Solução_Calculo_Stewart_6e

F.
508 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
57. (a)
(b)
(d)
n f (n) (x) f (n) (1)
0 x 1/2 1
1
1
2 x−1/2 1 2
2 − 1 4 x−3/2 − 1 4
3
3
8 x−5/2 3 8
4 − 15
16 x−7/2 − 15
16
.
.
.
√
x ≈ T3 (x)=1+ 1/2
1!
(x − 1) − 1/4
2!
(x − 1) 2 + 3/8
3!
=1+ 1 (x − 1) − 1 (x − 2 8 1)2 + 1 (x − 1)3
16
(x − 1) 3
(c) |R 3 (x)| ≤ M 4! |x −
1|4 ,wheref (4) (x) ≤ M with
f (4) (x) =− 15
16 x−7/2 .Now0.9 ≤ x ≤ 1.1 ⇒
−0.1 ≤ x − 1 ≤ 0.1 ⇒ (x − 1) 4 ≤ (0.1) 4 ,
and letting x =0.9 gives M =
|R 3(x)| ≤
15
,so
16(0.9)
7/2
15
16(0.9) 7/2 4! (0.1)4 ≈ 0.000 005 648
≈ 0.000 006 = 6 × 10 −6
From the graph of |R 3 (x)| = | √ x − T 3 (x)|,itappears
that the error is less than 5 × 10 −6 on [0.9, 1.1].
59. sin x = ∞
sin x − x
x 3
(−1) n x 2n+1
(2n +1)! = x − x3
3! + x5
5! − x7
x3
+ ···,sosin x − x = −
7! 3! + x5
5! − x7
+ ··· and
7!
= − 1
3! + x2
5! − x4
7! + ···.Thus,lim sin x − x
=lim − 1 ···
x→0 x 3 x→0 6 + x2
120 − x4
5040 + = − 1 6 .
n=0
61. f(x) = ∞ c n x n
⇒ f(−x) = ∞ c n(−x) n
= ∞ (−1) n c n x n
n=0
n=0
n=0
(a) If f is an odd function, then f(−x) =−f(x) ⇒ ∞
(−1) n c n x n = ∞ −c n x n . The coefficients of any power series
n=0
are uniquely determined (by Theorem 11.10.5), so (−1) n c n = −c n .
If n is even, then (−1) n =1,soc n = −c n ⇒ 2c n =0 ⇒ c n =0.Thus,allevencoefficients are 0, thatis,
c 0 = c 2 = c 4 = ···=0.
(b) If f is even, then f(−x) =f(x) ⇒ ∞ (−1) n c n x n
= ∞ c n x n ⇒ (−1) n c n = c n .
n=0
If n is odd, then (−1) n = −1,so−c n = c n ⇒ 2c n =0 ⇒ c n =0. Thus, all odd coefficients are 0,
that is, c 1 = c 3 = c 5 = ···=0.
n=0
n=0