Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS ¤ 499
19.
n f (n) (x) f (n) (0)
0 e x2 1
(a) f(x) =e x2 ≈ T 3(x) =1+ 2 2! x2 =1+x 2
(b) |R 3 (x)| ≤ M
4! |x|4 ,wheref (4) (x) ≤ M. Now0 ≤ x ≤ 0.1
⇒
1 e x2 (2x) 0
2 e x2 (2 + 4x 2 ) 2
3 e x2 (12x +8x 3 ) 0
4 e x2 (12 + 48x 2 +16x 4 )
x 4 ≤ (0.1) 4 , and letting x =0.1 gives
|R 3(x)| ≤ e0.01 (12 + 0.48 + 0.0016)
(0.1) 4 ≈ 0.00006.
24
(c)
From the graph of |R 3 (x)| = e x2 − T 3 (x) , it appears that the
error is less than 0.000 051 on [0, 0.1].
21.
n f (n) (x) f (n) (0)
0 x sin x 0
1 sin x + x cos x 0
2 2cosx − x sin x 2
3 −3sinx − x cos x 0
4 −4cosx + x sin x −4
5 5sinx + x cos x
(a) f(x) =x sin x ≈ T 4 (x) = 2 2! (x − 0)2 + −4
4! (x − 0)4 = x 2 − 1 6 x4
(b) |R 4 (x)| ≤ M 5! |x|5 ,where f (5) (x) ≤ M. Now−1 ≤ x ≤ 1
|x| ≤ 1, and a graph of f (5) (x) shows that f (5) (x) ≤ 5 for −1 ≤ x ≤ 1.
Thus, we can take M =5and get |R 4 (x)| ≤ 5 5! · 15 = 1
24 =0.0416.
⇒
(c)
From the graph of |R 4(x)| = |x sin x − T 4(x)|, it seems that the
error is less than 0.0082 on [−1, 1].
23. From Exercise 5, cos x = −
x − π 2 +
1
6 x −
π 3
2
+ R 3(x), where|R 3(x)| ≤ M x −
π 4
2
with
4!
f (4) (x) = |cos x| ≤ M =1.Nowx =80 ◦ =(90 ◦ − 10 ◦ )= π
− π
2 18 =
4π
radians, so the error is
9
R 4π
3
≤ 1 π
4
≈ 0.000 039, which means our estimate would not be accurate to five decimal places. However,
9 24 18
T 3 = T 4 ,sowecanuse R 4
4π
9
cos 80 ◦ ≈− − π 18
+
1
6 −
π 3
18
≈ 0.17365.
25. All derivatives of e x are e x ,so|R n(x)| ≤
≤ 1
120
π
18
5
≈ 0.000 001. Therefore, to five decimal places,
e x
(n +1)! |x|n+1 ,where0