10.06.2022 Views

James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

A five star textbook for college calculus

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

762 Chapter 11 Infinite Sequences and Series

In trying to show that lim n l ` R n sxd − 0 for a specific function f, we usually use the

following theorem.

9 Taylor’s Inequality If | f sn11d sxd | < M for | x 2 a |

R n sxd of the Taylor series satisfies the inequality

| R nsxd | < M

sn 1 1d! | x 2 a | n11

To see why this is true for n − 1, we assume that | f 0sxd |

f 0sxd < M, so for a < x < a 1 d we have

y x

a

f 0std dt < y x

M dt

a

< d, then the remainder

for | x 2 a | < d

< M. In particular, we have

Formulas for the

Taylor Remainder Term

As alternatives to Taylor’s Inequality,

we have the following formulas for the

remainder term. If f sn11d is continuous

on an interval I and x [ I, then

R nsxd − 1 n! yx sx 2 td n f sn11d std dt

a

This is called the integral form of the

remainder term. Another formula,

called Lagrange’s form of the remainder

term, states that there is a number z

between x and a such that

R nsxd − f sn11d szd

sx 2 adn11

sn 1 1d!

This version is an extension of the

Mean Value Theorem (which is the

case n − 0).

Proofs of these formulas, together

with discussions of how to use them to

solve the examples of Sections 11.10

and 11.11, are given on the website

www.stewartcalculus.com

Click on Additional Topics and then on

Formulas for the Remainder Term in

Taylor series.

An antiderivative of f 0 is f 9, so by Part 2 of the Fundamental Theorem of Calculus, we

have

Thus

f 9sxd 2 f 9sad < Msx 2 ad or f 9sxd < f 9sad 1 Msx 2 ad

y x

a

f 9std dt < y x

f f 9sad 1 Mst 2 adg dt

a

f sxd 2 f sad < f 9sadsx 2 ad 1 M

f sxd 2 f sad 2 f 9sadsx 2 ad < M 2

sx 2 ad2

But R 1 sxd − f sxd 2 T 1 sxd − f sxd 2 f sad 2 f 9sadsx 2 ad. So

R 1 sxd < M 2

sx 2 ad2

A similar argument, using f 0sxd > 2M, shows that

R 1 sxd > 2 M 2

sx 2 ad2

So | R 1sxd | < M 2 |x 2 a | 2

sx 2 ad2

2

Although we have assumed that x . a, similar calculations show that this inequality is

also true for x , a.

This proves Taylor’s Inequality for the case where n − 1. The result for any n is

proved in a similar way by integrating n 1 1 times. (See Exercise 83 for the case n − 2.)

Note In Section 11.11 we will explore the use of Taylor’s Inequality in approximating

functions. Our immediate use of it is in conjunction with Theorem 8.

In applying Theorems 8 and 9 it is often helpful to make use of the following fact.

10

x n

lim − 0 for every real number x

nl` n!

Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!