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

82 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
x 3 x 5 x 1 (0.2? (0.2) 5 (0.2) 7
33 By Example 7 arctan x = x - - + - - - + · · · so arctan 0 2 = 0 2 - --+ --- --+ ...
. ' 3 57' .. 3 5 7 .
Th . . I . 'f t1 1 · (0.2) 7
e senes 1s a ternatmg, so 1 we use 1ree terms, t 1e error ts at most - - ~ 0.000 002.
7
(0 2? (0 2?
Thus, to five decimal places, arctan0.2 ~ 0.2- - ·3-
+ -·5-
~ 0.197 40.
2 , 1 2
00
(- 1)"2n(2n- 1)x 2 "
00
(-1t2nx 2 "'
00
(- 1)"x 2 "+2
X Jo (x) + XJo(X) +X Jo(x) = n~l 22n(n!)2 + 1 ~ 1
22"(n!)2 + n~O 22" (n!)2
00
(- 1)"2n(2n-1)x 2 " oo (- 1)"2nx 2 '' oo (- 1t- 1 x 2 "
= n~l 2 2 "(n!) 2 + ,~1 2 2 "(n!J2 + ,~1 22 n- 2 [(n- 1)!] 2
_
00
(- 1)"2n(2n-1)x 2 "
00
(- 1t2nx 2 "
00
(-1)"(-1)- 1 2 2 n 2 x 2 "
- r~l 2 2 "(nl)2 + n~l 2 2 "(n!)2 + ,~1 2 2 "(n!)2
= ~ (- 1 )" [2n(2n - 1) + 2n - 2 2 n 2 ] x2"
n~l 2 2 "(n!)2 ,
= ~ (- 1 )" [4n 2 - 2n + 2n- 4n 2 ] 2n = 0
L- 22" ( 1)2 X
n=l n.
Since 16
.i 28
~ 0.000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places,
J 0
1
Jo (x) dx ~ 1 - 1 ; + 3 ~ 0 ~ 0.920.
00 Xn I 00 nxn- l 00 Xn-1
37. (a) f(x) = n~O nf ::} f (x) = •~1 ~ = •~1 (n- 1)!
oo x.,."
I: - 1
= f(x)
n = O n
(b) By Theorem 9.4.2, the only solution to the differential equation df(x)/d.'C = f(:c) is f(x) = Ke", but f(O) = 1, so
K = 1 and f(x) = e".
Or: We could solve the equation df(x)/ dx = f(x.) as a separable differential equation.
n I I I n+l 21 . ( ) 2 .
39. If a,.. = X 2' then by the Ratio Test, lim an+ I = lim (X )2 . ~ = lxl lim ....2::_1 = lx l < 1 for
n - n.-oo an n-oo n + 1 xn n----+oo n +
convergence, so R = 1. When x = ± 1, f I x: I = f ~which is a convergentp-series (p = 2 > 1), so the interval of
n = l n " =1 n
convergence for f is [-1, 1]. By Theorem 2, the radii of convergence off' and f" are both 1, so we need only check the
00 x" 00 nxn- 1 00 x"
endpoints. f(x) = I; 2 => . f'(x ) = I: - - 2 - = I; --,and this series diverges for x = 1 (harmonic series)
n=l n n=l n n=O n + 1
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