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

A five star textbook for college calculus

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768 Chapter 11 Infinite Sequences and Series

Table 1

Important Maclaurin

Series and Their Radii

of Convergence

1

1 2 x − ò x n − 1 1 x 1 x 2 1 x 3 1 ∙ ∙ ∙ R − 1

n−0

e x − ò

n−0

x n

n! − 1 1 x 1! 1 x 2

2! 1 x 3

3! 1 ∙ ∙ ∙ R − `

sin x − òs21d n

n−0

cos x − òs21d n

n−0

tan 21 x − òs21d n

n−0

x 2n11

s2n 1 1d! − x 2 x 3

x 2n

s2nd! − 1 2 x 2

x 2n11

2n 1 1 − x 2 x 3

3! 1 x 5

5! 2 x 7

2! 1 x 4

4! 2 x 6

3 1 x 5

5 2 x 7

7! 1 ∙ ∙ ∙ R − `

6! 1 ∙ ∙ ∙ R − `

7 1 ∙ ∙ ∙ R − 1

lns1 1 xd − òs21d x n

n21

n−1 n − x 2 x 2

2 1 x 3

3 2 x 4

4 1 ∙ ∙ ∙ R − 1

s1 1 xd k −

n−0S ò

nDx k n − 1 1 kx 1

ksk 2 1d ksk 2 1dsk 2 2d

x 2 1 x 3 1 ∙ ∙ ∙ R − 1

2!

3!

Example 10 Find the sum of the series

1

1 ? 2 2 1

2 ? 2 1 1

2 3 ? 2 2 1

3 4 ? 2 1 ∙ ∙ ∙.

4

SOLUTION With sigma notation we can write the given series as

òs21d n21

n−1

1

n ? 2 n

− òs21d n21 (1 2) n

n−1 n

Then from Table 1 we see that this series matches the entry for lns1 1 xd with x − 1 2 . So

òs21d n21

n−1

1

n ? 2 n − lns1 1 1 2d − ln 3 2 n

TEC Module 11.10/11.11 enables you

to see how successive Taylor polynomials

approach the original function.

One reason that Taylor series are important is that they enable us to integrate functions

that we couldn’t previously handle. In fact, in the introduction to this chapter we mentioned

that Newton often integrated functions by first expressing them as power series and then

integrating the series term by term. The function f sxd − e 2x 2 can’t be integrated by techniques

discussed so far because its antiderivative is not an elementary function (see Section

7.5). In the following example we use Newton’s idea to integrate this function.

ExamplE 11

(a) Evaluate y e 2x 2 dx as an infinite series.

(b) Evaluate y 1 0 e2x 2 dx correct to within an error of 0.001.

SOLUTION

(a) First we find the Maclaurin series for f sxd − e 2x 2 . Although it’s possible to use the

direct method, let’s find it simply by replacing x with 2x 2 in the series for e x given in

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