7: Strategy for Testing Series
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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.