Solução_Calculo_Stewart_6e

F.
490 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
9.
n f (n) (x) f (n) (0)
0 e 5x 1
1 5e 5x 5
2 5 2 e 5x 25
3 5 3 e 5x 125
4 5 4 e 5x 625
.
.
.
e 5x = ∞
n=0
lim
n→∞ an+1
a n
for all x,soR = ∞.
f (n) (0)
x n
= ∞ 5 n
n!
n=0 n! xn .
5 n+1 |x| n+1
= lim
·
n→∞ (n +1)!
n!
5 n |x| n = lim
n→∞
5 |x|
n +1 =0< 1
11.
n f (n) (x) f (n) (0)
0 sinh x 0
1 cosh x 1
2 sinh x 0
3 cosh x 1
4 sinh x 0
.
.
.
f (n) (0) =
0 if n is even
1 if n is odd
Use the Ratio Test to find R. Ifa n =
lim
n→∞ an+1
a n
= lim
n→∞
x 2n+3
(2n +3)!
so sinh x = ∞
n=0
x2n+1
(2n +1)! ,then
· (2n +1)!
x 2n+1
=0< 1 for all x,soR = ∞.
x 2n+1
(2n +1)! .
= x 2 · lim
n→∞
1
(2n + 3)(2n +2)
13.
n f (n) (x) f (n) (1)
0 x 4 − 3x 2 +1 −1
1 4x 3 − 6x −2
2 12x 2 − 6 6
3 24x 24
4 24 24
5 0 0
6 0 0
.
.
.
f (n) (x) =0for n ≥ 5,sof has a finite series expansion about a =1.
f(x)=x 4 − 3x 2 +1=
4
n=0
f (n) (1)
(x − 1) n
n!
= −1
0! (x − 1)0 + −2
1! (x − 1)1 + 6 2! (x − 1)2 + 24
3! (x − 1)3 + 24 (x − 1)4
4!
= −1 − 2(x − 1) + 3(x − 1) 2 +4(x − 1) 3 +(x − 1) 4
A finite series converges for all x,soR = ∞.
15. f(x) =e x ⇒ f (n) (x) =e x ,sof (n) (3) = e 3 and e x = ∞
lim
n→∞ an+1
a n
= lim
n→∞
e3 (x − 3) n+1
·
(n +1)!
n=0
e 3
n! (x − 3)n .Ifa n = e3
n! (x − 3)n ,then
n! |x − 3|
= lim =0< 1 for all x,soR = ∞.
e 3 (x − 3) n n→∞ n +1
17.
n f (n) (x) f (n) (π)
0 cos x −1
1 − sin x 0
2 − cos x 1
3 sin x 0
4 cos x −1
.
.
.
cos x = ∞
k=0
= ∞
n=0
lim
n→∞ an+1
a n
for all x,soR = ∞.
f (k) (π)
(x − π) k (x − π)2 (x − π)4 (x − π)6
= −1+ − + − ···
k!
2! 4! 6!
(−1) n+1 (x − π) 2n
(2n)!
|x − π|
2n+2
= lim
·
n→∞ (2n +2)!
(2n)!
|x − π| 2n = lim
n→∞
|x − π| 2
(2n + 2)(2n +1) =0< 1