Solução_Calculo_Stewart_6e

F.
t
23.
1 − t = t · 1
8 1 − t = t ∞ (t 8 ) n
= ∞ 8
n=0
t 8n+1
n=0
SECTION TX.10 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ¤ 487
⇒
t
1 − t dt = C + ∞
8
n=0
t 8n+2
8n +2 . Theseriesfor 1
1 − t 8 converges
when t
8 < 1 ⇔ |t| < 1,soR =1for that series and also the series for t/(1 − t 8 ). By Theorem 2, the series for
t
dt also has R =1.
1 − t8 25. By Example 7, tan −1
x = ∞ (−1) n x 2n+1
with R =1,so
n=0 2n +1
x − tan −1 x = x −
x ···
− x3
3 + x5
5 − x7
7 + = x3
3 − x5
5 + x7
7 − ··· = ∞ (−1) n+1 x 2n+1
2n +1 and
x − tan −1 x
x 3 =
∞
(−1) n+1 x 2n−2
2n +1 ,so
n=1
x − tan −1 x
dx = C + ∞ (−1) n+1 x 2n−1
x 3
(2n +1)(2n − 1) = C + ∞ (−1) n+1 x 2n−1
. By Theorem 2, R =1.
4n 2 − 1
n=1
1
27.
1+x = 1
5 1 − (−x 5 ) = ∞ −x
5 n
= ∞ (−1) n x 5n ⇒
n=0
n=0
1
1+x dx = ∞
(−1) n x 5n
dx = C + ∞ (−1) n x 5n+1
5 5n +1 . Thus,
0.2
I =
0
n=0
n=0
1
··· 0.2
1+x dx = x − x6
5 6 + x11
11 −
0
=0.2 − (0.2)6
6
n=1
+ (0.2)11
11
n=1
− ···. The series is alternating, so if we use
the first two terms, the error is at most (0.2) 11 /11 ≈ 1.9 × 10 −9 .SoI ≈ 0.2 − (0.2) 6 /6 ≈ 0.199989 to six decimal places.
29. We substitute 3x for x in Example 7, and find that
x arctan(3x) dx = x ∞ (−1) n (3x) 2n+1 ∞
n=0 2n +1 dx =
(−1) n 3 2n+1 x 2n+2
dx = C + ∞ (−1) n 3 2n+1 x 2n+3
n=0 2n +1
n=0 (2n +1)(2n +3)
So
0.1
0
3x
3
x arctan(3x) dx =
1 · 3 − 33 x 5
3 · 5 + 35 x 7
5 · 7 − 37 x ··· 9
0.1
7 · 9 + 0
The series is alternating, so if we use three terms, the error is at most
0.1
0
= 1
10 3 − 9
5 × 10 5 + 243
35 × 10 7 − 2187
63 × 10 9 + ···.
2187
63 × 10 9 ≈ 3.5 × 10−8 .So
x arctan(3x) dx ≈ 1
10 − 9
3 5 × 10 + 243 ≈ 0.000 983 to six decimal places.
5 35 × 107 31. Using the result of Example 6, ln(1 − x) =− ∞
n=1
x n
,withx = −0.1, wehave
n
ln 1.1 =ln[1− (−0.1)] = 0.1 − 0.01
2 + 0.001 − 0.0001 + 0.00001 − ···. The series is alternating, so if we use only
3 4 5
the first four terms, the error is at most 0.00001
5
=0.000002. Soln 1.1 ≈ 0.1 − 0.01
2 + 0.001 − 0.0001 ≈ 0.09531.
3 4