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

88 0 CHAPTER 11 INFINITE SEQUENCES AND SERIES
• oo x2n+l 00 x2n+ 4
51 . arctanx = 2: (- 1)" - 2 1
for lxl < 1, so x 3 arctanx = 2: ( -1)"- - for Jxl < 1 and
n = O n + n =O 2n+ 1
I
oo
x2n+5
x 3 arctanx dx = C + ,..;;:o(-1)"' ( 2
n + 1
)( 2
n + )' 5
Since ~ < 1, we have
(1/2) 5 (1/2r (1/ 2) 9 · · {1/ 2) 11
1.5 - ~ + ~ ~ 0.0059 and subtracting Til ~ 6.3 x 10- 6 does not affect the fourth decimal place,
so J112 0 x
3 arctan x dx ~ 0.0059 by the Alternating Series Estimation Theorem.'
(1/ 2) I oo ( 1/ 2) x4n+1
n =O n , n =O n 4n + .1
53. v'1 + x4 = (1 +.x 4 ) 1 1 2 00
= 2: (x 4 )", so }1 + x 4 dx = C + L: --and hence, since 0.4 < 1,
we have
1
I= } 1 +x 4 00
0.4
(1/ 2) (0.4) 4n+l
dx = E
o n = O n 4n + 1
(o.4) 1 ~ (o.4) 5 H-4) (o.4) 9 H-t)(-~) (o.4) 13 H-4)(-~)(-~) (o.4) 17
=(1)or+l!-5-+-2-,---9-+ 3! ~ + 4! ---rr- +· ··
(0.4) 5 (0.4) 0 (0.4) 13 5(0.4) 17
= 0.4 + ----w- - ----;:;2 + 208 - 2176 + ...
Now (O~~)o ~ 3.6 x 10- 6 < 5 x 10- 6 , so by the Alternating Series Esti~ation Theorem, I ~ 0.4 + (0~~) 5 ~ 0.40102
(correct to five decimal places).
x -ln(1 + x) x- (x - l x 2 + l x 3 - lx 4 + lx 5 - • • ·) lx 2 - lx 3 + l x 4 - l x 5 + · · ·
55. lim = lim 2 3 ., 4 fi = lim 2 :l 4 5
x-0 x2 x-.0 x· .,_,o X 2
= lim(! - l x + lx 2 - ix3 +: .. ) = !
m-+0 2 3 4 5 2
since power series are continuous functions.
sinx- x + l x 3
6
57. lim
x-o x5
since power series are continuous functions.
. 2 ~ ~ ~ ~ ~
_59. From Equation 1 I, we have e-x = 1 - I + - 21
- I + · · · and we know that cos x . = 1 - - 21
+ - 41
- · · · from
1. . 3. . .
Equation 16. Therefore, e-"' 2 cosx = (1 - x 2 + ~x 4 - ·- ·) (1 - ~x 2 + i4x 4 - - - • ). Writing only the tenns with -
degree < 4 we get e-"' 2 cos x = 1 - lx 2 + .l.x 4 - x 2 + lx" + lx 4 + . ·. = 1 - !!.x 2 + ~x 4 +- ·.
- ' 2 24 2 2 2 24 .
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