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97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 764

764

CHAPTER 11

INFINITE SEQUENCES AND SERIES

In the following examples we don’t work out all the details but simply indicate which

tests should be used.

ϱ

v

EXAMPLE 1

nϪ1

2n ϩ 1

͚

n1

Since a n l

0 as n l ϱ, we should use the Test for Divergence.

1

2

ϱ

͚

EXAMPLE 2

n1

sn 3 ϩ 1

3n ϩ 4n 2 ϩ 2

3

Since a n is an algebraic function of n, we compare the given series with a p-series. The

comparison series for the Limit Comparison Test is bn , where

bn

ϱ

v

EXAMPLE 3

n 3͞2

1

sn 3

3͞2

3n 3

3n 3

3n

͚ ne

Ϫn 2

n1

Since the integral x1ϱ xeϪx dx is easily evaluated, we use the Integral Test. The Ratio Test

also works.

2

ϱ

͚ ͑Ϫ1͒

n

EXAMPLE 4

n1

n3

n ϩ1

4

Since the series is alternating, we use the Alternating Series Test.

ϱ

v

EXAMPLE 5

͚

k1

2k

k!

Since the series involves k!, we use the Ratio Test.

ϱ

EXAMPLE 6

͚

n1

1

2 ϩ 3n

Since the series is closely related to the geometric series 1͞3 n, we use the Comparison

Test.

Exercises

11.7

1–38 Test the series for convergence or divergence.

ϱ

1.

͚

n1

1

n ϩ 3n

ϱ

n

3. ͚ ͑Ϫ1͒

nϩ2

n1

n

ϱ

5.

ϱ

8.

k1

2 Ϫk

e

n

n

4. ͚ ͑Ϫ1͒ 2

n ϩ2

n1

1

nsln n

͚k

͑2n ϩ 1͒

n 2n

n

ϱ

ϱ

ϱ

ϱ

6.

͚

͚

n1

͚

n2

9.

2.

n 2 2 nϪ1

͑Ϫ5͒ n

n1

7.

ϱ

͚

n1

ϱ

͚

k1

ϱ

10.

1

2n ϩ 1

2 k k!

͑k ϩ 2͒!

͚ne

n1

2 Ϫn 3

11.

͚

n1

ϱ

13.

͚

n1

ϱ

15.

͚

k1

ϱ

17.

͚

n1

ϱ

18.

͚

n2

ͩ

1

1

ϩ n

n3

3

ͪ

ϱ

12.

͚

k1

ϱ

3n n2

n!

14.

2 kϪ1 3 kϩ1

kk

16.

1

ksk 2 ϩ 1

͚

sin 2n

1 ϩ 2n

ϱ

n2 ϩ 1

n3 ϩ 1

n1

͚

n1

1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒

2 ؒ 5 ؒ 8 ؒ и и и ؒ ͑3n Ϫ 1͒

͑Ϫ1͒ nϪ1

sn Ϫ 1

Copyright 2010 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.

97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 765

POWER SERIES

SECTION 11.8

ϱ

19.

͚ ͑Ϫ1͒

n

n1

ϱ

21.

͚ ͑Ϫ1͒

n

ϱ

ln n

sn

20.

cos͑1͞n 2 ͒

22.

ϱ

ϱ

25.

͚

n1

ϱ

27.

͚

k1

͚

31.

k1

ϱ

͚ n sin͑1͞n͒

͚

33.

n1

n1

26.

͚

n1

k ln k

͑k ϩ 1͒3

ϱ

1

2 ϩ sin k

ϱ

n!

2

en

n1

ϱ

24.

n1

ϱ

͚

k1

͚ tan͑1͞n͒

ϱ

͚

29.

k1

n1

23.

3

k Ϫ1

s

͚ k (sk ϩ 1)

ϱ

28.

͚

n1

30.

5k

3 ϩ 4k

32.

n1

n

͚ (s2 Ϫ 1)

n

37.

ϱ

34.

͚

n1

ϱ

36.

1ϩ1͞n

ϱ

e 1͞n

n2

n

͚

n1

2

1

͚

35.

ϱ

ͩ ͪ

n

nϩ1

͚ ͑Ϫ1͒

j1

k

ϱ

n2 ϩ 1

5n

ϱ

͑Ϫ1͒ n

cosh n

͚

n2

j

765

sj

jϩ5

͑n!͒ n

n 4n

1

n ϩ n cos2 n

1

͑ln n͒ln n

ϱ

n

38.

n1

͚ (s2 Ϫ 1)

n

n1

Power Series

11.8

A power series is a series of the form

ϱ

͚cx

1

n

n

c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и

n0

where x is a variable and the cn’s are constants called the coefficients of the series. For each

fixed x, the series 1 is a series of constants that we can test for convergence or divergence.

A power series may converge for some values of x and diverge for other values of x. The

sum of the series is a function

f ͑x͒ c0 ϩ c1 x ϩ c2 x 2 ϩ и и и ϩ cn x n ϩ и и и

Trigonometric Series

A power series is a series in which each term is

a power function. A trigonometric series

ϱ

͚x

n

1 ϩ x ϩ x2 ϩ и и и ϩ xn ϩ и и и

n0

ϱ

͚ ͑a

whose domain is the set of all x for which the series converges. Notice that f resembles a

polynomial. The only difference is that f has infinitely many terms.

For instance, if we take cn 1 for all n, the power series becomes the geometric series

n

cos nx ϩ bn sin nx͒

n0

is a series whose terms are trigonometric

functions. This type of series is discussed on

the website

Խ Խ

which converges when Ϫ1 Ͻ x Ͻ 1 and diverges when x ജ 1. (See Equation 11.2.5.)

More generally, a series of the form

ϱ

2

n

n

c0 ϩ c1͑x Ϫ a͒ ϩ c2͑x Ϫ a͒2 ϩ и и и

n0

www.stewartcalculus.com

Click on Additional Topics and then on Fourier

Series.

͚ c ͑x Ϫ a͒

is called a power series in ͑x Ϫ a͒ or a power series centered at a or a power series about

a. Notice that in writing out the term corresponding to n 0 in Equations 1 and 2 we have

adopted the convention that ͑x Ϫ a͒0 1 even when x a. Notice also that when x a

all of the terms are 0 for n ജ 1 and so the power series 2 always converges when x a.

ϱ

v

͚ n!x

EXAMPLE 1 For what values of x is the series

n

convergent?

n0

SOLUTION We use the Ratio Test. If we let a n , as usual, denote the nth term of the series,

Notice that

͑n ϩ 1͒! ͑n ϩ 1͒n͑n Ϫ 1͒ ؒ . . . ؒ 3 ؒ 2 ؒ 1

͑n ϩ 1͒n!

then a n n! x n. If x

lim

nlϱ

0, we have

Ϳ Ϳ

Ϳ

Ϳ

a nϩ1

͑n ϩ 1͒!x nϩ1

lim

lim ͑n ϩ 1͒ x ϱ

nlϱ

nlϱ

an

n!x n

Խ Խ

Copyright 2010 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.

97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 766

766

CHAPTER 11

INFINITE SEQUENCES AND SERIES

By the Ratio Test, the series diverges when x

when x 0.

0. Thus the given series converges only

ϱ

v

͚

EXAMPLE 2 For what values of x does the series

n1

SOLUTION Let a n ͑x Ϫ 3͒ ͞n. Then

͑x Ϫ 3͒n

converge?

n

n

Ϳ Ϳ Ϳ

a nϩ1

͑x Ϫ 3͒ nϩ1

n

ؒ

an

nϩ1

͑x Ϫ 3͒ n

1

1

n

1ϩ

Խx Ϫ 3Խ

Ϳ

Խ

l xϪ3

Խ

as n l ϱ

By the Ratio Test, the given series is absolutely convergent, and therefore convergent,

when x Ϫ 3 Ͻ 1 and divergent when x Ϫ 3 Ͼ 1. Now

Խ

Խ

Խ

Խx Ϫ 3Խ Ͻ 1

&?

Խ

Ϫ1 Ͻ x Ϫ 3 Ͻ 1

2ϽxϽ4

&?

so the series converges when 2 Ͻ x Ͻ 4 and diverges when x Ͻ 2 or x Ͼ 4.

The Ratio Test gives no information when x Ϫ 3 1 so we must consider x 2

and x 4 separately. If we put x 4 in the series, it becomes 1͞n, the harmonic

series, which is divergent. If x 2, the series is ͑Ϫ1͒ n͞n , which converges by the

Alternating Series Test. Thus the given power series converges for 2 ഛ x Ͻ 4.

National Film Board of Canada

Խ

Խ

We will see that the main use of a power series is that it provides a way to represent

some of the most important functions that arise in mathematics, physics, and chemistry. In

particular, the sum of the power series in the next example is called a Bessel function, after

the German astronomer Friedrich Bessel (1784–1846), and the function given in Exercise 35

is another example of a Bessel function. In fact, these functions first arose when Bessel

solved Kepler’s equation for describing planetary motion. Since that time, these functions

have been applied in many different physical situations, including the temperature distribution in a circular plate and the shape of a vibrating drumhead.

EXAMPLE 3 Find the domain of the Bessel function of order 0 defined by

J0͑x͒

ϱ

͚

n0

͑Ϫ1͒ n x 2n

2 2n͑n!͒2

SOLUTION Let a n ͑Ϫ1͒ n x 2n͓͞2 2n͑n!͒2 ͔. Then

Notice how closely the computer-generated

model (which involves Bessel functions and

cosine functions) matches the photograph of a

vibrating rubber membrane.

Ϳ Ϳ Ϳ

a nϩ1

͑Ϫ1͒ nϩ1x 2͑nϩ1͒

2 2n͑n!͒2

2͑nϩ1͒

ؒ

2

an

2

͓͑n ϩ 1͒!͔

͑Ϫ1͒ nx 2n

x 2nϩ2

2 2n͑n!͒2

ؒ

2 2nϩ2͑n ϩ 1͒2͑n!͒2

x 2n

x2

l 0Ͻ1

4͑n ϩ 1͒2

Ϳ

for all x

Thus, by the Ratio Test, the given series converges for all values of x. In other words,

the domain of the Bessel function J0 is ͑Ϫϱ, ϱ͒ ޒ.

Copyright 2010 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.

97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 767

SECTION 11.8

y

sá

n

J0x lim snx

sÂ

0

nl

x

1

sĂ sÊ

767

Recall that the sum of a series is equal to the limit of the sequence of partial sums. So

when we define the Bessel function in Example 3 as the sum of a series we mean that, for

every real number x,

s™

1

POWER SERIES

sn͑x͒

where

͚

i0

͑Ϫ1͒ix 2i

2 2i͑i!͒2

The first few partial sums are

J¸

s0͑x͒ 1

s1͑x͒ 1 Ϫ

FIGURE 1

Partial sums of the Bessel function J¸

s3͑x͒ 1 Ϫ

y

x2

x4

x6

ϩ

Ϫ

4

64

2304

x2

4

s2͑x͒ 1 Ϫ

s4͑x͒ 1 Ϫ

x2

x4

ϩ

4

64

x2

x4

x6

x8

ϩ

Ϫ

ϩ

4

64

2304

147,456

1

y=J¸(x)

_10

10

0

x

FIGURE 2

Figure 1 shows the graphs of these partial sums, which are polynomials. They are all approximations to the function J0 , but notice that the approximations become better when more

terms are included. Figure 2 shows a more complete graph of the Bessel function.

For the power series that we have looked at so far, the set of values of x for which the

series is convergent has always turned out to be an interval [a finite interval for the geometric

series and the series in Example 2, the infinite interval ͑Ϫϱ, ϱ͒ in Example 3, and a collapsed interval ͓0, 0͔ ͕0͖ in Example 1]. The following theorem, proved in Appendix F,

says that this is true in general.

ϱ

Theorem For a given power series

͚ c ͑x Ϫ a͒

n

there are only three

possibilities:

(i) The series converges only when x a.

(ii) The series converges for all x.

(iii) There is a positive number R such that the series converges if x Ϫ a Ͻ R

and diverges if x Ϫ a Ͼ R.

3

n

n0

Խ

Խ

Խ

Խ

The number R in case (iii) is called the radius of convergence of the power series. By

convention, the radius of convergence is R 0 in case (i) and R ϱ in case (ii). The interval of convergence of a power series is the interval that consists of all values of x for which

the series converges. In case (i) the interval consists of just a single point a. In case (ii) the

interval is ͑Ϫϱ, ϱ͒. In case (iii) note that the inequality x Ϫ a Ͻ R can be rewritten as

a Ϫ R Ͻ x Ͻ a ϩ R. When x is an endpoint of the interval, that is, x a Ϯ R, anything

can happen—the series might converge at one or both endpoints or it might diverge at both

endpoints. Thus in case (iii) there are four possibilities for the interval of convergence:

Խ

͑a Ϫ R, a ϩ R͒

͑a Ϫ R, a ϩ R͔

͓a Ϫ R, a ϩ R͒

Խ

͓a Ϫ R, a ϩ R͔

The situation is illustrated in Figure 3.

convergence for |x-a|

a-R

FIGURE 3

a

a+R

divergence for |x-a|>R

Copyright 2010 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.

97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 768

768

CHAPTER 11

INFINITE SEQUENCES AND SERIES

We summarize here the radius and interval of convergence for each of the examples

already considered in this section.

Series

Radius of convergence

Interval of convergence

R1

͑Ϫ1, 1͒

n

R0

͕0͖

͑x Ϫ 3͒n

n

R1

͓2, 4͒

͑Ϫ1͒n x 2n

2 2n͑n!͒2

Rϱ

͑Ϫϱ, ϱ͒

ϱ

Geometric series

͚x

n

n0

ϱ

Example 1

͚ n! x

n0

ϱ

Example 2

͚

n1

ϱ

Example 3

͚

n0

In general, the Ratio Test (or sometimes the Root Test) should be used to determine the

radius of convergence R. The Ratio and Root Tests always fail when x is an endpoint of the

interval of convergence, so the endpoints must be checked with some other test.

EXAMPLE 4 Find the radius of convergence and interval of convergence of the series

ϱ

͚

n0

͑Ϫ3͒ n x n

sn ϩ 1

SOLUTION Let a n ͑Ϫ3͒ n x n͞sn ϩ 1. Then

Ϳ Ϳ Ϳ

Ϳ Ϳ ͱ Ϳ

a nϩ1

͑Ϫ3͒ nϩ1x nϩ1 sn ϩ 1

ؒ

Ϫ3x

an

͑Ϫ3͒ nx n

sn ϩ 2

ͱ

3

1 ϩ ͑1͞n͒

x l 3 x

1 ϩ ͑2͞n͒

Խ Խ

Խ Խ

Խ Խ

nϩ1

nϩ2

as n l ϱ

Խ Խ

By the Ratio Test, the given series converges if 3 x Ͻ 1 and diverges if 3 x Ͼ 1.

Thus it converges if x Ͻ 13 and diverges if x Ͼ 13 . This means that the radius of convergence is R 13 .

We know the series converges in the interval (Ϫ 13 , 13 ), but we must now test for convergence at the endpoints of this interval. If x Ϫ 13 , the series becomes

Խ Խ

ϱ

͚

n0

Խ Խ

n

ϱ

͑Ϫ3͒ n (Ϫ13 )

1

1

1

1

1

͚

ϩ

ϩ

ϩ

ϩ иии

ϩ

1

ϩ

1

n0

sn

sn

s1

s2

s3

s4

which diverges. (Use the Integral Test or simply observe that it is a p-series with

p 12 Ͻ 1.) If x 13 , the series is

ϱ

͚

n0

n

ϱ

͑Ϫ3͒ n ( 13 )

͑Ϫ1͒ n

͚

n0 sn ϩ 1

sn ϩ 1

which converges by the Alternating Series Test. Therefore the given power series converges when Ϫ13 Ͻ x ഛ 13 , so the interval of convergence is (Ϫ13 , 13 ].

Copyright 2010 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.

97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 769

SECTION 11.8

v

POWER SERIES

769

EXAMPLE 5 Find the radius of convergence and interval of convergence of the series

ϱ

͚

n0

n͑x ϩ 2͒ n

3 nϩ1

SOLUTION If a n n͑x ϩ 2͒ n͞3 nϩ1, then

Ϳ Ϳ Ϳ

ͩ ͪԽ

a nϩ1

͑n ϩ 1͒͑x ϩ 2͒ nϩ1

3 nϩ1

ؒ

an

3 nϩ2

n͑x ϩ 2͒ n

1ϩ

1

n

xϩ2

3

Խ

l

Ϳ

Խx ϩ 2Խ

as n l ϱ

3

Խ

Խ

Using the Ratio Test, we see that the series converges if x ϩ 2 ͞3 Ͻ 1 and it diverges

if x ϩ 2 ͞3 Ͼ 1. So it converges if x ϩ 2 Ͻ 3 and diverges if x ϩ 2 Ͼ 3. Thus the

radius of convergence is R 3.

The inequality x ϩ 2 Ͻ 3 can be written as Ϫ5 Ͻ x Ͻ 1, so we test the series at

the endpoints Ϫ5 and 1. When x Ϫ5, the series is

Խ

Խ

Խ

Խ

Խ

Խ

Խ

Խ

ϱ

͚

n0

ϱ

n͑Ϫ3͒ n

1

n

3 ͚ ͑Ϫ1͒ n

3 nϩ1

n0

which diverges by the Test for Divergence [͑Ϫ1͒nn doesn’t converge to 0]. When x 1,

the series is

ϱ

͚

n0

ϱ

n͑3͒ n

1

3 ͚ n

3 nϩ1

n0

which also diverges by the Test for Divergence. Thus the series converges only when

Ϫ5 Ͻ x Ͻ 1, so the interval of convergence is ͑Ϫ5, 1͒.

11.8

Exercises

ϱ

1. What is a power series?

2. (a) What is the radius of convergence of a power series?

How do you find it?

(b) What is the interval of convergence of a power series?

How do you find it?

3–28 Find the radius of convergence and interval of convergence

of the series.

͚ ͑Ϫ1͒ nx

n

ϱ

n

4.

n1

ϱ

5.

͚

n1

ϱ

7.

͚

n0

;

n

x

2n Ϫ 1

6.

͚

͑Ϫ1͒ nx n

3

n1

sn

ϱ

͑Ϫ1͒ x

n2

͚

n1

n

8.

͚n

ϱ

11.

͚

n1

13.

͚ ͑Ϫ1͒

ϱ

15.

n1

Graphing calculator or computer required

͚

n0

ϱ

n

17.

͚

n1

xn

19.

͚

n1

ϱ

n2 xn

2n

10.

ϱ

12.

x

4 ln n

n

3 ͑x ϩ 4͒

sn

͑x Ϫ 2͒

nn

CAS Computer algebra system required

10 n x n

n3

xn

n3 n

ϱ

14.

͚ ͑Ϫ1͒

n

n0

ϱ

͑x Ϫ 2͒ n

n2 ϩ 1

n

͚

n1

n

n

͚

n1

͑Ϫ3͒ n n

x

nsn

ϱ

ϱ

n

n

n1

ϱ

n

x

n!

͚ ͑Ϫ1͒

n2

ϱ

3.

9.

16.

͚ ͑Ϫ1͒

n0

ϱ

n

18.

͚

n1

ϱ

n

20.

͚

n1

n

x 2nϩ1

͑2n ϩ 1͒!

͑x Ϫ 3͒ n

2n ϩ 1

n

͑x ϩ 1͒ n

4n

͑2x Ϫ 1͒n

5 nsn

1. Homework Hints available at stewartcalculus.com

Copyright 2010 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.

97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 770

770

INFINITE SEQUENCES AND SERIES

CHAPTER 11

ϱ

21.

͚

n1

ϱ

22.

͚

n2

n

͑x Ϫ a͒ n ,

bn

35. The function J1 defined by

bϾ0

bn

͑x Ϫ a͒ n,

ln n

J1͑x͒

bϾ0

n0

ϱ

23.

͚ n!͑2x Ϫ 1͒

ϱ

n

25.

͚

n1

ϱ

27.

n1

ϱ

n

n2

ϱ

n! x n

1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒

͚

;

CAS

n

x

1 ؒ 3 ؒ 5 ؒ и и и ؒ ͑2n Ϫ 1͒

n1

x 2n

n͑ln n͒ 2

͚

26.

͚

n1

28.

͑5x Ϫ 4͒

n3

n2xn

2 ؒ 4 ؒ 6 ؒ и и и ؒ ͑2n͒

͚

24.

n1

ϱ

ϱ

n

ϱ

n

(b)

n0

͚ c ͑Ϫ4͒

n

n

;

n0

30. Suppose that ϱn0 cn x n converges when x Ϫ4 and diverges

when x 6. What can be said about the convergence or

divergence of the following series?

ϱ

(a)

ϱ

͚c

(b)

n

n0

͚ c ͑Ϫ3͒

n

͚c8

n

n

ϱ

n

(d)

n0

͚ ͑Ϫ1͒ c

n

n

9n

n0

31. If k is a positive integer, find the radius of convergence of

the series

ϱ

͚

n0

͑n!͒ k n

x

͑kn͒!

32. Let p and q be real numbers with p Ͻ q. Find a power series

whose interval of convergence is

(a) ͑ p, q͒

(b) ͑ p, q͔

(c) ͓ p, q͒

(d) ͓ p, q͔

33. Is it possible to find a power series whose interval of conver-

gence is ͓0, ϱ͒? Explain.

ϱ

n

; 34. Graph the first several partial sums sn͑x͒ of the series n0 x ,

together with the sum function f ͑x͒ 1͑͞1 Ϫ x͒, on a common screen. On what interval do these partial sums appear to

be converging to f ͑x͒?

11.9

CAS

x6

x9

x3

ϩ

ϩ

ϩ иии

2и3

2и3и5и6

2и3и5и6и8и9

is called an Airy function after the English mathematician

and astronomer Sir George Airy (1801–1892).

(a) Find the domain of the Airy function.

(b) Graph the first several partial sums on a common screen.

(c) If your CAS has built-in Airy functions, graph A on the

same screen as the partial sums in part (b) and observe

how the partial sums approximate A.

37. A function f is defined by

f ͑x͒ 1 ϩ 2x ϩ x 2 ϩ 2x 3 ϩ x 4 ϩ и и и

n0

ϱ

(c)

is called the Bessel function of order 1.

(a) Find its domain.

(b) Graph the first several partial sums on a common

screen.

(c) If your CAS has built-in Bessel functions, graph J1 on the

same screen as the partial sums in part (b) and observe

how the partial sums approximate J1.

A͑x͒ 1 ϩ

series are convergent?

͚ c ͑Ϫ2͒

͑Ϫ1͒ n x 2nϩ1

n!͑n ϩ 1͒! 2 2nϩ1

36. The function A defined by

29. If ϱn0 cn 4 n is convergent, does it follow that the following

(a)

ϱ

͚

that is, its coefficients are c2n 1 and c2nϩ1 2 for all

n ജ 0. Find the interval of convergence of the series and find

an explicit formula for f ͑x͒.

38. If f ͑x͒

ϱn0 cn x n, where cnϩ4 cn for all n ജ 0, find the

interval of convergence of the series and a formula for f ͑x͒.

Խ Խ

n

39. Show that if lim n l ϱ s

cn c , where c

0, then the

radius of convergence of the power series cn x n is R 1͞c.

40. Suppose that the power series cn ͑ x Ϫ a͒ n satisfies c n

0

for all n. Show that if lim n l ϱ cn ͞cnϩ1 exists, then it is equal

to the radius of convergence of the power series.

Խ

Խ

41. Suppose the series cn x n has radius of convergence 2 and

the series dn x n has radius of convergence 3. What is the

radius of convergence of the series ͑cn ϩ dn͒x n ?

42. Suppose that the radius of convergence of the power series

cn x n is R. What is the radius of convergence of the power

series cn x 2n ?

Representations of Functions as Power Series

In this section we learn how to represent certain types of functions as sums of power series

by manipulating geometric series or by differentiating or integrating such a series. You might

wonder why we would ever want to express a known function as a sum of infinitely many

terms. We will see later that this strategy is useful for integrating functions that don’t have

elementary antiderivatives, for solving differential equations, and for approximating func-

Copyright 2010 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.

97817_11_ch11_p762-771.qk_97817_11_ch11_p762-771 11/3/10 5:31 PM Page 771

SECTION 11.9

REPRESENTATIONS OF FUNCTIONS AS POWER SERIES

771

tions by polynomials. (Scientists do this to simplify the expressions they deal with; computer

scientists do this to represent functions on calculators and computers.)

We start with an equation that we have seen before:

1

A geometric illustration of Equation 1 is shown

in Figure 1. Because the sum of a series is the

limit of the sequence of partial sums, we have

where

1

lim sn͑x͒

nlϱ

1Ϫx

ϱ

1

1 ϩ x ϩ x2 ϩ x3 ϩ и и и ͚ xn

1Ϫx

n0

ԽxԽ Ͻ 1

We first encountered this equation in Example 6 in Section 11.2, where we obtained it by

observing that the series is a geometric series with a 1 and r x. But here our point of

view is different. We now regard Equation 1 as expressing the function f ͑x͒ 1͑͞1 Ϫ x͒

as a sum of a power series.

s¡¡

y

sˆ

s∞

sn͑x͒ 1 ϩ x ϩ x 2 ϩ и и и ϩ x n

f

is the nth partial sum. Notice that as n

increases, sn͑x͒ becomes a better approximation to f ͑x͒ for Ϫ1 Ͻ x Ͻ 1.

s™

FIGURE 1

0

_1

1

ƒ=

and some partial sums

1-x

v

x

1

EXAMPLE 1 Express 1͑͞1 ϩ x 2 ͒ as the sum of a power series and find the interval of

convergence.

SOLUTION Replacing x by Ϫx 2 in Equation 1, we have

ϱ

1

1

͚ ͑Ϫx 2 ͒n

2

2

1ϩx

1 Ϫ ͑Ϫx ͒

n0

ϱ

͚ ͑Ϫ1͒ x

n 2n

1 Ϫ x2 ϩ x4 Ϫ x6 ϩ x8 Ϫ и и и

n0

Խ

Խ

Because this is a geometric series, it converges when Ϫx 2 Ͻ 1, that is, x 2 Ͻ 1, or

x Ͻ 1. Therefore the interval of convergence is ͑Ϫ1, 1͒. (Of course, we could have

determined the radius of convergence by applying the Ratio Test, but that much work is

unnecessary here.)

Խ Խ

EXAMPLE 2 Find a power series representation for 1͑͞x ϩ 2͒.

SOLUTION In order to put this function in the form of the left side of Equation 1, we first

factor a 2 from the denominator:

1

2ϩx

Խ

1

1

ͩ ͪ ͫ ͩ ͪͬ

͚ͩ ͪ ͚

x

2 1ϩ

2

1

2

ϱ

n0

Խ

Ϫ

2 1Ϫ Ϫ

x

2

n

ϱ

n0

x

2

͑Ϫ1͒n n

x

2 nϩ1

Խ Խ

This series converges when Ϫx͞2 Ͻ 1, that is, x Ͻ 2. So the interval of convergence is ͑Ϫ2, 2͒.

Copyright 2010 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.

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