Solução_Calculo_Stewart_6e

F.
498 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
(c)
From the graph of |R 2 (x)| = | √ x − T 2 (x)|,itseemsthatthe
error is less than 1.52 × 10 −5 on [4, 4.2].
15.
(c)
(a) f(x) =x 2/3 ≈ T 3 (x)=1+ 2 2/9
(x − 1) −
n f (n) (x) f (n) 3
2! (x − 1)2 + 8/27 (x − 1) 3
3!
(1)
=1+ 2
0 x 2/3 1
(x − 1) − 1 (x − 3 9 1)2 + 4 81
(x − 1)3
2
1
3 x−1/3 2 (b) |R
3
3(x)| ≤ M 4! |x − 1|4 ,where f (4) (x) ≤ M. Now0.8 ≤ x ≤ 1.2 ⇒
2 − 2 9 x−4/3 − 2 9
|x − 1| ≤ 0.2 ⇒ |x − 1| 4
≤ 0.0016. Sincef (4) (x) is decreasing
8
3
27 x−7/3 8 27
4 − 56
on [0.8, 1.2],wecantakeM = f (4) (0.8) = 56
81 x−10/3 81 (0.8)−10/3 ,so
|R 3 (x)| ≤
56
81 (0.8)−10/3
24
(0.0016) ≈ 0.000 096 97.
From the graph of |R 3(x)| = x 2/3
− T 3(x) , it seems that the
errorislessthan0.000 053 3 on [0.8, 1.2].
17.
n f (n) (x) f (n) (0)
0 sec x 1
1 sec x tan x 0
2 sec x (2 sec 2 x − 1) 1
3 sec x tan x (6 sec 2 x − 1)
(a) f(x) =secx ≈ T 2 (x) =1+ 1 2 x2
(b) |R 2(x)| ≤ M 3! |x|3 ,where f (3) (x) ≤ M. Now−0.2 ≤ x ≤ 0.2 ⇒ |x| ≤ 0.2 ⇒ |x| 3 ≤ (0.2) 3 .
f (3) (x) is an odd function and it is increasing on [0, 0.2] since sec x and tan x are increasing on [0, 0.2],
so f (3) (x) ≤ f (3) (0.2) ≈ 1.085 158 892. Thus,|R 2 (x)| ≤ f (3) (0.2)
(0.2) 3 ≈ 0.001 447.
3!
(c)
From the graph of |R 2(x)| = |sec x − T 2(x)|, it seems that the
error is less than 0.000 339 on [−0.2, 0.2].