Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 489
39. By Example 7, tan −1
x = ∞ (−1) n x 2n+1
for |x| < 1. In particular, for x =
1 2n +1
n=0
have π
√ 2n+1
1√3
6 =tan−1 = ∞ 1/ 3
(−1) n 2n +1
π =
6 √
3
∞
n=0
n=0
(−1) n
(2n +1)3 =2√
3 ∞ n
n=0
(−1) n
(2n +1)3 n .
√
3
,we
=
n=0(−1) ∞ n 1 n 1 1 √3
3 2n +1 ,so
11.10 Taylor and Maclaurin Series
1. UsingTheorem5with ∞
n=0
b n(x − 5) n , b n = f (n) (a)
n!
,sob 8 = f (8) (5)
.
8!
3. Since f (n) (0) = (n +1)!, Equation 7 gives the Maclaurin series
∞
n=0
lim
n→∞
f (n) (0)
x n
= ∞
n!
n=0
= lim
a n+1
a n
n→∞
(n +1)!
n!
radius of convergence R =1.
x n
= ∞ (n +1)x n . Applying the Ratio Test with a n =(n +1)x n gives us
n=0
(n +2)xn+1 n +2
= |x| lim = |x|·1=|x|. For convergence, we must have |x| < 1,sothe
(n +1)x n n→∞ n +1
5.
n f (n) (x) f (n) (0)
0 (1 − x) −2 1
1 2(1 − x) −3 2
2 6(1 − x) −4 6
3 24(1 − x) −5 24
4 120(1 − x) −6 120
.
.
.
(1 − x) −2 = f(0) + f 0 (0)x + f 00 (0)
2!
a n
x 2 + f 000 (0)
3!
=1+2x + 6 2 x2 + 24 6 x3 + 120
24 x4 + ···
x 3 + f (4) (0)
x 4 + ···
4!
=1+2x +3x 2 +4x 3 +5x 4
+ ···= ∞ (n +1)x n
lim
n→∞ a
n+1 = lim
(n +2)xn+1 n +2
n→∞ = |x| lim = |x| (1) = |x| < 1
(n +1)x n n→∞ n +1
for convergence, so R =1.
n=0
7.
n f (n) (x) f (n) (0)
0 sin πx 0
1 π cos πx π
2 −π 2 sin πx 0
3 −π 3 cos πx −π 3
4 π 4 sin πx 0
5 π 5 cos πx π 5
.
.
.
sin πx = f(0) + f 0 (0)x + f 00 (0)
2!
n→∞
+ f (4) (0)
4!
x 2 + f 000 (0)
x 3
3!
x 4 + f (5) (0)
x 5 + ···
5!
=0+πx +0− π3
3! x3 +0+ π5
5! x5 + ···
= πx − π3
3! x3 + π5
5! x5 − π7
7! x7 + ···
= ∞
a n
n=0
(−1) n π 2n+1
(2n +1)! x2n+1
lim
a n+1 = lim
n→∞ π2n+3 x 2n+3
·
(2n +3)!
=0< 1 for all x,soR = ∞.
(2n +1)!
π 2n+1 x 2n+1
= lim
n→∞
π 2 x 2
(2n +3)(2n +2)