Solução_Calculo_Stewart_6e

F.
492 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES TX.10
1
27.
(2 + x) 3 = 1
[2(1 + x/2)] 3 = 1
1+ x −3 1 ∞ −3 x
n.
= The binomial coefficient is
8 2 8 n=0 n 2
Thus,
1
(2 + x) 3 = 1 8
−3
=
n
∞
n=0
(−3)(−4)(−5) ·····(−3 − n +1)
n!
= (−1)n · 2 · 3 · 4 · 5 ·····(n +1)(n +2)
2 · n!
=
(−3)(−4)(−5) ·····[−(n +2)]
n!
= (−1)n (n +1)(n +2)
2
(−1) n (n +1)(n +2) x n
2 2 = ∞ (−1) n (n +1)(n +2)x n
for
n n=0 2 n+4
x < 1 ⇔ |x| < 2,soR =2.
2
29. sin x = ∞ (−1) n x 2n+1
(2n +1)!
n=0
⇒
f(x) =sin(πx) = ∞ (−1) n (πx) 2n+1
(2n +1)! = ∞ (−1) n π 2n+1
(2n +1)! x2n+1 , R = ∞.
n=0
n=0
31. e x = ∞
n=0
R = ∞.
x n
n!
⇒
e 2x = ∞
n=0
(2x) n
n!
= ∞
n=0
2 n x n
,sof(x) =e x + e 2x
= ∞ 1
n!
n=0 n! xn + ∞ 2 n
n=0 n! xn = ∞ 2 n +1
x n ,
n=0 n!
33. cos x = ∞ (−1) n x 2n
(2n)!
n=0
⇒ cos
1
x2
= ∞ 1
(−1) n x2 2n
2
2
(2n)!
n=0
f(x) =x cos 1
x2
= ∞ (−1) n 1
2
2 2n (2n)! x4n+1 , R = ∞.
n=0
= ∞ (−1) n
n=0
x 4n
2 2n (2n)! ,so
35. We must write the binomial in the form (1+ expression), so we’ll factor out a 4.
x
√ = x
4+x
2 4(1 + x2 /4) =
x
2 1+x 2 /4 = x 2
= x 1+
− 1 x 2 −
1
2
2
4 + 2 −
3
2 x
2
2
+
2! 4
= x 2 + x 2
∞
(−1) n 1 · 3 · 5 ·····(2n − 1)
2 n · 4 n · n!
n=1
−1/2 1+ x2
= x 4 2
x 2n
−
1
2
−
3
2
3!
−
5
2
x
∞ − 1 2
2
n 4
n=0
x
2
3
+ ···
= x 2 + ∞ (−1) n 1 · 3 · 5 ·····(2n − 1)
x 2n+1 and x2
n!2 3n+1 4 < 1 ⇔ |x| < 1 ⇔ |x| < 2, soR =2.
2
n=1
37. sin 2 x = 1 2 (1 − cos 2x) =1 2
R = ∞
1 − ∞
n=0
(−1) n (2x) 2n
= 1
1 − 1 − ∞
(2n)! 2
n=1
4
n
(−1) n (2x) 2n
= ∞
(2n)!
n=1
(−1) n+1 2 2n−1 x 2n
,
(2n)!
39. cos x = ∞ (−1) n x 2n
(2n)!
n=0
⇒
f(x) =cos x 2 = ∞
n=0
(−1) n x 2 2n
(2n)!
= ∞
n=0
(−1) n x 4n
, R = ∞
(2n)!
Notice that, as n increases, T n (x)
becomes a better approximation to f(x).