SECTION 8.6
8.6
■ Find a power series representation for the function and
determine the interval of convergence.
1. f x x
1x
2. f x 1
4 x2
3. f x 1 x2
1 x2
4. f x 1
1 4x 2
5. f x 1
x 16
6. f x 7. f x 2
3x 4
4
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10 –11 ■ Find a power series representation for the function and
determine the radius of convergence.
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3x 2
2x 2 3x 1
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9. f x ■
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■
■
1x
1x
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Evaluate the indefinite integral as a power series.
1
12.
y 1x
14.
y
dx
4
13.
x
dx
1 x5
y
arctan x
dx
x
■
15.
■
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11. f x ln
■
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Use a power series to approximate the definite integral to
six decimal places.
x
x 2 3x 2
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15–16
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8. f x ■
12–14
x
x3
■
10. f x tan12x
Express the function as the sum of a power series by first
using partial fractions. Find the interval of convergence.
8 –9
1
S Click here for solutions.
1–7
■
■
REPRESENTING FUNCTIONS AS POWER SERIES
A Click here for answers.
■
REPRESENTING FUNCTIONS AS POWER SERIES
■
■
y
■
0.2
0
■
1
dx
1 x4
■
16.
■
■
■
y
12
0
■
tan1x 2 dx
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■
2
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SECTION 8.6
REPRESENTING FUNCTIONS AS POWER SERIES
8.6
ANSWERS
E Click here for exercises.
1.
∞
xn , (−1, 1)
S Click here for solutions.
9.
n=1
∞
2x2n , (−1, 1)
n=1
4.
∞
n=0
5.
(−1)n 4n x2n , − 12 , 12
∞
(−1)n x4n
, (−2, 2)
24n+4
n=0
6. −
∞ x n
n=1
7.
∞
∞
n=0
Copyright © 2013, Cengage Learning. All rights reserved.
3
, (−3, 3)
(−1) 3n xn 4 4 , −3, 3
22n+1
n=0
8. −
1 − 2−n xn , (−1, 1)
n=0
∞
(−1)n x2n
2.
, (−2, 2)
4n+1
n=0
3. 1 +
∞
n
(2n + 1) xn , − 12 , 12
10.
∞
(−1)n 22n+1 x2n+1 1
,
2n + 1
2
n=0
11.
∞
2x2n+1
,1
2n + 1
n=0
12. C +
∞
(−1)n x4n+1
4n + 1
n=0
13. C +
∞
(−1)n x5n+2
5n + 2
n=0
14. C +
∞
(−1)n
n=0
15. 0.199936
16. 0.041303
x2n+1
(2n + 1)2
SECTION 8.6
8.6
REPRESENTING FUNCTIONS AS POWER SERIES
■
3
SOLUTIONS
E Click here for exercises.
“R” stands for “radius of convergence” and “I” stands for “interval of convergence” in this section.
∞
∞
1
x
Another Method:
xn =
xn+1
1. f (x) =
=x
=x
1−x
1−x
∞
n=0
n=0
x
x x n
x
∞
=−
=−
f (x) =
n
x−3
3 (1 − x/3)
3 n=0 3
=
x
n=1
with R = 1 and I = (−1, 1).
=−
1
1 + x2 /4
n
∞
using Exercise 3
x2
1
=
(−1)n
4 n=0
4
from the text
2. f (x) =
1
1
=
4 + x2
4
1
1 + 3x/4
n ∞
∞
(−1)n 3n xn
3x
1
(−1)n
=
=
2 n=0
4
22n+1
n=0
7. f (x) =
∞
(−1)n x2n
4n+1
n=0
2
x with < 1 ⇔ x2 < 4 ⇔ |x| < 2, so R = 2 and
4
I = (−2, 2).
8.
∞ n
1 + x2
2x2
=1+
= 1 + 2x2
x2
2
2
1−x
1−x
n=0
∞
∞
=1+
2x2n+2 = 1 +
2x2n
n=0
n=1
with x2 < 1 ⇔ |x| < 1, so R = 1 and I = (−1, 1).
3. f (x) =
(−1)n 4n x2n
with 4x < 1 so x2 <
I = − 12 , 12 .
1
4
⇔ |x| < 12 , and so R =
1
2
Copyright © 2013, Cengage Learning. All rights reserved.
1
1
1
5. f (x) = 4
=
x + 16
16 1 + (x/2)4
∞
∞
x 4n 1 (−1)n x4n
=
(−1)n
=
16 n=0
2
24n+4
n=0
x
for < 1 ⇔ |x| < 2 so, R = 2 and I = (−2, 2).
2
x
3
1
6. f (x) =
=1+
=1−
x−3
x−3
1 − x/3
∞ ∞ x n
x n
=−
=1−
3
3
n=0
n=1
|x|
For convergence,
< 1 ⇔ |x| < 3, so R = 3 and
3
I = (−3, 3).
and I = − 43 , 43 .
3x − 2
3x − 2
A
B
=
=
+
2x2 − 3x + 1
(2x − 1) (x − 1)
2x − 1
x−1
⇔ A + 2B = 3 and −A − B = −2 ⇔ A = B = 1, so
3x − 2
1
1
f (x) =
=
+
2x2 − 3x + 1
2x − 1
x−1
∞
∞
∞
=−
(2x)n −
xn = −
(2n + 1) xn
n=0
using Exercise 3
1
n
2 n
4. f (x) =
=
(−1) 4x
1 + 4x2
from the text
n=0
n=0
2
4
3
n=0
n=0
with R = 12 . At x = ± 12 , the series diverges by the Test for
Divergence, so I = − 12 , 12 .
∞
∞
2
1
=
3x + 4
2
3x < 1 so R =
4 =
=
∞
∞
xn+1
xn
=−
n+1
3
3n
n=0
n=1
and
9.
x
x
A
B
=
=
+
⇔
x2 − 3x + 2
(x − 2) (x − 1)
x−2
x−1
A + B = 1 and −A − 2B = 0 ⇔ A = 2, B = −1, so
x
2
1
f (x) =
=
−
(x − 2) (x − 1)
x−2
x−1
∞ ∞
x n n
1
1
=−
+
x
+
=−
1 − x/2
1−x
2
n=0
n=0
=
∞ 1 − 2−n xn
n=0
with R = 1. At x = ±1, the series diverges by the Test for
Divergence, so I = (−1, 1).
dx
−1
10. f (x) = tan
2x = 2
1 + 4x2
∞
n
=2
(−1)n 4x2 dx
=2
n=0
∞
n=0
=C+2
(−1)n 4n x2n dx
∞
(−1)n 4n x2n+1
2n + 1
n=0
∞
(−1)n 22n+1 x2n+1
2n + 1
n=0
2
for 4x < 1 so |x| < 12 and R = 12 .
=
4
■
SECTION 8.6
REPRESENTING FUNCTIONS AS POWER SERIES
11. f (x) = ln (1 + x) − ln (1 − x) =
=
=
∞
n=0
(−1)n xn +
∞
n=0
dx
+
1+x
dx
1−x
15. We use the representation
xn dx
∞
2x2n+1
2x2n dx =
+C
2n + 1
n=0
n=0
∞
Problem 12 with C = 0. So
0.2
0.2
dx
x5
x9
x13
=
x
−
+
−
+
·
·
·
1 + x4
5
9
13
0
0
But f (0) = ln 1 − ln 1 = 0, so C = 0 and we have
∞
2x2n+1
f (x) =
with R = 1.
2n + 1
n=0
12.
dx
=
1 + x4
= 0.2 −
approximation is less than the first neglected term, by The
∞
(−1)n x4n+1
(−1)n x4n dx = C+
4n + 1
n=0
n=0
∞
Alternating Series Estimation Theorem. If we use only the
first two terms of the series, then the error is at most
0.29 /9 ≈ 5.7 × 10−8 . So, to six decimal places,
0.2
dx
0.25
≈ 0.2 −
= 0.199936.
4
1+x
5
0
∞
1
=
(−1)n x5n
1 + x5
n=0
⇒
16.
∞
x
=
(−1)n x5n+1 ⇒
1 + x5
n=0
∞
(−1)n x5n+2
x
dx
=
C
+
with R = 1.
1 + x5
5n + 2
n=0
Using Example 7 with  replaced by 2 , we have
∞


 
4+3
tan−1 2  =  +
(−1)
(2 + 1) (4 + 3)
=0
so
 12
 
tan−1 2 
0
=
14. By Example 7, arctan x =
∞
(−1)n
n=0
arctan x
dx =
x
∞
(−1)n
n=0
=C+
∞
n=0
Copyright © 2013, Cengage Learning. All rights reserved.
with R = 1.
0.25
0.29
0.213
+
−
+ ···
5
9
13
Since the series is alternating, the error in the nth-order
with R = 1.
13.
∞
dx
(−1)n x4n+1
=
C
+
from
1 + x4
4n + 1
n=0
x2n+1
, so
2n + 1
x2n
dx
2n + 1
(−1)n
x2n+1
(2n + 1)2

7
11
15
19
3
−
+
−
+
− ···
3
21
55
105
171
1
3
(05)3 −
12
0
057
0511
0515
0519
053
−
+
−
+
−···
=
3
21
55
105
171
The series is alternating, so if we use only the first four
terms of the series, then the error is at most
0519 171 ≈ 11 × 10−8 . So, to six decimal places,
 
 12
tan−1 2 
0
≈
≈ 0041303
1
21
(05)7 +
1
55
(05)11 −
1
105
(05)15