $PF$ , its distance from a given point $F$ , and by the angle which $PF$ makes with $AF$ a line given in position. We shall employ the same construction and notation here as in § 154, with the addition of drawing the chords $DD'$ , $PQ$ , $PP'$ , and putting the arch $AP=\dot{z}$ ; then it is manifest that the simultaneous increments of $v$ , $z$ , and $r$ will be the arches $DD'$ , $PP'$ , and the straight line $PQ$ respectively. Hence

\frac{\dot{z}}{\dot{v}} = \text{limit of } \frac{\text{arch } PP'}{\text{arch } DD'}

but it is evident from § 62, that the limiting ratio of those arches must be the same as that of their chords, therefore,

\frac{\dot{z}}{\dot{v}} = \text{limit of } \frac{\text{chord } PP'}{\text{chord } DD'}

now the limit of the angle $PQD$ being evidently a right angle, we have

\begin{aligned} \frac{PP'}{DD'} &= \text{lim.} \frac{\sqrt{(PQ^2 + QD^2)}}{DD'} \\ &= \text{lim.} \sqrt{\frac{PQ^2}{DD^2} + \frac{PQ^2}{DD^2}} \end{aligned}

but $\frac{PQ^2}{DD^2} = \frac{FP^2}{FD^2} = r^2$ , and $\text{lim.} \frac{PQ^2}{DD^2} = \frac{r^2}{v^2}$ therefore

\frac{\dot{z}}{\dot{v}} = \sqrt{\left( r^2 + \frac{r^2}{v^2} \right)}, \quad \text{and} \quad \dot{z} = \sqrt{\left( r^2 \dot{v}^2 + r^2 \right)}.

Let us apply this formula to the spiral of Archimedes, Fig. 27, the equation of which (§ 155.) is $amv = nr$ , and therefore

v = \frac{nr}{am}, \quad \text{and} \quad \dot{v}^2 = \frac{n^2 r^2}{a^2 m^2}; \quad \text{hence}

\dot{z} = \int \sqrt{\dot{v}^2 + r^2 \dot{v}^2} = \frac{1}{a} \int r \sqrt{a^2 + \frac{n^2}{m^2} r^2}.

This fluent may be found by formula B, § 131, and it is worthy of remark that the fluxion has the same form as that which we have found in § 161 for an arch of a parabola; thus the length of any portion of the spiral of Archimedes may be exhibited by means of an arch of a parabola.

To find the Contents of Solids.

163. If $AD$ the abscissa of a curve be denoted by $x$ , Fig. 44 and $PD$ the ordinate by $y$ , and the solid generated by the revolution of the curve $APD$ about $AD$ as an axis by $s$ , it has been shewn, in § 64, that $s = \pi y^2 \dot{x}$ , therefore the general formula for finding the content of a solid is

s = \pi \int y^2 \dot{x}.

Ex. 1. Suppose the solid to be a paraboloid, or that which is generated by the revolution of a parabola about its axis; in this case $y^2 = ax$ , and taking the fluent so that when $x=0$ , then $s=0$ ,

s = \pi \int y^2 \dot{x} = \pi a \int x \dot{x} = \frac{1}{2} \pi a x^2,

or $s = \frac{1}{2} \pi x y^2$ ; but $\pi x y^2$ is the content of a cylinder having $y$ for the radius of its base and $x$ for its altitude, therefore the content of a paraboloid is half that of a cylinder having the same base and altitude.

Ex. 2. Suppose the solid to be a parabolic spindle, which is generated by the revolution of $APB$ an arch of a parabola about $AC$ an ordinate to its axis. In this case let $AD=x$ , $DP=y$ , $AB=b$ , the parameter of the

Inverse Method. axis = $a$ , then from the nature of the parabola $AD \times DB$
$= a \times PD$ , that is $x(b-x) = ay$ , hence $y = \frac{(b-x)x}{a}$ , and
taking the fluent, so that $s$ and $x$ may vanish together;

\begin{aligned} s &= \pi \int y^2 \dot{x} = \frac{\pi}{a^2} \int x^2 (b-x)^2 \dot{x} \\ &= \frac{\pi}{a^2} \int (b^2 x^2 \dot{x} - 2bx^3 \dot{x} + x^4 \dot{x}) \\ &= \frac{\pi}{a^2} \left( \frac{b^2 x^3}{3} - \frac{bx^4}{2} + \frac{x^5}{5} \right) \end{aligned}

or, since $a^2 = \frac{(b-x)^2 x^2}{y^2}$ ,

s = \frac{\pi y^2}{(b-x)^2} \left\{ \frac{b^2 x^3}{3} - \frac{bx^4}{2} + \frac{x^5}{5} \right\},

which expression (by supposing $x = AC = \frac{1}{2}b$ , and putting $d$ for $CE$ , the greatest value of $y$ ) gives $\frac{4\pi d^3 b}{15}$ for the content of half the solid generated by the curve $AEB$ , therefore the entire spindle is $\frac{8\pi a^3 b}{15}$ , or (by observing that $\pi d^2 b$ is the content of a cylinder having $d$ for the radius of its base and $b$ for its length) it is $\frac{1}{3}$ of the circumscribing cylinder.

Ex. 3. Suppose the solid to be a spheroid produced by the revolution of an ellipse about either of its axes; put $a$ for $\frac{1}{2}AB$ the axis round which the curve revolves, $b$ for $\frac{1}{2}EF$ the other axis, $x$ for $AD$ the height of any segment made by a plane perpendicular to the axis of the solid, $y$ for $PD$ the radius of its base, and $s$ for its content. Then from the nature of the curve $y^2 = \frac{b^2}{a^2} (2ax - x^2)$ , therefore taking the fluent upon the supposition that $s$ and $x$ vanish together,

\begin{aligned} s &= \pi \int y^2 \dot{x} = \frac{\pi b^2}{a^2} \int (2ax - x^2) \dot{x} \\ &= \frac{\pi b^2}{a^2} \left( ax^2 - \frac{x^3}{3} \right). \end{aligned}

To find the content of the whole spheroid, we have only to take $x = 2a$ , thus the formula becomes $s = \frac{4\pi b^2 a}{3}$ , and as $2\pi b^2 a$ expresses the content of a cylinder having $2b$ for the diameter of its base, and $2a$ for its height, it follows that the contents of a spheroid is $\frac{2}{3}$ that of its circumscribing cylinder.

It is obvious that what has been found for the spheroid will apply also to the sphere, by supposing the axes equal, or $a = b$ .

164. If instead of supposing the solid $APQ\rho$ to be formed by the revolution of a curve round its axis (in which case it is called a solid of revolution) we had supposed it to have any figure whatever, then by referring the solid to some straight line $AC$ , given by position, as an axis, and in which $A$ is a given point, and supposing $PQ\rho$ to be a section of the solid made by a plane

perpendicular to that axis, meeting it in $D$ , and putting $AD = x$ , and the variable solid $APQ\rho$ (considered as a function of $x$ ) $= s$ , by proceeding as in § 64, we

would have found the limit of $\frac{\text{increment of } s}{\text{increment of } x}$ , and con-

sequently $\frac{s}{x}$ , equal to the area of the section of the solid

made by the plane $PQ\rho$ ; therefore putting $V$ for that function of $x$ which expresses the area of the section, we have $s = Vx$ , and $s = \int V \dot{x}$ .

Let us suppose for example that $AEFG$ is a solid Fig. 34. bounded by any plane figure $EFG$ as a base, and by the surface which will be generated if we suppose a straight line drawn from $A$ any given point above that plane to revolve in the circumference of the base.

Let $AC$ be a perpendicular drawn from the vertex of the figure to its base, and let $PQ\rho$ be a section of the solid by a plane parallel to the base, meeting the perpendicular in $D$ . Put $a$ the area of the base of the solid, $V$ the area of the section $PQ\rho$ , $b = AC$ the altitude of the whole solid, $x = AD$ the altitude of the part cut off by the plane $PQ\rho$ , and $s$ the content of that part; then, as from the nature of the solid it is pretty evident that the part of it cut off by the plane $PQ\rho$ is similar to the whole, and as the bases of similar solids are as the squares of their altitudes, we have

a : V :: b^2 : x^2, \text{ hence } V = \frac{ax^2}{b^2}, \text{ and}

s = \int V \dot{x} = \frac{a}{b^2} \int x^2 \dot{x} = \frac{ax^3}{3b^2},

this expression for $s$ does not require the addition of any constant quantity, for by putting $x = 0$ , we have $s = 0$ as it ought to be. Suppose now $x = b$ , then $s = \frac{ab^3}{3b^2} = \frac{1}{3}ab$ , from which it appears that the content of the whole solid is $\frac{1}{3}$ of the product of the base by the perpendicular. It is evident that pyramids and cones are solids of the kind we have been considering.

To find the Surfaces of Solids.

165. The altitude $AD$ of a solid, generated by re- Fig. 5. volution of a curve about $AD$ as an axis, being as before denoted by $x$ , and $PD$ the radius of its base by $y$ , let us now put $s$ to denote the curved surface of the solid, then, as it has been shewn, § 65, that $s = 2\pi y \sqrt{(x^2 + y^2)}$ , we have

s = 2\pi \int y \sqrt{(x^2 + y^2)} \dot{x}

as a general formula for the surface of a solid.

Ex. 1. Suppose the solid to be a sphere, generated by Fig. 35. the revolution of a circle about its diameter $AB$ , put the radius of the sphere $= a$ , then $AD$ being denoted by $x$ , and $PD$ by $y$ , we have from the nature of the curve $y^2 = 2ax - x^2$ , therefore

y = \sqrt{(2ax - x^2)}, \text{ and } \dot{y} = \frac{(a-x)\dot{x}}{\sqrt{(2ax - x^2)}}, \text{ and}

$x^2$

\dot{x}^2 + \dot{y}^2 = \dot{x}^2 \left( 1 + \frac{(a-x)^2}{2ax-x^2} \right) \\ = \frac{a^2 \dot{x}^2}{2ax-x^2} = \frac{a^2 \dot{x}^2}{y^2},

therefore, $y\sqrt{(\dot{x}^2 + \dot{y}^2)} = a\dot{x}$ , and taking the fluent, so that when $x=0$ , then $s=0$ ,

s = 2\pi \int y\sqrt{(\dot{x}^2 + \dot{y}^2)} = 2\pi a x s

now if it be considered that $2\pi a$ is the circumference of a great circle of the sphere, it will immediately appear that the surface of a segment of a sphere is equal to the circumference of a great circle of the sphere multiplied into the height of the segment. Hence it follows that the whole surface of the sphere is equal to four times the area of one of its great circles.

Fig. 5. Ex. 2. Suppose the curve to be a parabola, then putting $AD = x$ , $DP = y$ , the parameter of the axis $= a$ , we have (§ 161. example 1.)

\sqrt{(\dot{x}^2 + \dot{y}^2)} = \frac{1}{a} y \sqrt{(a^2 + 4y^2)}, \text{ therefore}

s = 2\pi \int y \sqrt{(\dot{x}^2 + \dot{y}^2)} \\ = \frac{2\pi}{a} \int y y \sqrt{(a^2 + 4y^2)} \\ = \frac{\pi}{6a} (a^2 + 4y^2)^{\frac{3}{2}} + C, \text{ by } \S 108.

To discover the value of the constant quantity $C$ , we must observe that when $x=0$ , then $y=0$ , and $s=0$ , therefore putting $0$ instead of $s$ and $y$ , the above equation becomes $0 = \frac{\pi a^2}{6} + C$ , hence $C = -\frac{\pi a^2}{6}$ , and

s = \pi \left\{ \frac{(a^2 + 4y^2)^{\frac{3}{2}}}{6a} - \frac{a^2}{6} \right\}.

To find the Centre of Gravity of any Line, Surface, or Solid.

166. It belongs to the theory of MECHANICS to explain what is meant by the centre of gravity, and to demonstrate its general properties, and here it is only necessary to show how the method of fluxions may be applied to deduce from some one of those properties rules for finding that centre in any proposed case.

The properties of centres of gravity which we shall employ as the foundation of the application of the method of fluxions to its determination may be enunciated shortly thus.

Let $C$ be the centre of gravity of a mass of matter denoted by $M$ , and $c$ the centre of gravity of another mass $m$ , and $D$ the centre of gravity of the two masses $M$ and $m$ , from these points let perpendiculars $CA$ , $ca$ , $DE$ be drawn to any straight line $PQ$ , then

M \times CA + m \times ca = (M+m) \times DE.

165. Let us now suppose that $AP$ is any curve line

(having weight), of which the centre of gravity is required, and that $PB$ , $PD$ are co-ordinates drawn from any point in the curve perpendicular to $AB$ , $AD$ two axes at right angles to each other; let the arch $AP$ receive any increment $Pp$ , let $C$ be the centre of gravity of $AP$ , $G$ the centre of gravity of $Pp$ , and $C'$ the centre of gravity of $APp$ . From $C$ and $G$ draw $CE$ , $CF$ , $GH$ , $GK$ perpendicular to the axes $AB$ , $AD$ . Put $PD=x$ , $PB=y$ , $CF=X$ , $CE=Y$ , $AP=\infty$ , also let the arch $APp=\infty'$ , and let the distances of $C'$ its centre of gravity from the axes $AD$ , $AB$ be denoted by $X'$ , and $Y'$ respectively; then, observing that the arch $Pp=\infty'-\infty$ , by the proposition in last §,

\infty X + (\infty' - \infty) \times GK = \infty' X';

\text{hence } \frac{\infty' X' - \infty X}{\infty' - \infty} = GK.

If we now suppose the arch $Pp$ to be continually diminished, and observe that $\infty' X' - \infty X$ , and $\infty' - \infty$ , are the simultaneous increments of $\infty X$ and $\infty$ , it will appear (§ 23.) that

\frac{\text{flux. of } (\infty X)}{\infty} = \text{limit of } GK.

By the very same way of reasoning we find

\frac{\text{flux. of } (\infty Y)}{\infty} = \text{limit of } GH,

but the point $p$ approaching to $P$ , it is manifest that the point $G$ will also approach to $P$ , so that the limit of $GK$ is $PD$ or $x$ , and the limit of $GH$ is $PB$ or $y$ , hence

\frac{\text{flux. } (\infty X)}{\infty} = x, \quad \frac{\text{flux. } (\infty Y)}{\infty} = y,

\text{flux. } (\infty X) = \infty x, \quad \text{flux. } (\infty Y) = \infty y.

Taking now the fluents of each side of these equations, and dividing by $\infty$ ,

X = \frac{\int x \dot{x}}{\infty}, \quad Y = \frac{\int y \dot{y}}{\infty};

it is evident that by these two equations the position of $C$ the centre of gravity is determined.

168. Let us next suppose that it is required to find $C$ Fig. 37, the centre of gravity of the plane area $APB$ . As the arch $AP$ was in last § supposed to receive the increment $Pp$ , so let the area $APB$ now receive the increment $BPpb$ , and let $C$ , $C'$ and $G$ (which in the former case were supposed to be the centres of gravity of the arches $AP$ , $APp$ , and $Pp$ respectively) now be supposed to be the centres of gravity of the areas $APB$ , $APpb$ , and $BPpb$ ; put the area $APB=s$ , the area $APpb=s'$ , and let $X$ , $Y$ , $X'$ , $Y'$ denote as before. Then, reasoning exactly as in last case, we have (by § 116.),

s X + (s' - s) \times GK = s' X' \\ s Y + (s' - s) \times GH = s' Y'

hence

\frac{s'X' - sX}{s' - s} = GK, \quad \frac{s'Y' - sY}{s' - s} = GH;

and the point $p$ being supposed to approach to $P$ , so that $s'X' - sX$ , $s'Y' - sY$ , and $s' - s$ , the simultaneous increments of $X$ , $Y$ and $s$ , may be continually diminished,

\frac{\text{flux. } (sX)}{s} = \lim. GK, \quad \frac{\text{flux. } (sY)}{s} = \lim. GH;

but as the ordinate $pb$ approaches to $PB$ , it is manifest that the ultimate position of $G$ , the centre of gravity of the area $BPpb$ , will be the middle of $PB$ , therefore the limit of $GK$ is $x$ , and the limit of $GH$ is $\frac{1}{2}y$ , thus we have

\frac{\text{flux. } (sX)}{s} = x, \quad \frac{\text{flux. } (sY)}{s} = \frac{1}{2}y,

and consequently,

X = \frac{\int x s}{s}, \quad Y = \frac{\int y s}{2s}

or since $s = yx$ , and $s = \int y \dot{x}$ (§ 61.),

X = \frac{\int y x \dot{x}}{\int y \dot{x}}, \quad Y = \frac{\int y^2 \dot{x}}{2 \int y \dot{x}}.

169. Let it next be required to find the centre of gravity of the surface of a solid generated by the revolution of the curve $AP$ about $AB$ as an axis. Let the surface of the solid be conceived to receive an increment generated by $Pp$ an arch of the curve. In this case it is evident that the centres of gravity of the surface generated by the curve $AP$ , the surface generated by the curve $APp$ , and the surface generated by the arch $Pp$ , will each be in $AB$ , the axis of the solid; suppose them to be at $E$ , $E'$ and $H$ respectively. Put $AE = X$ , $AE' = X'$ , also put $s$ for the surface generated by $AP$ , and $s'$ for the surface generated by $APp$ , then as before (from § 166.) we have

sX + (s' - s)AH = s'X',

\text{hence } \frac{s'X' - sX}{s' - s} = AH,

\text{and } \frac{\text{flux. } (sX)}{s} = \lim. AH;

but the point $p$ approaching to $P$ , the limit of $AH$ is manifestly $AB$ or $x$ , therefore

\frac{\text{flux. } (sX)}{s} = x, \quad \text{and } X = \frac{\int x s}{s}

or since $s = 2\pi yx$ (§ 65.),

X = \frac{2\pi \int x y \dot{x}}{2\pi \int y \dot{x}} = \frac{\int x y \dot{x}}{\int y \dot{x}}.

170. If instead of the centre of gravity of the surface generated by $AP$ , the centre of gravity of the solid generated by the revolution of the plane figure $APB$ about $AB$ as an axis be required, the reasoning will be the very same as in last §, substituting the solid generated by the plane figure instead of the surface generated by the curve line; so that putting $s$ for the content of the solid, and $X$ for $AE$ the distance of its centre of gravity from the vertex, we have also

X = \frac{\int x s}{s},

but in this case $s = \pi y^2 x$ (§ 64.), therefore

X = \frac{\pi \int y^2 x \dot{x}}{\pi \int y^2 x} = \frac{\int y^2 x \dot{x}}{\int y^2 x}.

171. We shall now apply these formulas to some examples.

Example 1. Let it be required to find the centre of gravity of $AP$ an arch of a circle. Suppose $AB$ to be a part of the diameter, and in addition to the notation of § 167. put $a$ for the radius of the circle, then from the nature of the curve, $y^2 = 2ax - x^2$ , hence

(proceeding as in § 165. Ex. 1.) we have $\dot{x} = \frac{ax}{y}$ and

therefore $\dot{y} = a \dot{x}$ , and $\dot{x} = \frac{ax \dot{x}}{y}$ , but from the equation $y^2 = 2ax - x^2$ , by taking the fluxions we get $2y \dot{y} = 2a \dot{x} - 2x \dot{x}$ , and hence $\frac{ax \dot{x}}{y} = \frac{a \dot{x}}{y} - \dot{x} = \dot{x} - \dot{x}$ ,

therefore $\dot{x} = a(\dot{x} - \dot{x})$ ; substituting now the values of $\dot{x}$ and $\dot{y}$ in the formula of § 167, we have

X = \frac{\int x \dot{x}}{x} = \frac{a}{x} \int (\dot{x} - \dot{x})

= \frac{a}{x} (x - y + c)

Y = \frac{\int y \dot{x}}{x} = \frac{a}{x} \int \dot{x}

= \frac{a}{x} (x + c').

To discover the values of the constant quantities $c$ , $c'$ , we have from the equations in which they occur,

a c = Xx - ax + ay, \quad ac' = Yx - ax;

but when $x = 0$ , then $x$ , $y$ , $X$ and $Y$ are each $= 0$ , therefore $c = 0$ , and $c' = 0$ , thus we have simply

X = \frac{a(x - y)}{x}, \quad Y = \frac{ax}{x}.

Ex. 2.

Inverse Method.
Ex. 2. Let it be required to find the centre of gravity of APB an area bounded by AP an arch of a circle and PB, BA its fine and versed fine. Let $a$ denote the radius, and let the remaining notation be as in § 168. Then, because $\dot{s} = y \dot{x}$ , we have $x \dot{s} = y x \dot{x}$ , but from the equation $y^2 = 2ax - x^2$ (which expresses the nature of the curve), we find

x \dot{x} = a \dot{x} - y \dot{y}, \text{ therefore}

x \dot{s} = ay \dot{x} - y^2 \dot{y} = a \dot{s} - y^2 \dot{y}.

We have also $y \dot{s} = y^2 \dot{x} = (2ax - x^2) \dot{x}$ , therefore,

X = \frac{\int x \dot{s}}{s} = \frac{\int (a \dot{s} - y^2 \dot{y})}{s}

= \frac{1}{s} (a s - \frac{1}{2} y^2 + c)

Y = \frac{\int y \dot{s}}{2s} = \frac{\int (2ax - x^2) \dot{x}}{2s}

= \frac{1}{2s} (a x^2 - \frac{1}{2} x^3 + c').

By proceeding as in the last example we find $c$ and $c'$ each $=0$ , thus we have

X = a - \frac{y^2}{3s}, \quad Y = \frac{3ax^2 - x^3}{6s}.

Fig. 37. Ex. 3. Suppose now the figure to be the surface generated by the revolution of AP an arch of a circle about the diameter AB, and that the centre of gravity of the generated surface is required. Then because from the nature of the circle $\dot{z} = \frac{ax}{y}$ we have $y \dot{z} = ax \dot{x}$ , and $x y \dot{z} = ax^2 \dot{x}$ , therefore, substituting these values in the formula of § 169. it becomes

X = \frac{\int x y \dot{z}}{\int y \dot{z}} = \frac{\int x^2 \dot{x}}{a \int \dot{x}}

= \frac{\frac{1}{3} x^3 + c}{x + c'}.

To find the values of the constant quantities $c, c'$ , we have

c = X(x + c') - \frac{1}{3} x^3,

c' = \frac{\frac{1}{3} x^3 + c}{X} - x,

but as when $x=0$ , then $X=0$ , it is manifest that $c$ and $c'$ are each $=0$ , thus we have

X = \frac{1}{3} x^2.

Fig. 37. Ex. 4. Let us now suppose that it is required to find the centre of gravity of the solid generated by the revolution of AP an arch of a circle about the diameter. In this case, because $y^2 = 2ax - x^2$ , we have from § 170,

X = \frac{\int y^2 x \dot{x}}{\int y^2 x} = \frac{\int x (2ax - x^2) \dot{x}}{\int x (2ax - x^2)}

= \frac{\frac{2}{3} ax^3 - \frac{1}{4} x^4 + c}{ax^2 - \frac{1}{3} x^3 + c'}

and reasoning as in the last example, we find $c=0$ , and $c'=0$ , and therefore

X = \frac{8ax - 3x^3}{12a - 4x}.

If the segment be a hemisphere, in which case $x=a$ , then $X = \frac{1}{2}a$ .

SECT. III. Of Fluxional Equations.

172. It has been shewn (§ 49.), how, from an equation being given, expressing the relation between $x$ a variable quantity, and $y$ a function of that quantity, we may deduce the equation that expresses the relation of their fluxions. We are now to show how from the latter, or fluxional equation, we may return to the equation of the fluents, which, relatively to the other, may be called its primitive equation.

173. As any primitive equation and the fluxional equation derived from it both hold true at the same time, and as the constant quantities which enter into the former retain the same values in the latter, it follows that by means of the two equations we may exterminate any one of the constant quantities, and thus from any proposed primitive equation deduce a fluxional equation, in which one of the constant quantities contained in that primitive shall not at all be found.

For example, let the primitive equation be $y + ax + b = 0$ , by taking the fluxions we have $\dot{y} + a \dot{x} = 0$ , a fluxional equation in which $b$ is not found; if, however, it be required to find an equation in which $a$ shall be wanting, we have only to eliminate $a$ by applying the common rules of Algebra (ALGEBRA, Sect. vii.) to the two equations

y + ax + b = 0, \quad \dot{y} + a \dot{x} = 0;

and hence we have $y \dot{x} - x \dot{y} + b \dot{x} = 0$ , thus it appears that from the primitive equation $y + ax + b = 0$ we may deduce a fluxional equation which may be expressed under either of these forms,

\dot{y} + a \dot{x} = 0, \quad y \dot{x} - x \dot{y} + b \dot{x} = 0;

these hold true at the same time as the primitive equation, they are alike related to, and any two of the three being given the other necessarily follows from them.

As a second example, suppose the primitive equation to be $x^2 - 2ay - a^2 - b = 0$ , by passing to the fluxions we immediately find $x \dot{x} - a \dot{y} = 0$ , an equation in which $b$ is not found. If, however, it be required that the fluxional equation shall want $a$ , we have only to apply the common rules of elimination to the two equations;

thus from the second we get $a = \frac{x \dot{x}}{\dot{y}}$ , and this being substituted in the first it becomes

x^2 - \frac{2xy\dot{x}}{y} - \frac{x^2x^2}{y^2} - b = 0

from which we have

(x^2 - b) \dot{y}^2 - 2xy\dot{y}\dot{x} - x^2\dot{x}^2 = 0

and taking the square root, having previously reduced the equation to a proper form,

\dot{y}\sqrt{(x^2 + y^2 - b) - y\dot{y} - x\dot{x}} = 0.

174. It is evident that by proceeding in this manner we shall, in some cases, arrive at a fluxional equation involving the second and higher powers of $\frac{\dot{y}}{x}$ , and when

this happens we can only find the value of $\frac{\dot{y}}{x}$ by the resolution of an equation; but this may be avoided by preparing the primitive equation in such a manner, that the constant quantity to be eliminated may be entirely separated from the variable quantities, so as to form one of the terms of the equation, then, upon taking the fluxions, this term being constant will vanish, and thus we shall obtain an equation entirely free from the constant quantity contained in that term. Thus the primitive equation $y + ax + b = 0$ has already such a form that by taking the fluxions we get $\dot{y} + a\dot{x} = 0$ an equation in which $b$ is not found. If it be required, that upon taking the fluxions, $a$ shall vanish; we must put the equation under this form $\frac{y+b}{x} + a = 0$ , and then taking the fluxion, we find immediately

\frac{x\dot{y} - (y+b)\dot{x}}{x^2} = 0,

an expression in which $a$ is not found, and which by rejecting the divisor $x^2$ becomes $y\dot{x} - x\dot{y} + b\dot{x} = 0$ , and these two forms of the fluxional equation are the very same as have been found in the last §. In the second example, viz. $x^2 - 2ay - a^2 - b = 0$ , the equation has already the form suited to the elimination of $b$ , for the fluxional equation is $x\dot{x} - a\dot{y} = 0$ , but in order that $a$ may vanish, we must resolve the equation with respect to $a$ , so as to give it this form

y - \sqrt{(x^2 + y^2 - b)} + a = 0;

passing now to the fluxional equation, $a$ disappears, and we have

\frac{\dot{y}\sqrt{(x^2 + y^2 - b)} - y\dot{y} - x\dot{x}}{\sqrt{(x^2 + y^2 - b)}} = c.

It is evident that we have only to reject the denominator to give the equation this form

\dot{y}\sqrt{(x^2 + y^2 - b)} - y\dot{y} - x\dot{x} = 0

the same as was found in the conclusion of last §.

175. From what has been now shewn we may infer that as from any proposed primitive equation we can deduce a fluxional equation that shall contain one constant quantity less than the primitive contains, so on the

contrary any fluxional equation being given, its primitive equation may contain one constant quantity more than the fluxional equation, but it can contain only one, for no more than one constant quantity can be made to disappear by returning from the primitive to its fluxional equation.

176. The fluxional equation expressing the value of $\frac{\dot{y}}{x}$ , which is derived from any primitive equation involving $x$ , and $y$ a function of $x$ , may be called a fluxional equation of the first order; and as from this equation considered as a primitive, we may in like manner

derive an equation that shall involve $\frac{\dot{y}}{x^2}$ (§ 50.), this last

may be called a fluxional equation of the second order, and the fluxional equation from which it is derived may be called its primitive equation of the first order, to distinguish it from the absolute primitive equation, from which all the others are conceived to be derived. A similar mode of definition is to be applied to the higher orders.

177. As any primitive equation and the fluxional equations of the first and second orders derived from it must all hold true at the same time, it is evident, that by means of the three equations, we may exterminate any two of the constant quantities contained in them that we please, and thus produce a fluxional equation of the second order that contains two constant quantities less than the primitive equation. There are however two other ways by which we may arrive at the very same fluxional equation of the second order. For as from the given primitive equation we may deduce two different fluxional equations of the first order, one of which shall contain one only of the two quantities to be eliminated, and the other shall contain the other quantity only, we may consider each of these equations in its turn as a primitive, and, by proceeding in the manner explained in § 173 and § 174, derive from it a fluxional equation, in which that particular constant quantity which remained in its primitive, but which was to be finally eliminated, shall not be found; thus, from each of these primitives we shall deduce the very same fluxional equation of the second order, that shall be freed from two of the constant quantities contained in the absolute primitive equation.

Let us take for example the equation

x^2 - 2ay + b^2 = 0;

by proceeding as explained in § 173, or § 174, we find these two fluxional equations of the first order,

x\dot{x} - a\dot{y} = 0, \quad (x^2 + b^2)\dot{y} - 2xy\dot{x} = 0,

in the one of these the constant quantity $a$ is wanting, and in the other $b$ is wanting. Taking the first equation $x\dot{x} - a\dot{y} = 0$ , and proceeding as in § 50 (observing that $x$ is constant) we find $\dot{x}^2 - a\dot{y} = 0$ , if from this equation we now eliminate $a$ by putting instead of it $\frac{x\dot{x}}{y}$

(deduced from the equation $x\dot{x} - a\dot{y} = 0$ ) we find after proper reduction

\dot{y}x

\ddot{y}x - \dot{x}\dot{y} = 0,

a fluxional equation of the second order, in which both $a$ and $b$ are wanting, and having $x^2 - 2ay + b^2$ for its absolute primitive equation.

Let us now take the other fluxional equation of the first order which involves $b$ , viz. $(x^2 + b^2)\dot{y} - 2xy\dot{x} = 0$ ; by proceeding with this as with the former we find $(x^2 + b^2)\ddot{y} - 2y\dot{x}\dot{y} = 0$ ; from the first of these equations we find $x^2 + b^2 = \frac{2xy\dot{x}}{\dot{y}}$ , and from the second $x^2 + b^2 = \frac{2y\dot{x}^2}{\dot{y}}$ ; therefore, $\frac{2xy\dot{x}}{\dot{y}} = \frac{2y\dot{x}^2}{\dot{y}}$ , and hence we have

\ddot{y}x - \dot{x}\dot{y} = 0

the same equation as before; and as we have arrived at the very same conclusion by considering each of these equations

\dot{x}\dot{x} - a\dot{y} = 0, \quad (x^2 + b^2)\dot{y} - 2xy\dot{x} = 0

as a primitive, it follows that both these are to be considered as primitive equations of the first order of the fluxional equation $\ddot{y}x - \dot{x}\dot{y} = 0$ .

178. In general, every fluxional equation of the second order has two primitive equations of the first order, and all three may be considered as originating from one and the same absolute primitive equation; and as a fluxional equation of the second order may contain two constant quantities less than its absolute primitive equation, and one less than either of its primitive equations of the first order; so, on the contrary, a primitive equation of the first order may contain one constant quantity more than the fluxional equation of the second order derived from it, and the absolute primitive may contain two constant quantities which are not found in the fluxional equation of the second order derived from it; and similar conclusions may be drawn relating to fluxional equations of the third or any higher order.

Of Fluxional Equations of the first order.

179. When it is required to find the primitive equation corresponding to a proposed fluxional equation of the first order, we may endeavour to separate the variable quantities, that is, to bring the equation to such a form, that it may be composed of two parts, one of which consists of $x$ multiplied or divided by a function of $x$ only, and the other of $y$ multiplied or divided by a function of $y$ only. When this separation of the variable quantities can be effected, we have only to take the fluents according to the methods explained in SECT. I. and put their sum $= 0$ , and we immediately have the primitive equation required.

Ex. Suppose the fluxional equation to be

my\dot{x} + nx\dot{y} = 0;

divide the terms of the equation by $xy$ , and it becomes

\frac{m\dot{x}}{x} + \frac{n\dot{y}}{y} = 0.

VOL. VIII. Part II.

Now the fluent of $\frac{m\dot{x}}{x}$ is $ml.x + c'$ (§ 103.), and in

like manner the fluent of $\frac{n\dot{y}}{y}$ is $nl.y + c''$ , therefore

ml.x + nl.y + c' + c'' = 0,

or, transposing $c' + c''$ , and putting a single constant quantity for their sum, which, to be homogeneous with the logarithmic quantities, may be $-\log. c$ , or $-l. c$ ,

ml.x + nl.y = l. c, \text{ or } l. (x^m) + l. (y^n) = l. c,

\text{or } l. (x^m y^n) = l. c, \text{ and hence } x^m y^n = c,

which last is the primitive equation required.

180. When a primitive equation is homogeneous, that is, when the sum of the exponents of the variable quantities $x$ and $y$ is the same in each term, as in this example,

ax\dot{x} + by\dot{y} + dx\dot{y} + ey\dot{x} = 0,

\text{or } (ax + by)\dot{x} + (dx + ey)\dot{y} = 0,

in which the variable part of each term is of the first degree, as also in this equation

\left. \begin{aligned} (ax^2 + bxy + dy^2)\dot{x} \\ + (ex^2 + fxy + gy^2)\dot{y} \end{aligned} \right\} = 0,

in which the variable part of each term is of the second degree, such an equation may be always transformed into another which will admit of the variable quantities being separated. To take a particular example, let us suppose the equation to be $x\dot{x} + y\dot{y} = nx$ , or $(x - ny)\dot{x} + y\dot{y} = 0$ . We assume $y = x$ (and the same assumption is to be made for any other homogeneous equation), then $\dot{y} = \dot{x}$ , thus the equation becomes transformed to

(x - nx)\dot{x} + x\dot{x} = 0,

but as the terms of this equation have a common factor $x$ , by leaving out that factor, it becomes

(1 - n)\dot{x} + \dot{x} = 0,

which also admits of being expressed thus

(1 - n\dot{x} + \dot{x})\dot{x} = 0,

and, by division,

\frac{\dot{x}}{x} + \frac{\dot{x}}{1 - n\dot{x} + \dot{x}} = 0,

and taking the fluents

\int \frac{\dot{x}}{x} + \int \frac{\dot{x}}{1 - n\dot{x} + \dot{x}} = C,

where $C$ denotes a constant quantity, or, since $\int \frac{\dot{x}}{x} = l. x$ ,

l. x + \int \frac{\dot{x}}{1 - n\dot{x} + \dot{x}} = C,

now

now the particular form of

\int \frac{x^2}{1-nx+x^2}

depends upon

the value of the number

n

; for if

\frac{n}{2} > 1

, it will be a

logarithmic function, and if

\frac{n}{2} < 1

, it will be expres-

sible by means of a circle, but if

\frac{n}{2} = 1

, then it is an al-

gebraic function, and in each case it may be found by
the methods delivered in SECT. I. for finding the fluent
of a rational fraction. It may however be simplified in
its form, by observing, that since

\frac{x^2}{1-nx+x^2} = \frac{2x^2-nx}{1-nx+x^2} + \frac{nx}{1-nx+x^2},

therefore (§ 103.)

\int \frac{x^2}{1-nx+x^2} =

\frac{1}{2} \ln(1-nx+x^2) + \frac{nx}{1-nx+x^2}.

If we limit our enquiry to the case of

n=2

, we have

\int \frac{x^2}{1-2x+x^2} = \int \frac{2x}{(1-x)^2} = \frac{2}{1-x},

let the terms be now collected into one expression, then
observing that

\frac{1}{2} \ln(1-2x+x^2) = \frac{1}{2} \ln(1-x)^2 =

\ln(1-x)

, we have

\ln x + \ln(1-x) + \frac{x}{1-x} = C;

and, substituting

\frac{y}{x}

instead of

x

\ln x + \ln\left(\frac{x-y}{x}\right) + \frac{x}{x-y} = C,

or, substituting

\ln c

instead of

C

, and collecting the lo-

\ln \frac{x-y}{c} = \frac{-x}{x-y},

therefore, passing from logarithms to numbers, by ob-

serving that, as when

a = \ln p

, we have by the nature of
logarithms

e^a = p

, where

e

denotes the number of which
the Napierian logarithm is 1, so in the present case we
have

\frac{x-y}{c} = e^{-\frac{x}{x-y}}

, and hence the primitive equation
is found to be

x-y-c e^{-\frac{x}{x-y}} = 0.

As a second example let the fluxional equation be

xy - yx = x \sqrt{x^2 + y^2}

which is also homogeneous. Assume as before

y = xz

,
then

y = xz + xz

, and substituting these values of

y

and

y

in the proposed equation, it becomes

x \sqrt{1+z^2} - xz = 0,

from which we get

\frac{x}{x} - \frac{z}{\sqrt{1+z^2}} = 0,

and taking the fluents of the terms, observing that each
being a logarithm function, their sum may be put equal
to a constant logarithm,

\ln x - \ln \left\{ z + \sqrt{1+z^2} \right\} = \ln c,

which expression, by substituting for

z

its value

\frac{y}{x}

,
becomes

\ln x = \ln c + \ln \left\{ \frac{y + \sqrt{x^2 + y^2}}{x} \right\}.

If we now consider that

(y + \sqrt{x^2 + y^2})(y - \sqrt{x^2 + y^2}) = -x^2,

and therefore that

\frac{y + \sqrt{x^2 + y^2}}{x} = \frac{-x}{y - \sqrt{x^2 + y^2}},

it will appear, that the above equation may be other-
wise expressed thus:

\ln x = \ln c + \ln \left\{ \frac{-x}{y - \sqrt{x^2 + y^2}} \right\},

from which, by passing from logarithms to their num-
bers, we find

y - \sqrt{x^2 + y^2} = -c

, and hence, by fo-
rdering the equation that the radical may disappear,
we get

x^2 = c^2 + 2cy

, which is the primitive equation
required.

181. An equation which is not homogeneous, may in
some cases, by proper transformations, be rendered ho-
mogeneous; this is the case in particular with the equa-
tion

(a + mx + ny)x + (b + px + qy)y = 0,

which is general of its kind; for this purpose we assume

x = t + \alpha

, and

y = u + \beta

, then

x = t

, and

y = u

; by substi-
tuting these values of

x, y, x, y

, in the proposed equa-
tion it becomes

\left. \begin{aligned} (a + m\alpha + n\beta + mt + nu)t \\ + (b + p\alpha + q\beta + pt + qu)u \end{aligned} \right\} = 0.

Let us now suppose

\alpha

and

\beta

such that

a + m\alpha + n\beta = 0,

b + p\alpha + q\beta = 0,

by these two equations the values of

\alpha

and

\beta

are deter-
mined, and the transformed equation is reduced to

(mt + nu)t + (pt + qu)u = 0,

an equation which is homogeneous, and therefore may
be treated in the manner explained in last §.

This transformation will not apply, however, when

mq

-np = 0

, because then the values of

\alpha

and

\beta

would be
infinite.

Inverse Method. infinite. In this case we have $q = \frac{np}{m}$ , and therefore

$px + qy = \frac{p}{m}(mx + ny)$ , hence the original equation may be expressed thus,

\left. \begin{aligned} a\dot{x} + b\dot{y} \\ + (mx + ny) \left( \dot{x} + \frac{p}{m}\dot{y} \right) \end{aligned} \right\} = 0.

Assume now $mx + ny$ , then $\dot{y} + \frac{\dot{x}}{n} = \frac{m}{n}\dot{x}$ ; the va-

lues of $mx + ny$ and $\dot{y}$ being now substituted in the equation, and the whole reduced to a proper form, it becomes

\dot{x} + \frac{(bm + p\dot{x})\dot{x}}{amn - bm^2 + (mn - pm)\dot{x}} = 0.

The fluent of the second term of this expression will involve logarithms, except that $mn - pm = 0$ , in which case the primitive equation is

x + \frac{2bm\dot{x} + p\dot{x}^2}{2(amn - bm^2)} = C.

182. When a fluxional equation has this form

\dot{y} + Py\dot{x} = Q\dot{x},

where $P$ and $Q$ denote any functions of $x$ , the variable quantities may be separated in the following manner. Assume $y = X\dot{x}$ , then, taking the fluxions, we have $\dot{y} = \dot{X}\dot{x} + X\ddot{x}$ , and by substitution, the proposed equation becomes

\dot{X}\dot{x} + X\ddot{x} + PX\dot{x} = Q\dot{x};

now as in this equation $X$ and $\dot{x}$ may be supposed to denote indeterminate functions of $x$ , we may divide it into two others, such, that the variable quantities in each may be separable; to effect this we assume

X\dot{x} + PX\dot{x} = 0, \quad \dot{X}\dot{x} = Q\dot{x};

hence, dividing the first equation by $X$ , we have

\dot{X} + P\dot{x} = 0, \quad \text{and} \quad \frac{\dot{X}}{\dot{x}} + P = 0, \quad \text{and taking the fluents,}

1. $X + \int P\dot{x} = 0$ , and hence, by passing from logarithms to their numbers,

X = e^{-\int P\dot{x}};

here no constant quantity is introduced, it being sufficient to add it at the end of the operation; let this value of $X$ be substituted in the second equation, then by deducing from it the value of $X$ , we have

\dot{X} = e^{-\int P\dot{x}} Q\dot{x}

\text{and } X = \int e^{-\int P\dot{x}} Q\dot{x} + c;

and since $y = X\dot{x}$ , therefore

y = e^{-\int P\dot{x}} \left\{ \int e^{\int P\dot{x}} Q\dot{x} + c \right\}.

Let us take a particular case, and suppose the equation to be $\dot{y} + y\dot{x} = x\dot{x}$ , then we have $P=1$ , $Q=x$ , and $\int P\dot{x} = x$ , hence in this case the general formula becomes

y = e^{-x} \left\{ \int e^x x\dot{x} + c \right\}.

The fluent $\int e^x x\dot{x}$ may be found by § 143; let us suppose for example that $n=2$ , then we have

\int e^x x^2\dot{x} = e^x (x^2 - 2x - 2),

so that the fluxional equation being

\dot{y} + y\dot{x} = x^2\dot{x},

the primitive equation is

y = x^2 - 2x + 2 + ce^{-x}.

The general equation $\dot{y} + Py\dot{x} = Q\dot{x}$ , which involves the simple power only of the variable quantity $y$ , and its fluxion, has been called a linear equation of the first order; it has also, with more propriety, been called a fluxional equation of the first degree, and of the first order.

183. The equation

\dot{y} + Py\dot{x} = Qy\dot{x},

where $P$ and $Q$ as before denote any functions of $x$ , is easily reduced to the form we have been considering; for assume $y^{1-n} = (1-n)\dot{x}$ , then $y^{-n}\dot{y} = \dot{x}$ , and $\dot{y} = \dot{x}y^n$ , and $y = (1-n)\dot{x}y^n$ ; if we now substitute the values of $\dot{y}$ and $y$ in the equation, it becomes

y^n\dot{x} + (1-n)Py^n\dot{x} = Qy^n\dot{x};

let the terms of this equation be divided by $y^n$ , then, including the factor $(1-n)$ in the indeterminate function $P$ , the result is

\dot{x} + Px\dot{x} = Q\dot{x}

an equation of the very same form as that which has been considered in last §.

184. The most general form that can be given to a fluxional equation of the first order, and consisting of three terms only, is

y^k\dot{x}^l\dot{y} + \beta u^s\dot{x}^b\dot{u} = \gamma u^r\dot{x}^f\dot{u};

to give this equation a more simple form, let all its terms be divided by $y^k\dot{x}^l$ , it then becomes

\dot{x}^k\dot{x}^l + \frac{\beta}{y}u^s\dot{x}^{b-l}\dot{u} = \frac{\gamma}{y}u^r\dot{x}^{f-l}\dot{u}.

Suppose now

\dot{x}^{k-l}\dot{x}^l = \frac{\dot{x}}{k-l+1}, \quad u^{s-l}\dot{u} = \frac{\dot{u}}{s-l+1},

\text{then } \dot{x}^{k-l+1} = y, \quad u^{s-l+1} = u,

\text{and } \dot{y} + \frac{(k-f+1)\beta}{(g-i+1)\gamma} \frac{b-f}{k-f+1} \dot{x};

= \frac{(k-f+1)\alpha}{(g-i+1)\gamma} \frac{e-g}{g-i+1} \dot{x};

Let us in order to abridge put

\frac{(k-f+1)\beta}{(g-i+1)\gamma} = b, \quad \frac{(k-f+1)\alpha}{(g-i+1)\gamma} = a,

\frac{k-f}{k-f+1} = n, \quad \frac{e-g}{g-i+1} = m,

then the equation becomes

\dot{y} + b y^m \dot{x} = a x^n \dot{x}.

If we suppose $n=1$ , the resulting equation $\dot{y} + b y^m \dot{x} = a x^n \dot{x}$ may have its variable quantities separated by the method explained in § 183; but if we go only one step farther, and suppose $n=2$ , so that the equation is

\dot{y} + b y^m \dot{x} = a x^n \dot{x},

the difficulty of separating the variable quantities generally is so great as to have hitherto baffled the utmost efforts of the most expert analysts. This equation is commonly called Riccati's equation, on account of its having been first treated of by an Italian mathematician of that name, who succeeded in separating the variable quantities in some particular cases, namely, when $m$ is equal to $\frac{-4p}{2p-1}$ , where $p$ denotes any whole positive number.

185. If the separating of the variable quantities generally be a problem of insurmountable difficulty when the equation consists of only three terms, its solution can much less be expected, when the equation consists of four, or any greater number. There are, however, particular cases in which some of the most skilful analysts have, by employing happy and peculiar artifices, succeeded in resolving the problem, but the methods of proceeding are, generally speaking, not reducible to any determinate rules.

186. When the expression which constitutes a fluxional equation is such as would be produced by taking the fluxion of some function of $x$ and $y$ , in which case it may be said to be a complete fluxion, then, without attempting to separate the variable quantities, we have only to add a constant quantity to that function, and the result put $=0$ , will evidently be the primitive equation required.

If, for example, the equation be $x\dot{y} + y\dot{x} = 0$ , it is obvious that the expression $x\dot{y} + y\dot{x}$ is immediately produced by taking the fluxion of the function $xy$ ( $y$ being also considered as a function of $x$ ), therefore the primitive equation is $xy + c = 0$ .

From the view which has been given in § 174. of the origin of fluxional equations it appears, that in passing from a primitive equation to its fluxional equation, the terms of the latter in many cases will not constitute a complete fluxion, by reason of some multiplier, or di-

visor, which was common to them all, having disappeared. In such cases, however, if we can by any means discover that factor, by restoring it we shall immediately have a complete fluxion, the fluent of which, with the addition of a constant quantity, when put $=0$ , will be the primitive equation.

For example, if the equation be $x\dot{y} - y\dot{x} = 0$ , here $x\dot{y} - y\dot{x}$ cannot be immediately produced by taking the fluxion of a function of $x$ and $y$ ; but, if we divide the equation by $x^2$ so as to give it this form $\frac{x\dot{y} - y\dot{x}}{x^2} = 0$ ,

we obtain the expression $\frac{x\dot{y} - y\dot{x}}{x^2}$ which is a complete

fluxion, viz. that of the fraction $\frac{y}{x}$ , therefore $\frac{y}{x} + c = 0$ , or $y + cx = 0$ , is the primitive equation.

In like manner, the equation $mx\dot{y} + ny\dot{x} = 0$ , which does not in its present form express a complete fluxion, yet becomes so when multiplied by $x^{m-1} y^{n-1}$ , for then it is

m x^n y^{m-1} \dot{y} + n x^{m-1} y^n \dot{x} = 0,

from which it appears that the primitive equation in this case must be $x^n y^m + c = 0$ .

187. That we may be able to discover whether the terms of any proposed fluxional equation constitute a complete fluxion, and also from what expression such a fluxion has been derived, we must attend to the process, by which we find the fluxion of an expression composed of two variable quantities, one of which is a function of the other.

To avoid very general reasoning, we shall take for granted what is evidently possible, that any function of $x$ and $y$ may be generally expressed by a formula of this nature,

A x^m y^n = B x^p y^q + C x^r y^s + \&c.

where $A, B, C, \&c.$ denote constant quantities, and the exponents $m, n, \&c.$ given numbers, the number of terms being supposed either finite or infinite. Now the fluxion of the whole expression is the sum of the fluxion of its terms, but in taking the fluxion of each term, beginning with the first $A x^m y^n$ , the fluxion of which is

m A x^{m-1} y^n \dot{x} + n A x^m y^{n-1} \dot{y},

it is evident that the result is composed of two parts, one of which is the expression we would find for its fluxion, if $x$ only were considered as variable, and $y$ as constant, and the other is the expression for its fluxion, if $y$ only were considered as variable and $x$ as constant; hence it follows, that the sum of the fluxions of all the terms will have the very same property; so that, if $u$ be put for the whole expression, we shall in every case have

\dot{u} = M \dot{x} + N \dot{y},

where $M \dot{x}$ denotes the result that will be found if the fluxion of $u$ be taken upon the hypothesis that $x$ alone

Inverse Method. is variable, and $Ny$ is the fluxion of $u$ , supposing $y$ alone to be variable.

188. Refuming the consideration of the general expression

Ax^m y^n + Bx^k y^l + Cx^r y^s + \dots

let the fluxion of any one of its terms, for example, $Ax^m y^n$ , be taken, supposing $x$ alone variable, and the result is $mAx^{m-1}y^n$ . Again, let the fluxion of this result be taken, supposing $y$ alone variable, and we find it to be $mnAx^{m-1}y^{n-1}xy$ . Now, if we first take the fluxion of $Ax^m y^n$ , supposing $y$ variable, we get $nAx^m y^{n-1}y$ , and then, the fluxion of this result, considering $x$ alone as variable, we get $mnAx^{m-1}y^{n-1}xy$ , which is the very same expression as was found by proceeding in a contrary order; and as the same must hold true of all the terms, we may conclude, that if the fluxion of $u$ any function of $x$ and $y$ be taken, considering $x$ only as variable, and then the fluxion of that result, considering $y$ only as variable, the very same final result will be obtained as if we were first to take the fluxion of $u$ supposing $y$ variable, and then the fluxion of that result, supposing $x$ variable; but the fluxion of $u$ being expressed thus $Mx + Ny$ , it has been shewn that $Mx$ is the fluxion of $u$ , if $x$ only be supposed variable and $Ny$ is its fluxion, if $y$ only be variable, therefore, if we take the fluxion of $Mx$ upon the supposition that $y$ only is variable, also the fluxion of $Ny$ upon the supposition that $x$ only is variable, the results must be identical. This property affords the following rule, by which we may always determine whether any proposed expression constitutes an exact fluxion or not. Let the expression be put under this form, $Mx + Ny$ ; let $M'x$ be the fluxion of $M$ , supposing $y$ alone variable, and $N'x$ the fluxion of $N$ , supposing $x$ alone variable, then, if $M'$ and $N'$ are identical, $Mx + Ny$ is a complete fluxion; and if they are not, $Mx + Ny$ is not a complete fluxion.

189. It is easy to see, how, from a complete fluxion $u = Mx + Ny$ we may determine $u$ its fluent; for as $Mx$ has been deduced from $u$ by considering $x$ as variable, and $y$ as constant, on which account all the terms of $u$ that involved $y$ only must have vanished, it follows on the contrary, that if we put $Y$ to denote those terms, we shall have

u = \int Mx + Y,

the fluent of $Mx$ being taken, regarding $x$ only as variable. The function $Y$ may be determined, by comparing the fluxion of the expression thus obtained with the given fluxion $Mx + Ny$ .

Ex. 1. Let the fluxion be $\frac{ay + 2x^2y + y^2x + xy}{2\sqrt{(ay + x^2 + xy)}}$ , this expression when reduced to the form $u = Mx + Ny$ is

u = \frac{(2x + y)x}{2\sqrt{(ay + x^2 + xy)}} + \frac{(a + x)y}{2\sqrt{(ay + x^2 + xy)}}

\text{hence } M = \frac{2x + y}{2\sqrt{(ay + x^2 + xy)}}, N = \frac{a + x}{2\sqrt{(ay + x^2 + xy)}}.

Inverse Method. the fluxion of $M$ , supposing $y$ only variable, gives us

M'x = \frac{(ay + xy - 2ax)y}{4(ay + x^2 + xy)^{\frac{3}{2}}}

and in like manner the fluxion of $N$ , supposing $x$ only variable, gives

N'x = \frac{(ay + xy - 2ax)x}{4(ay + x^2 + xy)^{\frac{3}{2}}}

hence it appears that $M' = N'$ , and therefore that the proposed expression is an exact fluxion. To determine its fluent, the formula $u = \int Mx + Y$ gives us

u = \sqrt{(ay + x^2 + xy)} + Y;

the fluxion of this expression taken, upon the supposition that both $x$ and $y$ are variable, is

\dot{u} = \frac{ay + 2x^2y + y^2x + xy}{2\sqrt{(ay + x^2 + xy)}} + \dot{Y};

this result, compared with the original fluxion, shews that $\dot{Y} = 0$ , and $Y = c$ , a constant quantity.

Ex. 2. Suppose the fluxion to be

\dot{x}\sqrt{(a^2 + y^2)} + y\frac{(a^2 + xy + 2y^2)}{\sqrt{(a^2 + y^2)}}.

Here $M = \sqrt{(a^2 + y^2)}$ , $N = \frac{a^2 + xy + 2y^2}{\sqrt{(a^2 + y^2)}}$ , and by proceeding as in last example, we shall find $M' = N' = \frac{y}{\sqrt{(a^2 + y^2)}}$ , hence it follows that the expression is a complete fluxion, and the formula $u = \int Mx + Y$ shews

u = \int x\sqrt{(a^2 + y^2)} + Y

= x\sqrt{(a^2 + y^2)} + Y.

To determine $Y$ , we take the fluxion of this expression, supposing $x$ and $y$ both variable, and find it to be

\dot{u} = \dot{x}\sqrt{(a^2 + y^2)} + \frac{xy\dot{y}}{\sqrt{(a^2 + y^2)}} + \dot{Y},

and this compared with the original fluxion

\dot{u} = \dot{x}\sqrt{(a^2 + y^2)} + \frac{(a^2 + xy + 2y^2)\dot{y}}{\sqrt{(a^2 + y^2)}},

shews that $\dot{Y} = \frac{(a^2 + 2y^2)\dot{y}}{\sqrt{(a^2 + y^2)}}$ , hence

Y = \int \frac{(a^2 + 2y^2)\dot{y}}{\sqrt{(a^2 + y^2)}} = y\sqrt{(a^2 + y^2)};

therefore the fluent required is

u =

u = x \sqrt{(a^2 + y^2)} + y \sqrt{(a^2 + y^2)} + C \\ = (x + y) \sqrt{(a^2 + y^2)} + C,

where $C$ denotes a constant quantity.

190. It may be demonstrated, that as often as a fluxional equation does not constitute a complete fluxion, there is always an infinite number of factors, such, that if the equation were multiplied by any one of them, the result would be a complete fluxion. A general method of determining some one of these factors, however, seems to be a problem of such difficulty, that its solution, except in some particular cases, is not to be expected.

191. When a fluxional equation involves the second or higher powers of $\dot{x}$ and $\dot{y}$ , as in this example,

\dot{y}^2 - a^2 \dot{x}^2 = 0,

which may be put under this form,

\frac{\dot{y}^2}{\dot{x}^2} - a^2 = 0,

we may, by the theory of algebraic equations, deduce from it the values of $\frac{\dot{y}}{\dot{x}}$ , considering this quantity as a root of the equation; thus, in the present example, by resolving the quadratic equation $\frac{\dot{y}^2}{\dot{x}^2} - a^2 = 0$ , we have

\frac{\dot{y}}{\dot{x}} = \pm a, \text{ so that } \dot{y} - a\dot{x} = 0, \text{ and } \dot{y} + a\dot{x} = 0, \text{ hence}

y - ax + c = 0, \quad y + ax + c' = 0,

are two primitive equations, from either of which the fluxional equation $\dot{y}^2 - a^2 \dot{x}^2 = 0$ may be derived, and it may also be deduced from their product

(y - ax + c)(y + ax + c') = 0.

192. As often as the equation contains only one of the two variable quantities, for example $x$ , by the resolution of the equation we may obtain $\frac{\dot{y}}{\dot{x}} = X$ (where $X$

denotes some function of $x$ ), and hence $y = \int X \dot{x}$ , but if it be more easy to resolve the equation with respect to $x$ than to $\frac{\dot{y}}{\dot{x}}$ , which we shall denote by $p$ , then, instead

of seeking the values of $p$ from the equation, we may find that of $x$ , thus we shall have $x = P$ , some function of $p$ , and hence $\dot{x} = \dot{P}$ , and since $\dot{y} = p\dot{x}$ , therefore, $\dot{y} = p\dot{P}$ , and $y = \int p \dot{P} = Pp - \int P \dot{p}$ . The relation between $x$ and $y$ is now to be found by eliminating $p$ by means of the two equations

x = P, \quad y = Pp - \int P \dot{p}.

As a particular example, let us suppose the equation to be $x \dot{x} + a \dot{y} = b \sqrt{(x^2 + y^2)}$ , from which, by putting

\frac{\dot{y}}{\dot{x}} = p, \text{ we find}

x = b \sqrt{(1 + p^2)} - ap = P,

y = bp \sqrt{(1 + p^2)} - \frac{1}{2}ap^2 - b \int p \sqrt{(1 + p^2)}.

the fluent of $p \sqrt{(1 + p^2)}$ may be found by the formulas given in § 130, and § 131.

193. When we cannot by any means find an expression for the relation between $x$ and $y$ in finite terms, then we must, as a last resource, have recourse to approximation, that is, we must express the value of $y$ in terms of $x$ by means of a series.

When the form of the series is known, we may determine the coefficients of its terms, by substituting the series and its fluxion instead of $y$ and $\dot{y}$ in the proposed equation.

Suppose, for example, that the equation is

\dot{y} + y \dot{x} - m x^n \dot{x} = 0,

we may assume

y = Ax^n + Bx^{n+1} + Cx^{n+2} + \&c.

\text{then } \dot{y} = Ax^{n-1} \dot{x} + (n+1)Bx^{n+1} \dot{x}

+ (n+2)Cx^{n+2} \dot{x} + \&c.

Substituting now the values of $y$ and $\dot{y}$ in the equation, and dividing the whole by $\dot{x}$ , it becomes

Ax^{n-1} + (n+1)Bx^n + (n+2)Cx^{n+1} \\ - m x^n + A \int + B \int + C \int + \&c. = 0.

This equation becomes identical, if we assume $n-1=n$ , or $n=n+1$ , and

A = \frac{m}{n}, \quad B = \frac{-m}{n(n+1)}, \quad C = \frac{m}{n(n+1)(n+2)},

D = \frac{-m}{n(n+1)(n+2)(n+3)}, \quad \&c.

Hence we have

y = m \left\{ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{(n+1)(n+2)} \right. \\ \left. + \frac{x^{n+3}}{n(n+1)(n+2)(n+3)} - \&c. \right\}

In order that a primitive equation may be general, it ought to contain an indeterminate constant quantity more than is found in the fluxional equation, therefore, this series which contains no such quantity, must be considered as incomplete, or as exhibiting the value of $y$ , upon the supposition, that, when $x=0$ , then $y=0$ . However, we may obtain a value of $y$ that shall be general,

Inverse Method. neral, by proceeding as follows. Let us suppose we know that when $x=a$ , then $y=b$ ; assume $x=a+t$ , and $y=b+u$ , then it is manifest, that if the value of $u$ be found by a series involving $t$ , all the terms of the series ought to vanish when $t=0$ . From the assumed values of $x$ and $y$ the equation

\dot{y} + y\dot{x} - mx^n\dot{x} = 0

\dot{u} + (b+u)\dot{t} - m(a+t)t^2 = 0.

u = At^2 + Bt^3 + Ct^4 + \dots

then, proceeding as before, we find

\left. \begin{aligned} & A t^{n-1} + (a+t)B t^{n-2} + (a+2t)C t^{n-3} + \dots \\ & + b A t^n + B t^{n+1} + \dots \\ & - m a^n - m \frac{n}{1} a^{n-1} t - m \frac{n(n-1)}{1 \cdot 2} a^{n-2} t^2 - \dots \end{aligned} \right\} = 0.

It is necessary, in this equation, to assume $a=1=0$ , or $a=1$ , and hence we find

\begin{aligned} A &= m a^n - b, & B &= \frac{m n a^{n-1} - m a^n + b}{2}, \\ C &= \frac{m n (n-1) a^{n-2} - m n a^{n-1} + m a^n - b}{2 \cdot 3}, \\ & \dots \end{aligned}

If we now substitute $x=a$ , and $y=b$ for $t$ and $u$ respectively, the result will have all the generality that belongs to a primitive equation, expressing the relation between $x$ and $y$ .

Of Fluxional Equations of the second or higher orders.

194. Whatever difficulties occur in finding the primitive equation of a fluxional equation of the first order, it will easily be conceived, that these difficulties must be greater and more numerous when we have to consider fluxional equations of the second and higher orders.

One of the most simple cases of an equation of the second order is this,

\ddot{y} - X\dot{x}^2 = 0, \text{ or } \frac{\ddot{y}}{\dot{x}^2} = X,

where $X$ denotes a function of $x$ , the variable quantity whose fluxion is supposed to be constant; in this case, because $\frac{\ddot{y}}{\dot{x}^2} = X\dot{x}$ , we have $\frac{\ddot{y}}{\dot{x}^2} = \int X\dot{x}$ . Let $P$ denote that function of which $X\dot{x}$ is the fluxion, and $c$ , as usual, an indeterminate constant quantity, then $\frac{\ddot{y}}{\dot{x}^2} = P+c$ , and $\dot{y} = P\dot{x} + c\dot{x}$ , and taking the fluents a second time

y = \int P\dot{x} + c\dot{x} + c'

where $c'$ denotes a second indeterminate constant quantity.

\text{As } \int P\dot{x} = Px - \int \dot{P}x = x \int X\dot{x} - \int X\dot{x}x,

we have also

y = x \int X\dot{x} - \int X\dot{x}x + cx + c'.

Suppose, for example, that the equation is $\ddot{y} - ax\dot{x}^2 = 0$ , so that $\frac{\ddot{y}}{\dot{x}^2} = ax$ ; here $X=ax$ , and therefore

\begin{aligned} y &= x \int ax\dot{x} - \int ax^2\dot{x} + cx + c' \\ &= \frac{1}{2}ax^3 - \frac{1}{3}ax^3 + cx + c' \\ &= \frac{1}{6}ax^3 + cx + c'. \end{aligned}

In the very same manner we may deduce from the equation of the third order

\ddot{\dot{y}} - X\dot{x}^3 = 0, \text{ or, } \frac{\ddot{\dot{y}}}{\dot{x}^3} = X

its primitive equation; thus we have

\frac{\ddot{\dot{y}}}{\dot{x}^3} = X\dot{x}, \quad \frac{\ddot{\dot{y}}}{\dot{x}^3} = \int X\dot{x} = P+c,

where $P$ denotes such a function of $x$ , that its fluxion is $X\dot{x}$ , and $c$ represents a constant quantity. Again

\begin{aligned} \frac{\ddot{\dot{y}}}{\dot{x}^2} &= P\dot{x} + c\dot{x}, \\ \frac{\ddot{\dot{y}}}{\dot{x}} &= \int P\dot{x} + c\dot{x} + c' = Q + cx + c'; \end{aligned}

here $Q$ is put for $\int P\dot{x}$ , and $c'$ for a second constant quantity. In like manner we have

\begin{aligned} \dot{y} &= Q\dot{x} + cx\dot{x} + c'\dot{x}, \\ y &= \int Q\dot{x} + \frac{1}{2}cx^2 + c'x + c''; \end{aligned}

and as $P$ and $Q$ are functions of $x$ , the fluents of $P\dot{x}$ and $Q\dot{x}$ may be found by the methods formerly explained.

195. Let us next consider such equations as involve only $\frac{\ddot{y}}{\dot{x}^2}$ and constant quantities. In order to abridge let us put $\frac{\ddot{y}}{\dot{x}^2} = p$ , then such an equation may be generally expressed thus $\frac{\ddot{y}}{\dot{x}^2} = P$ , where $P$ denotes some known

known function of $p$ ; now as $\frac{\dot{y}}{x} = p$ , by taking the fluxions, and observing that $\dot{x}$ is constant, we have $\frac{\ddot{y}}{x} = \frac{\dot{p}}{x}$ , hence $\frac{\dot{p}}{x} = P$ , and $\dot{x} = \frac{\dot{p}}{P}$ , and $x = \int \frac{\dot{p}}{P}$ ; let the value of $\dot{x}$ be substituted instead of it in the equation $\dot{y} = p\dot{x}$ , and it becomes $\dot{y} = \frac{p\dot{p}}{P}$ , and hence $y = \int \frac{p\dot{p}}{P}$ ; thus it appears that if we can find the fluents $\int \frac{\dot{p}}{P}$ and $\int \frac{p\dot{p}}{P}$ we shall have the primitive equation when we eliminate $p$ by means of these two equations

x = c + \int \frac{\dot{p}}{P}, \quad y = c' + \int \frac{p\dot{p}}{P},

where $c$ and $c'$ denote the two indeterminate constant quantities that ought to enter into the primitive equation.

Suppose for example that the equation is

\frac{(\dot{x}^2 + \dot{y}^2)^{\frac{1}{2}}}{-x\dot{y}} = a;

which, by putting $p$ for $\frac{\dot{y}}{x}$ , and $\frac{\dot{p}}{x}$ for $\frac{\ddot{y}}{x^2}$ becomes transformed to

-\frac{(1+p^2)^{\frac{1}{2}} \dot{x}}{\dot{p}} = a;

hence we have

\dot{x} = \frac{-a\dot{p}}{(1+p^2)^{\frac{1}{2}}}, \quad \dot{y} = p\dot{x} = \frac{-ap\dot{p}}{(1+p^2)^{\frac{1}{2}}},

x = c - \frac{ap}{\sqrt{(1+p^2)}}, \quad y = c' + \frac{a}{\sqrt{(1+p^2)}};

when by means of these equations we eliminate $p$ , we obtain $(x-c)^2 + (y-c')^2 = a^2$ .

The fluxional equation is evidently formed by putting the general expression for the radius of curvature (given in § 97.) equal to a constant quantity, and the primitive equation is accordingly an equation to a circle having that constant quantity for its radius, as it ought to be.

196. Suppose now that the equation has this form

\frac{\ddot{y}}{x^2} = Y,

where $Y$ denotes a function of $y$ , then putting as before $\frac{\dot{y}}{x} = p$ , we have $\frac{\ddot{y}}{x^2} = \frac{\dot{p}}{x} = \frac{p\dot{p}}{y}$ , hence the equation $\frac{\ddot{y}}{x^2} = Y$ becomes $\frac{p\dot{p}}{y} = Y$ , and $p\dot{p} = Yy$ , and $p^2 =$

$2 \int Yy + c$ , hence $p = \frac{\dot{y}}{x} = \sqrt{c + 2 \int Yy}$ and

x = \int \frac{\dot{y}}{\sqrt{c + 2 \int Yy}} + c',

where $c$ and $c'$ denote two constant quantities.

To take a particular example, let us suppose the equation to be $\ddot{y} - y\dot{x}^2 = a\dot{x}^2$ , or $\frac{\ddot{y}}{x^2} = a + y$ , here $Y = a + y$ , and $2 \int Yy = 2ay + y^2$ , hence (and by § 127.)

x = \int \frac{\dot{y}}{\sqrt{c + 2ay + y^2}} + c'

= \int \frac{\dot{y}}{\sqrt{a + y + \sqrt{(a + 2ay + y^2)}}} + c'.

197. When the equation contains $\frac{\dot{y}}{x}$ , $\frac{\ddot{y}}{x^2}$ and $x$ , it may be transformed to a fluxional equation of the first order, by substituting in it $p\dot{x}$ and $\dot{p}\dot{x}$ , instead of $\dot{y}$ and $\ddot{y}$ ; if we can find the primitive of that fluxional equation, and thence the value of $p$ in terms of $x$ , we shall have the value of $y$ from the formula $y = \int p\dot{x}$ , or if we have the value of $x$ in terms of $p$ , then, because $\int p\dot{x} = p\dot{x} - \int \dot{x}p$ , we shall have

y = p\dot{x} - \int \dot{x}p.

Suppose the equation to be

\frac{(\dot{x}^2 + \dot{y}^2)^{\frac{1}{2}}}{-x\dot{y}} = X, \quad \text{or} \quad -\frac{(1+p^2)^{\frac{1}{2}} \dot{x}}{\dot{p}} = X,

where $X$ denotes any function of $x$ , then,

\frac{\dot{x}}{X} = \frac{-\dot{p}}{(1+p^2)^{\frac{1}{2}}}, \quad \text{and} \quad \int \frac{\dot{x}}{X} = \frac{-p}{\sqrt{(1+p^2)}};

Let us represent $\int \frac{\dot{x}}{X}$ by $V$ , then $p = \frac{V}{\sqrt{(1-V^2)}}$ , and

y = \int p\dot{x} = \int \frac{V\dot{x}}{\sqrt{(1-V^2)}}.

This equation evidently expresses the nature of a curve such, that $\frac{(\dot{x}^2 + \dot{y}^2)^{\frac{1}{2}}}{-x\dot{y}}$ , its radius of curvature (§ 97.), is equal to $X$ a function of $x$ one of its co-ordinates.

198. If the proposed fluxional equation of the second order contains $\frac{\dot{y}}{x}$ , $\frac{\ddot{y}}{x^2}$ and $y$ , to transform it we must terminate $\dot{x}$ by means of its value $\frac{\dot{y}}{p}$ deduced from the equation $\dot{y} = p\dot{x}$ , thus we shall have

\frac{\ddot{y}}{x^2}

\frac{\ddot{y}}{x^2} = \frac{\dot{p}}{x} = \frac{p\dot{p}}{y}

and the result will be an equation of the first order containing only $p$ , $\dot{p}$ and $\dot{y}$ ; when its primitive equation can be found, and thence the value of $p$ in terms of $y$ , we may find $x$ by the formula $x = \int \frac{y}{p}$ , and by the formula $x = \frac{y}{p} = \int \frac{y\dot{p}}{p^2}$ when $y$ is expressed by means of $p$ .

199. As an example of the manner in which fluxional equations of the second order are to be resolved by approximation, we shall take the particular equation

\ddot{y} + ax^n y x^2 = 0.

If the value of $y$ which satisfies the equation be supposed to have this form

Ax^n + Bx^{n+2} + Cx^{n+4} + \dots

and that the series of exponents goes on increasing, or that $\delta$ is positive, we may, by supposing $x$ to be a very small quantity, conceive that the expression for $y$ is reduced to its first term, because in that case each of the following terms will be inconsiderable in respect of that term. According to this hypothesis we shall have

y = Ax^n, \quad \ddot{y} = n(n-1)Ax^{n-2}x^2

and thus the proposed equation becomes

n(n-1)Ax^{n-2} + ax^n x^2 = 0.

It will not be possible to give to $x$ such a value that the two exponents $n-2$ and $n+2$ shall become equal, except in the particular case of $n = -2$ ; but if we suppose $x$ very small, the equation may be satisfied in two ways, namely, by taking $x = 0$ , and $x = 1$ , because upon either supposition the term $n(n-1)Ax^{n-2}$ , which is the greatest, vanishes, and therefore $A$ is left indeterminate; thus we have two series, one beginning with $A$ , and the other with $Ax$ .

Assuming therefore successively

y = A + Bx^2 + Cx^4 + \dots

y = Ax + Bx^{1+2} + Cx^{1+4} + \dots

and substituting these values as well as their corresponding values of $\dot{y}$ in the proposed equation, we shall find by arranging the terms, that $\delta$ ought to be $=2$ ; afterwards by determining the coefficients $A$ , $B$ , $C$ , &c. in the usual manner (ALGEBRA, § 261.) we obtain two series, one of these is

A - \frac{aAx^{n+2}}{(n+1)(n+2)} + \frac{a^2Ax^{n+4}}{(n+1)(n+2)(2n+3)(2n+4)} - \frac{a^3Ax^{n+6}}{(n+1)(n+2)(2n+3)(2n+4)(3n+5)(3n+6)} + \dots

VOL. VIII. Part II.

and the other

Ax - \frac{aAx^{n+3}}{(n+2)(n+3)} + \frac{a^2Ax^{n+5}}{(n+2)(n+3)(2n+4)(2n+5)} - \frac{a^3Ax^{n+7}}{(n+2)(n+3)(2n+4)(2n+5)(3n+6)(3n+7)} + \dots

As a primitive equation in its general form ought to contain two constant quantities which do not appear in the fluxional equation of the second order derived from it (§ 177.), the value of $y$ to be complete ought to contain two arbitrary constant quantities, but as each of these series contains only one such quantity, namely $A$ , it must be considered as expressing only a particular value of $y$ . The fluxional equation $\ddot{y} + ax^n y x^2 = 0$ is however of such a nature that from two particular values of $y$ we may deduce its general value; for let us denote these values by $x$ and $Z$ , then, as each of them must satisfy the fluxional equation, we have

\ddot{x} + ax^n x^2 = 0, \quad \ddot{Z} + ax^n Z^2 = 0;

let $c$ and $C$ denote two arbitrary constant quantities, then we have also

c\ddot{x} + cax^n x^2 = 0, \quad C\ddot{Z} + Cax^n Z^2 = 0,

and as each of these equations is identical, their sum must also be identical, that is

c\ddot{x} + C\ddot{Z} + ax^n(cx + CZ)x^2 = 0;

but the very same result will be obtained if we substitute $cx + CZ$ instead of $y$ in the proposed fluxional equation, therefore $cx + CZ$ is also a value of $y$ , and as it involves two arbitrary constant quantities, $c$ and $C$ , it possesses all the generality of which the value of $y$ is susceptible. Hence it follows that if $c$ be put instead of $A$ in one of the two series which we have found for the value of $y$ , and $C$ instead of $A$ in the other series, the sum of the two results will be a general expression for the value of $y$ .

200. Having now explained the theory of fluxional equations at as great length as we conceive to be compatible with the nature of this work, we shall conclude this treatise by resolving a few problems which produce fluxional equations.

Prob. 1. Having given any hyperbolic, or, as it may more properly be called, Napierian logarithm, it is required to find a general expression for its corresponding natural number.

Let the number be denoted by $1+x$ , and its logarithm by $y$ , then $\dot{y} = \frac{\dot{x}}{1+x}$ (§ 57.), or

\dot{y} + x\dot{y} = \dot{x},

and the problem requires that from this equation we deduce an expression for $x$ .

As when $y = 0$ , then $x = 0$ , we may assume

x = Ay + By^2 + Cy^3 + \dots

\text{then } \dot{x} = A\dot{y} + 2B\dot{y}y + 3C\dot{y}y^2 + \dots

and our equation becomes

5 F \quad \dot{y} + Ay\dot{y}

\left. \begin{aligned} & \dot{y} + A\dot{y}y + B\dot{y}y^2 + C\dot{y}y^3 + \dots c. \\ & - A\dot{y}y - 2B\dot{y}y^2 - 3C\dot{y}y^3 - 4D\dot{y}y^4 - \dots c. \end{aligned} \right\} = 0.

Hence, by comparing the coefficients of the like terms, it appears that $A=1$ , $2B=A$ , $3C=B$ , $4D=C$ , &c. so

\text{that } A=1, B=\frac{1}{2}, C=\frac{1}{2 \cdot 3}, D=\frac{1}{2 \cdot 3 \cdot 4}, \&c.

\text{therefore } x=y + \frac{y^2}{2} + \frac{y^3}{2 \cdot 3} + \frac{y^4}{2 \cdot 3 \cdot 4} + \dots c. \text{ and}

1+x=1+y + \frac{y^2}{2} + \frac{y^3}{2 \cdot 3} + \frac{y^4}{2 \cdot 3 \cdot 4} + \dots c.

Prob. 2. Let AB, AC, be two straight lines given by position meeting each other at right angles in A, let C be a given point in AC one of the lines, and let a straight line PQ meet them in P and Q, and cut off from them equal segments AP, CQ adjacent to the given points A, C, it is required to find the nature of the curve to which PQ is a tangent.

Let D be the point in which the tangent PQ meets the curve, draw DE perpendicular to AC, and DF to AP, put CA=a, CE=x, ED=y, then AE or DF

=a-x, \text{ and since } EQ=\frac{y^2}{x} \text{ (§ 73.) and } EQ:ED::

DF:FP, \text{ therefore } FP=\frac{(a-x)y}{x}; \text{ hence } PA=

(PF+FA)=\frac{(a-x)y}{x}+y, \text{ and } CQ=(CE-EQ)=

x-\frac{y^2}{x}, \text{ and as by hypothesis } AP=CQ, \text{ therefore}

y + \frac{(a-x)y}{x} = x - \frac{y^2}{x}.

This expression belongs to a class of fluxional equations which have the singular property of being more easily resolved by first taking their fluxion, considering the fluxion of one of the variable quantities as constant; thus, in the present case, making $x$ constant, we find

\dot{y} \frac{\dot{x}y - (a-x)\dot{y}}{x} = \dot{x} - \frac{\dot{x}y^2 - y\dot{x}y}{y^2}

\text{or } \frac{(a-x)\dot{y}}{x} = \frac{\dot{x}y\dot{y}}{y^2};

hence dividing by $\dot{y}$ , the equation is easily reduced to

\frac{\dot{y}}{\sqrt{y}} = \frac{\dot{x}}{\sqrt{(a-x)}},

and taking the fluents

\sqrt{y} = c - \sqrt{(a-x)}

but when $x=0$ , then $y=0$ , therefore $c=\sqrt{a}$ , and

\sqrt{y} = \sqrt{a} - \sqrt{(a-x)}, \text{ or } x=2\sqrt{ay} - y,

which equation belongs to the common parabola.

Prob. 3. Let APQ be one of any number of curves of the parabolic kind, having the same vertex A, and axis AE, and the nature of which is defined by the

equation $px^m=y^n$ , where $x$ denotes the abscissa AB, and $y$ the ordinate PB, and $p$ an indeterminate quantity which is the same for the whole of any one of the parabolas, but different for different parabolas; it is required to find the nature of a curve that shall intersect them all in a given angle.

Let the curve whose nature is required meet any one of the parabolas in P, let PT, Pt tangents to the two curves meet the axis in T and t; then, from the nature of the problem, the lines PT, Pt must contain a given angle, let $a$ denote its numerical tangent.

Because PT touches the parabola, the tangent of the angle PTB will be equal to $\frac{y}{x}$ (§ 75.) the value of

this expression being supposed deduced from the equation $px^m=y^n$ ; but taking the fluxion of this equation, and eliminating the indeterminate quantity $p$ by means of the two equations, we find $\frac{\dot{y}}{x} = \frac{my}{nx}$ , therefore $\tan.$

T = \frac{my}{nx}.

Again, by considering $x$ and $y$ as the abscissa and ordinate of the curve whose equation is sought, and to which Pt is a tangent, we have the tangent of the difference of two angles, (ALGEBRA, § 368.) that

a = \frac{\frac{\dot{y}}{x} - \frac{my}{nx}}{1 + \frac{my}{nx}}

hence we have

a(nxx + myy) + myx - nx y = 0,

a fluxional equation expressing the nature of the curve, which being homogeneous may be treated according to the method explained in § 180.

If the curves be supposed to cut each other at right angles, then, $a$ being infinite, the part of the equation which is not multiplied by $a$ vanishes in respect of the other, which is multiplied by it; hence we have

nxx + myy = 0,

and taking the fluents

nx^2 + my^2 = c

where $c$ is put for a constant quantity. This equation shews that the curve is an ellipse, the centre of which is at A the common vertex of all the parabolas.

The problem which we have here resolved is only a particular case of one more general, and which has for its object To determine the nature of the curve which intersects all other curves of a given kind in a given angle. The problem thus generalised is known by the name of the Problem of Trajectories; it was originally proposed by Leibnitz as a challenge to the English mathematicians, and resolved by Newton, on the day he received it.

FLY,

FLUXIONS.

PLATE CCXIX.

Fig. 1.

Geometric diagram showing a curve segment with points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z. A vertical line segment connects points C and M, and another connects D and N. A horizontal line segment connects points F and G.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

Fig. 9.

Fig. 10.

Fig. 11.

Fig. 12.

Fig. 14.

Fig. 15.

Fig. 13.

Fig. 16.

Fig. 17.

Fig. 18.

Fig. 19. N^o 1.

V. Archiv. math. 1847.