1549901369-Elements_of_Real_Analysis__Denlinger_

336 Chapter 6 m Differentiable Functions

Also, letting t = x in Equation (6), we find

n J(kl(x) (x xr+1

G(x) = f(x) + 2::-k-! -(x - x)k + Rn(x) (x = a)n+l = f(x) + 0 + 0 = f(x).

k=l

Thus, G(a) = G(x). Hence, G satisfies all the hypotheses of Rolle's theorem

on the closed interval between a and x. Therefore, by Rolle's theorem, 3 c

between a and x such that

G'(c) = 0.

By Equation (7) this means

j(n+ll(c)( _ )n_ (n+l)(x-c)nR ( )=O

n! x c (x - a)n+l n x.

Solving for Rn ( x), we have

(x - ar+l J(n+ll(c) n

Rn(x) = · (x-c)

(n + l)(x - c)n n!

J(n+ll(c) n+l

( n+ 1. )' (x - a). •

APPLICATIONS OF TAYLOR'S THEOREM

Example 6.5.12 Use Taylor's theorem to prove that 'v'x > 0,

x2 x3 x2 x3

1 + x + - + - < ex < 1 + x + - + - ex.

2 3! 2 3!

Solution: Let x > 0. We use the Taylor polynomial T 2 (x) for ex about 0.

Using the notation of Definition 6.5.9, 'v'n E N,

2
That is, ex = 1 + x + ~ + R 2 ( x). By Taylor's theorem, 3 c between 0 and
ec
x such that R2(x) =
3
! x^3. Thus, for some c between 0 and x,

x2 ec

ex = 1 + x + - + - x^3.

2 3!