Thomas Calculus 13th [Solutions]

772 Chapter 10 Infinite Sequences and Series
29.
2 3
3 5 7
x
e 1 x x x and sin x x x x x
2! 3!
3! 5! 7!
x
2 3 3 5 7
2 1 3 1 5
e sin x 1 x x x x x x x x x x x .
2! 3! 3! 5! 7! 3 30
30.
1 2 1 3 1 4
ln(1 x)
x x x x and 1
2 3
1 x x x
2 3 4
1 x
ln(1 x) 1 1 2 1 3 1 4 2 3 1 2 5 3 7 4
ln(1 x) x x x x 1 x x x x x x x .
1 x
1 x 2 3 4 2 6 12
31.
1 3 5 7 1
2
1 1 1
1 1
tan x x x x x tan x tan x tan x
3 5 7
1 3 1 5 1 7 1 3 1 5 1 7 2 2 4 23 6 44 8
x x x x x x x x x x x x
3 5 7 3 5 7 3 45 105
.
32.
3 5 7
2 4 6
2
sin x x x x x and cos x 1 x x x cos x sin x cos x cos x sin x
3! 5! 7!
2! 4! 6!
3 5 7
2 4 6
1 1
(2 x) (2 x) (2 x) 7 3 61 5 1247 7
cos x sin 2x 1 x x x 2x x x x x
2 2 2! 4! 6! 3! 5! 7! 6 120 5040
33.
34.
3 5 7
2 3
sin x x x x x and e
x 1 x x x
3! 5! 7!
2! 3!
3 5 7 3 5 7 2 3 5 7 3
sin x
e 1 x x x x 1 x x x x 1 x x x x
3! 5! 7! 2 3! 5! 7! 6 3! 5! 7!
1 2 1 4
1 x x x .
2 8
3 5 7
sin x x x x x and
3! 5! 7!
1 1 3 1 5 1 7 1
tan x x x x x sin tan x
3 5 7
3 5 7 3 5 7
3
3 5 7
5
x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x
3 5 7 6 3 5 7 120 3 5 7
3 5 7
7
1 1 1 1 1 3 3 5 5 7
x x x x x x x x
5040 3 5 7 2 8 16
35. Since n 3, then
(4) (4)
f ( x) sin x, f ( x)
M on [0,0.1] sin x 1 on [0,0.1] M 1.
Then
4
0.1 0 6
R 3(0.1) 1 4.2 10 error
4!
4.2 10
6
36. Since n 4, then
(5) x (5)
f ( x) e , f ( x)
M on [0,0.5]
x
e e on [0,0.5] M 2.7.
Then
5
0.5 0 4
R 4(0.5) 2.7 7.03 10 error
5!
7.03 10
4
37. By the Alternating Series Estimation Theorem, the error is less than
5 4 5 2
x 600 10 x 6 10 0.56968
5
x
5!
5 4
x (5!) 5 10
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