x

1

In these theorems, if we make f(y) = y, we find

b 1

1.2

d

dz

Ex. (1) Let yay+b= 0, or y =

y3 ;

a

b

dx2

$ = 9.8.

ぴ

Whence y = (1++3+12 +55 +8

Expand y in terms of b.

0,

Here f(y) = y, $(y) = y",

= a { 1 + a* −1 b + 2 n.a2n −2

12

b

a9

or y

= a + by".

≈ = a. Then

+ &c.)

(4) Let y = a + x log y.

Expand y in terms of x.

Here ƒ (*) = ≈, f'(x) = 1, x = a, f(x) = log ≈.

There

b

2n

y = a + (a" + ca") { + { 2 na2"-' + 2c (n+r) a"+r- ' + c°2 ra2r−1}

+ &c.

In the preceding examples it will be seen that the expansion of y in terms of b is the solution of an equation either algebraic or transcendental, and Lagrange has shewn that the series always gives the least root of the equation.

Lagrange has shown* that if by his theorem we develop the nth negative power of the root of the equation

y = x + xq (y),

of x,

and if we only retain the terms involving negative powers the result gives us the sum of the nth negative powers of the roots; while, as has just been stated, the whole series gives the nth negative power of the least root.

the series only continuing so long as there are positive powers

Then if we represent the sum of the inverse nth

powers of

c2

r-1

a

с

n (n − 2r + 1) (a\

+

b

1.2

(1)

2r-2

b

b?

n

b

n (n − 3r + 1) (n − 3r + 2)

1.2. 3

the series being continued only so long as it involves positive

find the sum of the inverse nth powers of the roots of the transformed equation, we obtain a series for the direct th powers of the roots of the original equation.

(11) If we thus transform the equation in Ex. 10, it becomes

c - by + ay2 = 0;

and if a, ẞ be the same quantities as before,

(13)

We might employ Lagrange's Theorem to express

h in terms of u from the equation

but the following method is

more convenient, as it gives at

once the law of the series. It is easily seen that the series is the development of some function of a + h, which when

h

=

0 becomes u.

Let u = f(x), then f(x + h) = 0.

But since u = f(x),

x = f(u), and if we call k the increment of u due to the increment h of x,

-1

x + h = ƒ ̃ 1 (u + k),

or, expanding by Taylor's Theorem,

expression which is given by

Paoli, Elementi d'Algebra, Vol. 11. p. 40.