Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

90 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
00
75. (a) g(x) = :L; 00 (k) x"' => g'(x) = :L; (k) nx"- 1 , so
n~ n n~ n
00
00
(1 + x)g'(x) = (1 + x) :L; 00 (k) nxn- (k) (k)
1 = :L; nx"'- 1 + :L; nxn
n=1 n n=1 n n = 1 n
. .
I
Replace n with n
= f: ( k ) (n 1
+ 1)xn + f: (k)nxn
+ 1 ]
n=O n+ n = O n [ in the first series
00 00
_ k(k-1)(k-2) .. ·(k-n+1)(k-n) n
[ k(k- 1)(k - 2)···(k-n+ 1)] ,,
- n~o (n + 1 ) (n + 1)! x + n~O (n) nl x
00
(n + l )k(k - '1)(k - 2) ... (k- n + 1) n
= n~O (n+ 1)! . ((k -n)_+n]X
= k n~o k(k - 1)(k- 2l! ... (k- n + 1) x" = k fo ( ~) x" = kg(x)
Thus, g'(x) = k
1
g(x).
+x
(b) h(x) = (1 + x)-k g(x) =>
h'(x) = - k(l + x)- k- 1 g{x) + (1 + x)-k g'(x)
= -k(l + x)-i.:- 1 g(x) + (1 + x)- k kg(x)
1+x
= -k(l + x)-k- 1 g(x) + k{1 + x)-"'- 1 g(x) = 0
[Product Rule]
[from part (a)]
(c) From part (b) we see that h(x) must be constant for x E ( -1, 1), so h(x) = h(O) = 1 for x E ( -1, 1).
Thus, h{x) = 1 = (1 + x)-k g{x) g(x) = {1 + x)k for x E ( -1, 1).
11.11 Applications of Taylor Polynomials
1. (a)
J (x) t