Section 9.8 Power Series Power Seriesstaff.katyisd.org/sites/thscalculusap/Larson... · 660 CHAPTER...
Transcript of Section 9.8 Power Series Power Seriesstaff.katyisd.org/sites/thscalculusap/Larson... · 660 CHAPTER...
SECTION 9.8 Power Series 659
Section 9.8 Power Series
• Understand the definition of a power series.• Find the radius and interval of convergence of a power series.• Determine the endpoint convergence of a power series.• Differentiate and integrate a power series.
Power Series
In Section 9.7, you were introduced to the concept of approximating functions byTaylor polynomials. For instance, the function can be approximated by itsMaclaurin polynomials as follows.
1st-degree polynomial
2nd-degree polynomial
3rd-degree polynomial
4th-degree polynomial
5th-degree polynomial
In that section, you saw that the higher the degree of the approximating polynomial,the better the approximation becomes.
In this and the next two sections, you will see that several important types offunctions, including
can be represented exactly by an infinite series called a power series. For example,the power series representation for is
For each real number it can be shown that the infinite series on the right convergesto the number Before doing this, however, some preliminary results dealing withpower series will be discussed—beginning with the following definition.
NOTE To simplify the notation for power series, we agree that even if x � c.�x � c�0 � 1,
ex.x,
ex � 1 � x �x2
2!�
x3
3!� . . . �
xn
n!� . . . .
ex
f�x� � ex
ex � 1 � x �x2
2!�
x3
3!�
x4
4!�
x5
5!
ex � 1 � x �x2
2!�
x3
3!�
x4
4!
ex � 1 � x �x2
2!�
x3
3!
ex � 1 � x �x2
2!
ex � 1 � x
f�x� � ex
Definition of Power Series
If is a variable, then an infinite series of the form
is called a power series. More generally, an infinite series of the form
is called a power series centered at c, where is a constant.c
��
n�0 an�x � c�n � a0 � a1�x � c� � a2�x � c�2 � . . . � an�x � c�n � . . .
��
n�0 anxn � a0 � a1x � a2x
2 � a3x3 � . . . � anxn � . . .
x
E X P L O R A T I O N
Graphical Reasoning Use a graphingutility to approximate the graph ofeach power series near (Usethe first several terms of each series.)Each series represents a well-knownfunction. What is the function?
a.
b.
c.
d.
e. ��
n�0 2n xn
n!
��
n�0 ��1�n x2n�1
2n � 1
��
n�0 ��1�n x2n�1
�2n � 1�!
��
n�0 ��1�n x2n
�2n�!
��
n�0 ��1�nxn
n!
x � 0.
332460_0908.qxd 11/4/04 3:30 PM Page 659
660 CHAPTER 9 Infinite Series
EXAMPLE 1 Power Series
a. The following power series is centered at 0.
b. The following power series is centered at
c. The following power series is centered at 1.
Radius and Interval of Convergence
A power series in can be viewed as a function of
where the domain of is the set of all for which the power series converges.Determination of the domain of a power series is the primary concern in this section.Of course, every power series converges at its center because
So, always lies in the domain of The following important theorem states that thedomain of a power series can take three basic forms: a single point, an intervalcentered at or the entire real line, as shown in Figure 9.17. A proof is given inAppendix A.
c,
f.c
� a0.
� a0�1� � 0 � 0 � . . . � 0 � . . .
f�c� � ��
n�0 an�c � c�n
c
xf
f�x� � ��
n�0 an�x � c�n
xx
��
n�1 1n
�x � 1�n � �x � 1� �12
�x � 1�2 �13
�x � 1�3 � . . .
��
n�0 ��1�n �x � 1�n � 1 � �x � 1� � �x � 1�2 � �x � 1�3 � . . .
�1.
��
n�0 xn
n!� 1 � x �
x2
2�
x3
3!� . . .
cx
A single point
The domain of a power series has only three basic forms: a single point, an intervalcentered at or the entire real line.Figure 9.17
c,
cx
R R
An interval
cx
The real line
THEOREM 9.20 Convergence of a Power Series
For a power series centered at precisely one of the following is true.
1. The series converges only at
2. There exists a real number such that the series converges absolutely forand diverges for
3. The series converges absolutely for all
The number is the radius of convergence of the power series. If the seriesconverges only at the radius of convergence is and if the seriesconverges for all the radius of convergence is The set of all values of
for which the power series converges is the interval of convergence of thepower series.x
R � �.x,R � 0,c,
R
x.�x � c� > R.�x � c� < R,
R > 0
c.
c,
332460_0908.qxd 11/4/04 3:30 PM Page 660
SECTION 9.8 Power Series 661
EXAMPLE 2 Finding the Radius of Convergence
Find the radius of convergence of
Solution For you obtain
For any fixed value of such that let Then
Therefore, by the Ratio Test, the series diverges for and converges only at itscenter, 0. So, the radius of convergence is
EXAMPLE 3 Finding the Radius of Convergence
Find the radius of convergence of
Solution For let Then
By the Ratio Test, the series converges if and diverges if Therefore, the radius of convergence of the series is
EXAMPLE 4 Finding the Radius of Convergence
Find the radius of convergence of
Solution Let Then
For any value of this limit is 0. So, by the Ratio Test, the series converges forall Therefore, the radius of convergence is R � �.x.
x,fixed
� limn→�
x2
�2n � 3��2n � 2�.
limn→�
�un�1
un � � limn→�
� ��1�n�1 x2n�3
�2n � 3�!
��1�n x2n�1 �2n � 1�! �
un � ��1�nx2n�1��2n � 1�!.
��
n�0 ��1�nx2n�1
�2n � 1�! .
R � 1.�x � 2� > 1.�x � 2� < 1
� �x � 2�. � lim
n→� �x � 2�
limn→�
�un�1
un � � limn→�
�3�x � 2�n�1
3�x � 2�n �un � 3�x � 2�n.x � 2,
��
n�0 3�x � 2�n.
R � 0.�x� > 0
� �.
� �x� limn→�
�n � 1�
limn→�
�un�1
un � � limn→�
��n � 1�!xn�1
n!xn �un � n!xn.�x� > 0,x
f�0� � ��
n�0 n!0n � 1 � 0 � 0 � . . . � 1.
x � 0,
��
n�0 n!xn.
STUDY TIP To determine the radius ofconvergence of a power series, use theRatio Test, as demonstrated in Examples2, 3, and 4.
332460_0908_pg 661.qxd 6/9/05 8:38 AM Page 661
662 CHAPTER 9 Infinite Series
Endpoint Convergence
Note that for a power series whose radius of convergence is a finite number Theorem 9.20 says nothing about the convergence at the endpoints of the interval ofconvergence. Each endpoint must be tested separately for convergence or divergence.As a result, the interval of convergence of a power series can take any one of the sixforms shown in Figure 9.18.
EXAMPLE 5 Finding the Interval of Convergence
Find the interval of convergence of
Solution Letting produces
So, by the Ratio Test, the radius of convergence is Moreover, because theseries is centered at 0, it converges in the interval This interval,however, is not necessarily the interval of convergence. To determine this, you musttest for convergence at each endpoint. When you obtain the divergent harmonicseries
Diverges when
When you obtain the convergent alternating harmonic series
Converges when
So, the interval of convergence for the series is as shown in Figure 9.19.��1, 1�,
x � �1��
n�1 ��1�n
n� �1 �
12
�13
�14
� . . . .
x � �1,
x � 1��
n�1 1n
�11
�12
�13
� . . . .
x � 1,
��1, 1�.R � 1.
� �x�. � lim
n→� � nx
n � 1� limn→�
�un�1
un � � limn→�
� xn�1
�n � 1�
xn
n �un � xn�n
��
n�1 xn
n.
R,
Radius: R = 1
c = 0x
−1 1
Interval: [−1, 1)
Figure 9.19
cx
Radius: 0
cx
R
(c − R, c + R)
RRadius:
cx
(c − R, c + R]
R
cx
[c − R, c + R)
R
cx
[c − R, c + R]
R
Intervals of convergenceFigure 9.18
Radius: �
cx
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SECTION 9.8 Power Series 663
EXAMPLE 6 Finding the Interval of Convergence
Find the interval of convergence of
Solution Letting produces
By the Ratio Test, the series converges if or So, theradius of convergence is Because the series is centered at it willconverge in the interval Furthermore, at the endpoints you have
Diverges when
and
Diverges when
both of which diverge. So, the interval of convergence is as shown in Figure9.20.
EXAMPLE 7 Finding the Interval of Convergence
Find the interval of convergence of
Solution Letting produces
So, the radius of convergence is Because the series is centered at itconverges in the interval When you obtain the convergent series
Converges when
When you obtain the convergent alternating series
Converges when
Therefore, the interval of convergence for the given series is ��1, 1�.
x � �1��
n�1 ��1�n
n2 � �112 �
122 �
132 �
142 � . . . .
x � �1,
x � 1��
n�1 1n2 �
112 �
122 �
132 �
142 � . . . .
p-x � 1,��1, 1�.x � 0,R � 1.
� limn→�
� n2x�n � 1�2� � �x�.
limn→�
�un�1
un � � limn→�
�xn�1��n � 1�2
xn�n2 �un � xn�n2
��
n�1 xn
n2.
��3, 1�,
x � 1��
n�0 ��1�n�2�n
2n � ��
n�0 ��1�n
x � �3��
n�0 ��1�n ��2�n
2n � ��
n�0 2n
2n � ��
n�0 1
��3, 1�.x � �1,R � 2.
�x � 1� < 2.��x � 1��2� < 1
� �x � 12 �.
� limn→�
�2n�x � 1�2n�1 �
limn→�
�un�1
un � � limn→�
� ��1�n�1�x � 1�n�1
2n�1
��1�n�x � 1�n
2n �un � ��1�n�x � 1�n�2n
��
n�0 ��1�n�x � 1�n
2n .
Radius: R = 2
c = −1x
−3 −2 10
Interval: (−3, 1)
Figure 9.20
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664 CHAPTER 9 Infinite Series
Differentiation and Integration of Power Series
Power series representation of functions has played an important role in the develop-ment of calculus. In fact, much of Newton’s work with differentiation and integrationwas done in the context of power series—especially his work with complicatedalgebraic functions and transcendental functions. Euler, Lagrange, Leibniz, and theBernoullis all used power series extensively in calculus.
Once you have defined a function with a power series, it is natural to wonder howyou can determine the characteristics of the function. Is it continuous? Differentiable?Theorem 9.21, which is stated without proof, answers these questions.
Theorem 9.21 states that, in many ways, a function defined by a power seriesbehaves like a polynomial. It is continuous in its interval of convergence, and both itsderivative and its antiderivative can be determined by differentiating and integratingeach term of the given power series. For instance, the derivative of the power series
is
Notice that Do you recognize this function?f��x� � f�x�.
� f�x�.
� 1 � x �x2
2�
x3
3!�
x4
4!� . . .
f��x� � 1 � �2� x2
� �3� x2
3!� �4� x
3
4!� . . .
� 1 � x �x2
2�
x3
3!�
x4
4!� . . .
f�x� � ��
n�0 xn
n!
THEOREM 9.21 Properties of Functions Defined by Power Series
If the function given by
has a radius of convergence of then, on the interval isdifferentiable (and therefore continuous). Moreover, the derivative and anti-derivative of are as follows.
1.
2.
The radius of convergence of the series obtained by differentiating or integratinga power series is the same as that of the original power series. The interval ofconvergence, however, may differ as a result of the behavior at the endpoints.
� C � a0�x � c� � a1 �x � c�2
2� a2
�x � c�3
3� . . .
f �x� dx � C � ��
n�0 an
�x � c�n�1
n � 1
� a1 � 2a2�x � c� � 3a3�x � c�2 � . . .
f��x� � ��
n�1 nan�x � c�n�1
f
f�c � R, c � R�,R > 0,
� a0 � a1�x � c� � a2�x � c�2 � a3�x � c�3 � . . .
f�x� � ��
n�0 an�x � c�n
JAMES GREGORY (1638–1675)
One of the earliest mathematicians to workwith power series was a Scotsman, JamesGregory. He developed a power series methodfor interpolating table values—a method thatwas later used by Brook Taylor in the develop-ment of Taylor polynomials and Taylor series.
The
Gra
nger
Col
lect
ion
332460_0908.qxd 11/4/04 3:30 PM Page 664
SECTION 9.8 Power Series 665
EXAMPLE 8 Intervals of Convergence for and
Consider the function given by
Find the intervals of convergence for each of the following.
a. b. c.
Solution By Theorem 9.21, you have
and
By the Ratio Test, you can show that each series has a radius of convergence of Considering the interval you have the following.
a. For the series
Interval of convergence:
converges for and its interval of convergence is See Figure9.21(a).
b. For the series
Interval of convergence:
converges for and diverges for So, its interval of convergence isSee Figure 9.21(b).
c. For the series
Interval of convergence:
diverges for and its interval of convergence is See Figure 9.21(c).
From Example 8, it appears that of the three series, the one for the derivative,is the least likely to converge at the endpoints. In fact, it can be shown that if the
series for converges at the endpoints the series for will also converge there.
f�x�x � c ± R,f��x�f��x�,
��1, 1�.x � ±1,
��1, 1���
n�1 xn�1
f��x�,��1, 1�.
x � 1.x � �1
��1, 1���
n�1 xn
n
f�x�,
��1, 1�.x � ±1,
��1, 1���
n�1
xn�1
n�n � 1�
f�x� dx,
��1, 1�,R � 1.
� C �x2
1 � 2�
x3
2 � 3�
x4
3 � 4� . . . .
f �x� dx � C � ��
n�1
xn�1
n�n � 1�
� 1 � x � x2 � x3 � . . .
f��x� � ��
n�1 xn�1
f��x�f�x� f�x� dx
f�x� � ��
n�1 xn
n� x �
x2
2�
x3
3� . . . .
f�x� dxf��x�,f�x�,
Radius: R = 1
c = 0
x
−1 1
Interval: [−1, 1]Radius: R = 1
c = 0
x
−1 1
Interval: [−1, 1)Radius: R = 1
c = 0
x
−1 1
Interval: (−1, 1)
(a) (b) (c)Figure 9.21
332460_0908.qxd 11/4/04 3:30 PM Page 665
666 CHAPTER 9 Infinite Series
E x e r c i s e s f o r S e c t i o n 9 . 8 See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
In Exercises 1–4, state where the power series is centered.
1. 2.
3. 4.
In Exercises 5–10, find the radius of convergence of the powerseries.
5. 6.
7. 8.
9. 10.
In Exercises 11–34, find the interval of convergence of thepower series. (Be sure to include a check for convergence at theendpoints of the interval.)
11. 12.
13. 14.
15. 16.
17. 18.
19. 20.
21. 22.
23. 24.
25. 26.
27. 28.
29. 30.
31.
32.
33.
34.
In Exercises 35 and 36, find the radius of convergence of thepower series, where and is a positive integer.
35. 36.
In Exercises 37– 40, find the interval of convergence of thepower series. (Be sure to include a check for convergence at theendpoints of the interval.)
37. 38.
39.
40.
In Exercises 41–44, write an equivalent series with the index ofsummation beginning at
41. 42.
43. 44.
In Exercises 45–48, find the intervals of convergence of (a) (b) (c) and (d) Include a check for convergence at the endpoints of the interval.
45.
46.
47.
48.
Writing In Exercises 49–52, match the graph of the first 10terms of the sequence of partial sums of the series
with the indicated value of the function. [The graphs are labeled(a), (b), (c), and (d).] Explain how you made your choice.
(a) (b) Sn
n2
2468
1012
4 6 8
Sn
n2
2
4 6 8
1
3
g�x� � ��
n�0 x
3�n
f �x� � ��
n�1 ��1�n�1�x � 2�n
n
f �x� � ��
n�0 ��1�n�1�x � 1�n�1
n � 1
f �x� � ��
n�1 ��1�n�1 �x � 5�n
n5n
f �x� � ��
n�0 �x
2�n
f �x� dx.f � �x�,f��x�,f �x�,
��
n�0 ��1�n x2n�1
2n � 1��
n�0
x2n�1
�2n � 1�!
��
n�0 ��1�n�1�n � 1�xn�
�
n�0 xn
n!
n � 1.
��
n�1
n!�x � c�n
1 � 3 � 5 . . . �2n � 1�
k ≥ 1��
n�1 k�k � 1��k � 2� . . . �k � n � 1�xn
n!,
��
n�1 ��1�n�1�x � c�n
ncnk > 0��
n�0 �x
k�n
,
��
n�0 �n!�k xn
�kn�!��
n�1 �x � c�n�1
c n�1
kc > 0
��
n�1
n!�x � 1�n
1 � 3 � 5 . . . �2n � 1�
��
n�1 ��1�n�1 3 � 7 � 11 . . . �4n � 1��x � 3�n
4n
��
n�1 � 2 � 4 � 6 . . . 2n
3 � 5 � 7 . . . �2n � 1�� x2n�1
��
n�1 2 � 3 � 4 . . . �n � 1�xn
n!
��
n�1 n!xn
�2n�!��
n�0
x2n�1
�2n � 1�!
��
n�0 ��1�n x2n
n!��
n�1
nn � 1
��2x�n�1
��
n�0 ��1�n x2n�1
2n � 1��
n�1 �x � 3�n�1
3n�1
��
n�1 (�1�n�1�x � 2�n
n2n��
n�0 ��1�n�1�x � 1�n�1
n � 1
��
n�0
�x � 2�n�1
�n � 1�4n�1��
n�1 ��1�n�1�x � 5�n
n5n
��
n�0 ��1�n n!�x � 4�n
3n��
n�1 ��1�n�1xn
4n
��
n�0
��1�n xn
�n � 1��n � 2���
n�0 �2n�! �x
2�n
��
n�0 �3x�n
�2n�!��
n�0 xn
n!
��
n�0 ��1�n�1�n � 1�xn�
�
n�1 ��1�n xn
n
��
n�0 �x
5�n
��
n�0 �x
2�n
��
n�0 �2n�!x2n
n!��
n�0 �2x�2n
�2n�!
��
n�0 ��1�n xn
2n��
n�1 �2x�n
n2
��
n�0 �2x�n�
�
n�0 ��1�n
xn
n � 1
��
n�0
��1�n�x � ��2n
�2n�!��
n�1
�x � 2�n
n3
��
n�1
��1�n1 � 3 . . . �2n � 1�2nn!
xn��
n�0nxn
332460_0908.qxd 11/4/04 3:30 PM Page 666
SECTION 9.8 Power Series 667
(c) (d)
49. 50.
51. 52.
Writing In Exercises 53–56, match the graph of the first 10terms of the sequence of partial sums of the series
with the indicated value of the function. [The graphs are labeled(a), (b), (c), and (d).] Explain how you made your choice.
(a) (b)
(c) (d)
53. 54.
55. 56.
63. Let and
(a) Find the intervals of convergence of and
(b) Show that
(c) Show that
(d) Identify the functions and
64. Let
(a) Find the interval of convergence of
(b) Show that
(c) Show that
(d) Identify the function
In Exercises 65–70, show that the function represented by thepower series is a solution of the differential equation.
65.
66.
67.
68.
69.
70.
71. Bessel Function The Bessel function of order 0 is
(a) Show that the series converges for all
(b) Show that the series is a solution of the differential equation
(c) Use a graphing utility to graph the polynomial composed ofthe first four terms of
(d) Approximate accurate to two decimal places.10 J0 dx
J0.
x2 J0 � x J0� � x2 J0 � 0.
x.
J0�x� � ��
k�0 ��1�k x2k
22k �k!�2 .
y � x 2 y � 0y � 1 � ��
n�1
��1�n x 4n
22n n! � 3 � 7 � 11 . . . �4n � 1�,
y � xy� � y � 0y � ��
n�0
x2n
2n n!,
y � y � 0y � ��
n�0
x2n
�2n�!,
y � y � 0y � ��
n�0
x2n�1
�2n � 1�!,
y � y � 0y � ��
n�0 ��1�n x2n
�2n�!
y � y � 0y � ��
n�0 ��1�n x2n�1
�2n � 1�! ,
f.
f �0� � 1.
f��x� � f �x�.f.
f �x� � ��
n�0 xn
n!.
g.f
g��x� � �f �x�.f��x� � g�x�.
g.f
g�x� � ��
n�0 ��1�n x2n
�2n�! .f �x� � ��
n�0 ��1�n x2n�1
�2n � 1�!
g�38�g� 9
16�
g��18�g�1
8�
1 2 3 4 5 6 7 8 9
3
6
9
12
15
18
Sn
n−1 1 2 3 4 5 6 7 8 9
0.25
0.50
0.75
1.00
Sn
n
Sn
n−1 1 2 3 4 5 6 7 8 9
0.5
1.0
1.5
2.0
1 2 3 4 5 6 7 8 9
1
2
3
4
Sn
n
g�x� � ��
n�0 �2x�n
g��2�g�3.1�g�2�g�1�
Sn
n2 4 6 8
1
1
1
3
2
4
4
Sn
n2
2
4 6 8
1
Writing About Concepts (continued)60. Describe how to differentiate and integrate a power series
with a radius of convergence Will the series resultingfrom the operations of differentiation and integration havea different radius of convergence? Explain.
61. Give examples that show that the convergence of a powerseries at an endpoint of its interval of convergence may beeither conditional or absolute. Explain your reasoning.
62. Write a power series that has the indicated interval of convergence. Explain your reasoning.
(a) (b) (c) (d) ��2, 6���1, 0���1, 1���2, 2�
R.
Writing About Concepts57. Define a power series centered at
58. Describe the radius of convergence of a power series.Describe the interval of convergence of a power series.
59. Describe the three basic forms of the domain of a powerseries.
c.
332460_0908.qxd 11/4/04 3:30 PM Page 667
668 CHAPTER 9 Infinite Series
72. Bessel Function The Bessel function of order 1 is
(a) Show that the series converges for all
(b) Show that the series is a solution of the differential equation
(c) Use a graphing utility to graph the polynomial composed ofthe first four terms of
(d) Show that
In Exercises 73–76, the series represents a well-known function.Use a computer algebra system to graph the partial sum andidentify the function from the graph.
73. 74.
75.
76.
77. Investigation In Exercise 11 you found that the interval of
convergence of the geometric series is
(a) Find the sum of the series when Use a graphing utility to graph the first six terms of the sequence of partialsums and the horizontal line representing the sum of theseries.
(b) Repeat part (a) for
(c) Write a short paragraph comparing the rate of convergenceof the partial sums with the sum of the series in parts (a)and (b). How do the plots of the partial sums differ as theyconverge toward the sum of the series?
(d) Given any positive real number there exists a positiveinteger such that the partial sum
Use a graphing utility to complete the table.
78. Investigation The interval of convergence of the series
is
(a) Find the sum of the series when Use a graphing utili-ty to graph the first six terms of the sequence of partial sumsand the horizontal line representing the sum of the series.
(b) Repeat part (a) for
(c) Write a short paragraph comparing the rate of convergenceof the partial sums with the sum of the series in parts (a)and (b). How do the plots of the partial sums differ as theyconverge toward the sum of the series?
(d) Given any positive real number M, there exists a positiveinteger N such that the partial sum
Use a graphing utility to complete the table.
True or False? In Exercises 79–82, determine whether thestatement is true or false. If it is false, explain why or give anexample that shows it is false.
79. If the power series converges for then it also
converges for
80. If the power series converges for then it also
converges for
81. If the interval of convergence for is then the
interval of convergence for is
82. If converges for then
83. Prove that the power series
has a radius of convergence of if p and q are positiveintegers.
84. Let where the coef-ficients are and for
(a) Find the interval of convergence of the series.
(b) Find an explicit formula for
85. Let where for
(a) Find the interval of convergence of the series.
(b) Find an explicit formula for
86. Prove that if the power series has a radius of conver-
gence of R, then has a radius of convergence of
87. For let and Prove that if the interval of
convergence of the series is
then the series converges conditionally at x0 � R.
�x0 � R, x0 � R�,��
n�0 cn�x � x0�n
cn > 0.R > 0n > 0,
�R.��
n�0 cn x2n
��
n�0 cn x n
f �x�.
n ≥ 0.cn�3 � cnf �x� � ��
n�0 cn xn,
g�x�.
n ≥ 0.c2n�1 � 2c2n � 1g�x� � 1 � 2x � x2 � 2x3 � x4 � . . . ,
R � �
��
n�0
�n � p�!n!�n � q�! xn
��
n�0
an
n � 1.1
0 f �x� dx �
�x� < 2,f �x� � ��
n�0 an x
n
�0, 2�.��
n�0 an �x � 1�n
��1, 1�,��
n�0 an x
n
x � �1.
x � 2,��
n�0 an x
n
x � �2.
x � 2,��
n�0 an x
n
�N
n�0 �3 �
23�
n
> M.
x � �16.
x �16.
��13, 13�.�
�
n�0 �3x�n
�N
n�0 �3
2�n
> M.
NM,
x � �34.
x �34.
��2, 2�.��
n�0 �x
2�n
�1 ≤ x ≤ 1f �x� � ��
n�0 ��1�n
x2n�1
2n � 1,
�1 < x < 1f �x� � ��
n�0 ��1�n xn,
f �x� � ��
n�0 ��1�n
x2n�1
�2n � 1�!f �x� � ��
n�0 ��1�n
x2n
�2n�!
S10
J0��x� � �J1�x�.J1.
x2 J1 � x J1� � �x2 � 1� J1 � 0.
x.
J1�x� � x ��
k�0
��1�k x2k
22k�1 k!�k � 1�!.
10 100 1000 10,000
N
M
10 100 1000 10,000
N
M
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