1549901369-Elements_of_Real_Analysis__Denlinger_

6.5 Taylor's Theorem 337

Since 0 < c < x, we have 1 < ec < ex. Thus, from the above equation, we

have

x2 x3 x2 x3

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

2 3! 2 3!

Example 6.5.13 Use Taylor's theorem to prove that

Solution: Let x E JR.. We use the Taylor polynomials Tn(x) for ex about

0. Using the notation of Definition 6.5.9, \:/n E N,

By Taylor's theorem, for x =!=-0, :Jc between 0 and x such that

-lf(n+l)(c) n+ll-lecllxn+ll

IRn(x)I - (n + 1)! x - (n + 1)!

elxllxn+ll

< (n + 1)! ·

xn+l

By Corollary 2 .3.11, lim ( )' = 0. Thus, for each x, lim Rn(x) = 0.

n-><XJ n + 1. n-><XJ

n k

Therefore, lim L xkl = lim Tn(x) = lim [ex -Rn(x)]

n-+oo. n-+oo n--+oo

k=O

=ex- lim Rn(x) =ex. D

n-><XJ

Example 6.5.13 is a special case of a general theorem, which we now state.

Theorem 6.5.14 Suppose f and all its derivatives exist on an open interval
I containing a. If :JM > 0 3 \:/x E I and \:/n E N, if(n)(x)I ~ Mn, then
lim Tn ( x) = f ( x), where Tn ( x) denotes the nth Taylor polynomial for f about
n-><XJ
= f(n)(a)
a. That is, L 1 (x - a)n = f(x).
n.
n=O
Proof. Exercise 15. •

Taylor's theorem has some unexpected applications. For example, it pro-
vides a way to prove the "Second Derivative Test,'' so familiar to all students
who have studied maximum/minimum problems in elementary calculus. In fact,
the theorem we are going to prove goes beyond this test to a more general "nth
Derivative Test."