Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 495
65.
∞
n=0
(−1) n π 2n+1
4 2n+1 (2n +1)! = ∞
n=0
(−1) n
π 2n+1
4
=sin π 4
(2n +1)!
= √ 1
2
, by (15).
67. 3+ 9 2! + 27
3! + 81 31
+ ···=
4! 1! + 32
2! + 33
3! + 34
4! + ···= ∞ 3 n
n=1 n! = ∞ 3 n
n=0 n! − 1=e3 − 1, by (11).
69. Assume that |f 000 (x)| ≤ M, sof 000 (x) ≤ M for a ≤ x ≤ a + d. Now x
a f 000 (t) dt ≤ x
a Mdt
f 00 (x) − f 00 (a) ≤ M(x − a) ⇒ f 00 (x) ≤ f 00 (a)+M(x − a). Thus, x
a f 00 (t) dt ≤ x
a [f 00 (a)+M(t − a)] dt
f 0 (x) − f 0 (a) ≤ f 00 (a)(x − a)+ 1 M(x − 2 a)2 ⇒ f 0 (x) ≤ f 0 (a)+f 00 (a)(x − a)+ 1 M(x − 2 a)2 ⇒
x f 0 (t) dt ≤ x
a a f 0 (a)+f 00 (a)(t − a)+ 1 M(t − a)2 dt ⇒
2
f(x) − f(a) ≤ f 0 (a)(x − a) + 1 2 f 00 (a)(x − a) 2 + 1 6 M(x − a)3 .So
⇒
⇒
f(x) − f(a) − f 0 (a)(x − a) − 1 2 f 00 (a)(x − a) 2 ≤ 1 6 M(x − a)3 .But
R 2 (x) =f(x) − T 2 (x) =f(x) − f(a) − f 0 (a)(x − a) − 1 2 f 00 (a)(x − a) 2 ,soR 2 (x) ≤ 1 6 M(x − a)3 .
A similar argument using f 000 (x) ≥−M shows that R 2 (x) ≥− 1 6 M(x − a)3 .So|R 2 (x 2 )| ≤ 1 6 M |x − a|3 .
Although we have assumed that x>a, a similar calculation shows that this inequality is also true if x