Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 493
41. e x (11)
= ∞
n=0
x n
n! ,soe−x = ∞
n=0
(−x) n
n!
f(x)=xe −x
= ∞ (−1) n 1
n=0
= ∞ (−1) n x n
n! ,so
n=0
n! xn+1
= x − x 2 + 1 2 x3 − 1 6 x4 + 1 24 x5 − 1
120 x6 + ···
= ∞ (−1) n−1 x n
(n − 1)!
n=1
The series for e x converges for all x, so the same is true of the series
for f(x);thatis,R = ∞. From the graphs of f and the first few Taylor
polynomials, we see that T n(x) provides a closer fittof(x) near 0 as n increases.
43. e x = ∞
n=0
x n
n! ,soe−0.2 = ∞
n=0
(−0.2) n
n!
=1− 0.2+ 1 2! (0.2)2 − 1 3! (0.2)3 + 1 4! (0.2)4 − 1 5! (0.2)5 + 1 6! (0.2)6 − ···.
But 1 6! (0.2)6 =8.8 × 10 −8 , so by the Alternating Series Estimation Theorem, e −0.2 ≈
5
n=0
(−0.2) n
n!
≈ 0.81873,correctto
five decimal places.
45. (a) 1/ √ 1 − x 2 = 1+ −x 2−1/2 =1+
−
1
2 −x
2 −
1
2 −
3
2
+
−x
2 2 +
2!
=1+ ∞ 1 · 3 · 5 ·····(2n − 1)
x 2n
2 n · n!
(b) sin −1 x =
= x + ∞
n=1
1
√ dx = C + x + ∞
1 − x
2
n=1
n=1
1 · 3 · 5 ·····(2n − 1)
x 2n+1
(2n +1)2 n · n!
1 · 3 · 5 ·····(2n − 1)
x 2n+1
(2n +1)2 n · n!
since 0=sin −1 0=C.
−
1
2
−
3
2
3!
−
5
2
−x
2 3 + ···
47. cos x (16)
= ∞ (−1) n x 2n
(2n)!
n=0
x cos(x 3
)= ∞ (−1) n x 6n+1
(2n)!
n=0
⇒
cos(x 3 )= ∞ (−1) n (x 3 ) 2n
= ∞ (−1) n x 6n
⇒
n=0 (2n)! n=0 (2n)!
⇒ x cos(x 3
) dx = C + ∞ (−1) n x 6n+2
,withR = ∞.
(6n + 2)(2n)!
n=0
49. cos x (16)
= ∞ (−1) n x 2n
(2n)!
n=0
⇒
cos x − 1= ∞ (−1) n x 2n
(2n)!
n=1
⇒ cos x − 1
x
= ∞ (−1) n x 2n−1
(2n)!
n=1
⇒
cos x − 1
dx = C + ∞ (−1) n x 2n
,withR = ∞.
x
2n · (2n)!
51. By Exercise 47,
1
0
n=1
x cos(x 3
) dx = C + ∞ (−1) n x 6n+2
(6n +2)(2n)! ,so
n=0
∞
x cos(x 3
) dx = (−1) n x 6n+2 1
= ∞ (−1) n
n=0 (6n + 2)(2n)!
0 n=0 (6n + 2)(2n)! = 1 2 − 1
8 · 2! + 1
14 · 4! − 1
20 · 6! + ···,but
1
20 · 6! = 1
14,400 ≈ 0.000 069, so 1
Alternating Series Estimation Theorem.
0
x cos(x 3 ) dx ≈ 1 2 − 1 16 + 1 ≈ 0.440 (correct to three decimal places) by the
336