(7) Let utan æ ;

ƒ (0) = 2", ƒ'(0) = n2"-1, ƒ"(0) = n2"−2 (n + 1),

Maclaurin's Theorem may also be applied to the development of implicit functions, the differentiations being effected by the methods required in such cases.

=

When a 0, u=1; therefore ƒ (0) = ± 1.
Differentiating the implicit function we have

when ∞ =

du

0, 2 - 10, therefore ƒ"" (0) = 0.

dx

Since the given function is a quadratic in u it involves really two different functions of a, which in the development are given by means of the double sign.

The possible root of this is 2, and if we take it, we find by the same method as in the last example the series

The other series for u would be found by taking the impossible values of the cube root of 8.

(12) Let u3 - a2u + a xu — x3 = 0.

When a = 0, u3 - a'u = 0, which gives

u = 0, u = ±a.

Taking the first of these values, we find the series

(13)

If sin ya sin (a + y),

expand y in terms of æ.

When a = = 0, sin y = 0; therefore y=rπ, r being 0, or

any positive integer.

In a similar manner we should find

ƒ"" (0) = 2 sin a - 4 (sin a)*},

and so on; therefore, substituting in Maclaurin's Theorem,

(14) If u" logu = ax, expand u in terms of x.

When ∞ = 0, one value of u is 1, as log 1 = 0; therefore taking ƒ (0) = 1, we find

ƒ'(0) = a, ƒ"(0) = (2n − 1) a2, ƒ"" (0) = (3n − 1)2 a3,

(15) Let y = 1 + xe" expand y in terms of a.

As the calculation of the high differential coefficients of implicit functions is necessarily very tedious, this application of Maclaurin's Theorem is not of much use; and a better means of expanding implicit functions, is to be found in the Theorems of Lagrange and Laplace, to which we now proceed.

SECT. 3. Theorems of Lagrange and Laplace.

If y be given in an equation of the form

y = x + xp (y),

and if u = ƒ (y), ƒ and being any functions whatever, then u may be expanded in ascending powers of a by the theorem.

This is Lagrange's Theorem. See Equations Numériques, Note XI; Mémoires de Berlin, 1768, p. 251.

The Theorem of Laplace is an extension of the preceding, made by assuming the given equation in y to be

d

y = F{x+x+(y)}.

Then if u = f (y), and if we put ƒ F(x) = ƒ1 (*), and

~_ƒF(x) = ƒ'(x), and μF (~) = $.(~),

dx

Mémoires de l'Académie des Sciences, 1777, p. 99.