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then, since x-z+fx)=l(x+a,)+l(≈+a2) + . . . + 1(z+αn),

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Substitute F=1-f's; then, by actual division on both sides of the equation, if _, represent the sum of the reciprocals of the nth powers of the roots of the equation, we shall have (Alg. 346.)

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1

& S−(n+1)= q(1 − p) + (pq (1−p') )' + √9 (P°q(1−p'))"

1

+

1.2

1.2.3 (p3q(1 − p') )'" + &c.

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1

z (p2q)" = (pqp')' + 1⁄2 (p°q')' or -(pqp')'+12(p°q)"..

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1

p2q')'

1

S— (n+1) = q+pq' + = q(p?q)' + 1.2.§(p3q')'+ &c.

1.2

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In this equation, change n+1 into n and there results 1 n (ƒ) x n ((f)x

S-n=

X

which

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the sum of all the terms in the developement of 2" that contain negative powers of a (Art. 41. Praxis, Ex. 6).

Ex. 1. Required to demonstrate the formula of Art. 41. Ex. 12.

1

The roots of 2-2 cos.x. ≈+1=0, are of the form a, α

(Alg. 325); hence we may assume ɑ +

and consequently (2. 37) 2 cos.nx = a" + article, part of the developement of

1

a

1

= 2 cos. x = P,

==, by the

This has been developed in Art. 42, Ex. 6, if we make a=1, c=1, b=p and n-n; hence by that example

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The transformed equation is 1-p1 ≈ +P2 ≈2 -... ± p"x"=0,

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fx=

P. Pi

1

Pi

1

p; here

5, { P ̧¤2 — P ̧x3 + .... ± Prx"}, and by the article,

Pi

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2

: { (n−3 ) p22x−n+2 — 2 ( n −4)P2P ̧x¬n+3 +&c. }

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This is Waring's series, who in the Meditationes Algebraicæ, Ch. 1. has calculated it as far as p"-8.

The examples of this chapter are only a few of the numerous instances in which the principle of successive differentiation is applicable to the developement of functions. The subject itself is of the greatest utility in the mixed mathematics, and it has been handled by almost all scientific writers of any note from the days of Sir I. Newton, who was the first to see its importance. It would be inconsistent with the plan of this work to give a fuller account of the developement of functions into series, or even to enter into further details concerning Lagrange's theorem; for information on the latter subject I must refer the reader to the work from which this and the preceding article have been taken," Resolution des Equations Numeriques, Note 11."

Before we conclude this subject with the cases in which these developements fail, we shall subjoin two theorems which will be required in the second volume, the one due to Euler and the other to John Bernoulli.

45. Let u=fx be such a function of x that when x=a, q u = b; then shall u = b + (x − a) —— − ( x − u)2 1, 2

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Ρ

1

1.2

&c., where p, q, r, &c. are the fluxional

coefficients of u = fx.

For, since u =fx, therefore (ex. hyp.) b =ƒa, and by

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fa, which b, up· 1

Suppose a-x=h, or a=x+h; hence, by substitution,

x-a (x − a)2 r(x − a)3
+q

+ &c.

1.2

1.2.3

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It is remarked of this theorem, that it has the same relation to the integral calculus which that of Taylor has to the fluxional.

CHAPTER V.

The cases in which the developement of functions fails.Fractions whose numerator and denominator vanish at the same time.

1. Ir frequently happens in algebraick calculations, that by assigning a particular value to the variable the result x2-a2 becomes vain and nugatory. Thus,

x-a

when x = a,

becomes ; from which we are not to infer that the value 0

of the function equals nothing, the only strict inference being, that in this particular case the algebraick rule of division ceases to be applicable; and the reason is obvious, that in the operation we suppose x2. a2 and x -a to represent real quantity, so that this particular value of x is excluded by the supposition. In such cases the rule, though the phrase is certainly incorrect, is said to fail.

It is upon the same principle that we explain, what at first sight appears paradoxical and inconsistent with the definition Ch. 1. Art. 7., that the fluxion of a quantity which vanishes is not necessarily equal to nothing.

Let OP=x, OA=a; then if p describe the line OAP uniformly, the fluxion of Ap

+ + +

P др

or ax must be a constant quantity; a negative quantity before p coincides with A, and afterwards positive. It does not vanish, then, when P is at A, though AP = 0.

To take another example;

du 1 therefore =

let u =

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dx X

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