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

SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES 0 81
2x 2x 3 2x 5
The partial sums are s1 = l' s2 = s1 + 3' S3 = s2 + 5 ' · · · ·
As n increases, s,.(x) approximates f better on the interval of
convergence, which is {-1, 1).
25. _ t_ B = t . _1_ 8 = t f: ( tB )" = f: t8n+l
1 - t 1 - t n = O n =O
=> j . t 00 tB·n+2 1
--dt = C + I: ---. The series for -- 8
converges
1 - t 8 n =O 8n + 2 1 - t
when lt 8 l < 1 ¢} It I < 1, so R = 1 for that series and also the series for t/{1 - t 8 ). By Theorem 2, the series for
I
_!_a dt also has R = 1.
1 - t
oo x" oo x"+2
27. From Example 6,ln{1 + x) = E ( - 1)"- 1 - for lx l < 1, so x 2 ln{1 + x) = I: {- 1t- 1 - -and
n =1 n n = 1 n
I
x 2 ln(1 + x) dx = C + I: ( -1)"' (
n=1 n n+ 3
oo x"+3 .
)' R = 1 for the series for ln{1 + x), so R = 1 for the series representing
x 2 ln{1 + x) as well. By Theorem 2, the series for I x 2 ln{l + x) dx also has R = 1.
29
1 1
f;(-x 5 )"=f:{-1)"xu" =>
· 1 + x 5 = 1 - (- x·5 ) = o
I
n =O n =
1 ;· oo oo ,&n+l
--.. dx = 2:{- l )"'x 5 " dx = C + 2:(-1)"-x
5 1 . Thus,
1 + x~ ,.=o n=O n +
0.2 1 [ x6 x lt . ] 0.2 {0.2)6 {0.2) u . .
I= - - dx = x- - + - - ··· = 0.2---+ - ---· ··. The senes IS alternating so if we use
. £ 0 1 + x 5 6 11 0 6 11 . '
the first two terms, the error is at most (0.2) 11 /11 ~ 1.9 x 10- 9 . So I ~ 0.2 - {0.2) 6 /6 ~ 0.199. 989 to six decimal places.
31. We substitute 3x for x in Example 7, and find that
j . I 00 "' (3x?"+l I 00 ".32.. +1 x2n+2 00 . .. 32n+l x2n+3
xarctan(3:c)d:v = x 2:(-1) dx= 2:: (- 1)
dx = C+ E(- 1) ( )( )
n = O 2 n -1- 1
n = O 2 n + 1
n =O 2n -1- 1 2n -j- 3
1
0 .1 [
3
So x arctan(3x) dx = .....:::.__ 3 33 G 35 7 37 !) ] 0 .1
-~ +~-__E._+ ···
0 1 ·3 3·5 5·7 7 · 9 0
1 9 243 2187
= 103 - 5 X 10° + 35 X 107 - 63 X 1Q9 + .. ..
• • 1 . 'f h 1 •
2187
3 8
The senes IS a ternatmg, so 1 we use t ree terms, t 1e error IS at most x ~ .5 x 10- . s o
63 109
r ·l 1 9 243
J
0
x arctan(3x) dx ~ 103
-
5 x 105 + 35 x ~ 0.000 983 to six decimal places.
107