Solução_Calculo_Stewart_6e

F.
SECTION TX.10 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES ¤ 485
11. f(x) =
3
x 2 − x − 2 = 3
(x − 2)(x +1) = A
x − 2 + B
x +1
⇒
3=A(x +1)+B(x − 2). Letx =2to get A =1and
x = −1 to get B = −1. Thus
3
x 2 − x − 2 = 1
x − 2 − 1
x +1 = 1
1
−2 1 − (x/2)
= ∞ − 1 n
1
− 1(−1) n x n
= ∞
2 2
n=0
−
n=0
1
1 − (−x) = − 1 2
(−1) n+1 − 1
2 n+1
x n
∞ x
n ∞ − (−x) n
n=0 2 n=0
We represented f as the sum of two geometric series; the first converges for x ∈ (−2, 2) and the second converges for (−1, 1).
Thus, the sum converges for x ∈ (−1, 1) = I.
1
13. (a) f(x) =
(1 + x) 2 = d −1
= − d ∞
(−1) n x n [from Exercise 3]
dx 1+x dx n=0
= ∞ (−1) n+1 nx n−1
[from Theorem 2(i)] = ∞ (−1) n (n +1)x n with R =1.
n=1
n=0
In the last step, note that we decreased the initial value of the summation variable n by 1, andthenincreased each
occurrence of n in the term by 1 [also note that (−1) n+2 =(−1) n ].
1
(b) f(x) =
(1 + x) 3 = − 1
d 1
2 dx (1 + x) 2 = − 1
d ∞
(−1) n (n +1)x n [from part (a)]
2 dx n=0
∞
∞
= − 1 (−1) n (n +1)nx n−1 = 1 (−1) n (n +2)(n +1)x n with R =1.
2
2
n=1
n=0
x 2
(c) f(x) =
(1 + x) = 1
3 x2 ·
(1 + x) = 3 x2 · 1
2
= 1 ∞
(−1) n (n +2)(n +1)x n+2
2
n=0
∞
(−1) n (n +2)(n +1)x n
n=0
[from part (b)]
To write the power series with x n rather than x n+2 ,wewilldecrease each occurrence of n in the term by 2 and increase
the initial value of the summation variable by 2. Thisgivesus 1 2
15. f(x) =ln(5− x) =−
17.
dx
5 − x = −1 5
∞
(−1) n (n)(n − 1)x n with R =1.
n=2
dx
1 − x/5 = −1 ∞ x
n
dx = C −
5 n=0 1 5
5
Putting x =0,wegetC =ln5. The series converges for |x/5| < 1 ⇔ |x| < 5,soR =5.
1
2 − x = 1
2(1 − x/2) = 1 ∞ x
2 n=0
2
1
(x − 2) = d 1
= d
∞ 2 dx 2 − x dx
f(x) =
x 3
(x − 2) 2 = ∞
x3
n=0
n
=
∞
n=0
n=0
n +1
2 n+2 xn = ∞
1
2 n+1 xn for x < 1 ⇔ |x| < 2. Now
2
1
2 n+1 xn = ∞
n=0
n=1
n +1
2 n+2 xn+3 or ∞
n
2 n+1 xn−1 = ∞
n=3
n=0
∞
n=0
n +1
2 n+2 xn .So
x n+1
5 n (n +1) = C − ∞
n=1
n − 2
2 n−1 xn for |x| < 2. Thus,R =2and I =(−2, 2).
x n
n 5 n