Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS ¤ 497
7.
n f (n) (x) f (n) (0)
0 arcsin x 0
1
1
√
1 − x
2
1
2
3
x
(1 − x 2 ) 3/2 0
2x 2 +1
(1 − x 2 ) 5/2 1
T 3(x) =
3
n=0
f (n) (0)
n!
x n = x + x3
6
9.
n f (n) (x) f (n) (0)
0 xe −2x 0
1 (1 − 2x)e −2x 1
2 4(x − 1)e −2x −4
3 4(3 − 2x)e −2x 12
T 3 (x) =
3
n=0
f (n) (0)
x n = 0 1
n! · 1+ 1 1 x1 + −4
2 x2 + 12 6 x3 = x − 2x 2 +2x 3
11. You may be able to simply find the Taylor polynomials for
f(x) =cotx using your CAS. We will list the values of f (n) (π/4)
for n =0to n =5.
n 0 1 2 3 4 5
f (n) (π/4) 1 −2 4 −16 80 −512
T 5 (x)=
5
n=0
f (n) (π/4)
x −
π n
n!
4
=1− 2 x − π 4
+2
x −
π
4
2
− 8 3
x −
π
4
3
+ 10
3
x −
π
4
4
− 64
15
x −
π 5
4
For n =2to n =5, T n(x) is the polynomial consisting of all the terms up to and including the x − π 4
n
term.
13.
(a) f(x) = √ x ≈ T 2 (x)=2+ 1 1/32
(x − 4) − (x − 4) 2
n f (n) (x) f (n) 4 2!
(4)
3
3
8 x−5/2 on [4, 4.2], we can take M = |f 000 (4)| = 3 8 4−5/2 = 3 256
0
√ x 2
=2+ 1 1
(x − 4) − (x − 4)2
4 64
1
1
2 x−1/2 1 4
(b) |R 2 (x)| ≤ M 3! |x − 4|3 ,where|f 000 (x)| ≤ M. Now 4 ≤ x ≤ 4.2 ⇒
2 − 1 4 x−3/2 − 1 32
|x − 4| ≤ 0.2 ⇒ |x − 4| 3 ≤ 0.008. Sincef 000 (x) is decreasing
|R 2 (x)| ≤ 3/256 (0.008) = 0.008 =0.000 015 625.
6
512