quantities within them are to be taken, viz. when x = a. And replacing t by p(x) and (x) by its value given in (149), we have

(x) 1 Ldx

d.f(x) (x-a

{d.f(x)

+

d2

{

dx2 dx φ (π)

dx (x)

{(x)}3

1.2.3

{$(x)}*

1.2

where the square brackets indicate particular values of the quantities within them, viz. those which correspond to x = a.

If (x)=x-ah, (151) becomes Taylor's series.

As an example of (151), let f(x) = ex, $(x) = (x −1) (x−2), and let a = 2; then

which gives the expansion of e* in ascending powers of (x −1)(x-2). Also if a = 1, we have

exe-e (x-1) (x-2)+3e

(x-1)2(x-2)2
1.2

+ ...,

which gives another expansion of ex in powers of (x-1) (x-2).

93.] A still more general form of expansion than that of Lagrange was discovered by Laplace, and is known by the name of Laplace's Theorem.

Given y = F{+x (y)}; it is required to find ƒ(y).

Using the same notation of Maclaurin's Theorem as heretofore, we have

And by a process similar to that employed in the proof of
Lagrange's Theorem, it may be shewn that

and so on for other and for the nth terms; whence

As an example of this Theorem, let it be required to find e",

whence, substituting in the formula (152), we have

Laplace's Theorem, it will be observed, becomes Lagrange's, when F = 1; and Taylor's Series is also a particular case of Laplace's; for as

94.] Another form of function which it is often necessary to expand by Maclaurin's Theorem is that in which a subsidiary variable z is introduced; and where we have two equations of the form

y = f(z),

and z =

$(x);

(153)

and wherein it is required to expand y in ascending powers of x.

Using Lagrange's notation of derived-functions, we have

day dx4

=

(154)

(155)

(156)

ƒ'''(z) {$'(x)}3 +3ƒ"(z) p′(x) p′′ (x) +ƒ'(z) q′′(x), (157)

· ƒ'ˇ (≈) {p′(x)}1+6ƒ''' (z) {Þ′(x)}2 p′′ (x) +3ƒ"(z) {p′′(x)}2 +4ƒ"(z) p′(x) p′′''(x) + ƒ′(z) p′ˇ(x); (158)

and so on. Now substituting these quantities in the several terms of Maclaurin's Series, (13), Art. 58, and putting x = 0, and introducing the corresponding value of z, we shall have the required series. To take a simple case, let

and so on; but when x= 0, 2= 0; therefore using the square brackets in the same signification as heretofore,

dy

[v] = 1, [dv] = 1, [d24] = 1, [d] = 0, [14].

dx

0

dx2

0

0

dx4

= −3, ...

95.] But in the case wherein z is a series of terms in ascending powers of x, the preceding expansion takes a particular form which deserves much attention; and gives rise to a process which has been called Derivation *; and on which Arbogast

*This process, though called by the same name, is essentially different from that explained in Art. 18. The title of Arbogast's work is, Du Calcul des Dérivations; it was published at Strasbourg, An VIII. (1800.)

has constructed his Calcul des Dérivations. It will be convenient to take the exponential series for the base-form of the series, and I shall accordingly assume

so that the problem is, the expansion in ascending powers of

In all these equations let x = 0; then z = αo,

= a1,

= α2,

dx

dx2

dy day

values of

...

y = f(a); in whatever manner therefore the

given in (155), (156), (157), are com

dx' dx2' posed of f(z), f'(z), ƒ"(z), ... combined with

...

dz d2z d3z

dx' dx2' dx3

" ...

;

O, be composed .... From this

in the same manner will these values, when x = of ƒ (ao), f'(ao), ƒ"(a), ... combined with a1, a2, aз, peculiarity we may deduce the following process; [d] is the

differential of f(a.) on the supposition that dao = a1;

́dy

the second differential of f(a) on the supposition that da。= a1, is the nth differential of ƒ(a) on the

and da1 = a2; and [d"

=

Lươn 0 supposition that da。 a1, da1 = a2, da2 = a3, ... this power of substituting new constants for the differentials of other constants does Arbogast's method of derivation consist. If therefore we replace the successive coefficients of the powers

of x in Maclaurin's Series, equation (13), Art. 58, by their values determined as above, we have

x2
y = f(ao) + [dƒ(ao)] }2 + [d2 ƒ(a)] +

1.2

the square brackets in this case indicating that particular values of the functions enclosed in them are to be taken, viz. when we replace dao by a1, da1 by a2, ......

And if we perform the several operations of derivation and introduce Lagrange's notation of derived-functions, we have

Xx
1

x2

1.2

= f(ao)+ƒ′(ao) a1 1⁄2 + {ƒ” (ao) ar3 +ƒ′(ao) a2}

+ {ƒ''' (ao) aï3 +3 ƒ" (ao) a1 a2+ƒ′ (ao) a3}

Of this process we propose to give a few examples.

Ex. 1. It is required to expand in ascending powers of x

(163)

=

...