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

SECTION 11.8 POWER SERIES 0 n
(b) It does not follow that I:::'=o en( - 4)" is necessarily convergent. [See the comments after Theorem 3 about convergence at
the endpoint of an interval. An example is Cn = ( - 1 t / ( n4").]
(n!)k n
31. If a,.=-( )' x , then
kn.
lim I an+ll = lim [(n + 1)!]k (kn)! lxl = lim - (n + 1)k lxl
....... oo a., n ..... ex> (n!)k [k(n + 1)]! ,.....,CX> (kn + k)(kn + k- 1) · · · (kn + 2)(kn + 1)
. [ (n+1) (n+1) (n+1)]
= J~ (kn + 1) (kn + 2) ... (kn + k) lxl
- l. t.m [--- n + 1 ] lim [--- n + 1 ] · · · 1. liD [--- n + 1 ] I X I
- n-+ex> kn + 1 ,...... ex> kn + 2 n-oo kn + k
= ( ~ J lxl < 1 ¢:} lxl < kk for convergence, and the radius of convergence is R = kk.
33. No. If a power series is centered at a, its interval of convergence is symmetric about a. lf a power series has an infinite radius
of convergence, then its interval of convergence must be (- oo, oo), not [0, oo).
(-1t X2n+l
35. (a) If a,. = I(
1 ) 122n+l, then
n. n+ .
I
a I I
r n+l li
x2n+3 . nl(n
.
+ 1)'22n+l
.
l (x)2
lim
1
,.:.,~ ---a,: = n ..... ~ (n + 1)!(n + 2)! 22n+3 . x2n+l = 2 ,..._
So J 1 ( x) converges for all x and its domain is (- oo, oo).
(b), (c) The initial terms of J1(x) upton = 5 are ao = ~·
..... oo (n + 1)(n + 2) = 0 for all x.
xu
and a(j = - 176
, 94 7
, 200
. The partial sums seem to
approximate J1 ( x) well near the origin, but as lx l increases,
we need to take a large number of terms to get a good
approximation.
37. S2n-l = 1 + 2x + x 2 + 2x 3 + x 4 + 2x 5 + · · · + x 2 "- 2 + 2x 2 "- 1
= 1(1 + 2x) + x 2 (1 + 2x) + x 4 (1 + 2x) + · · · + x 2 " - 2 (1 + 2x) = (1 + 2x)(1 + x 2 + x 4 + ... + x 2 ~ - 2 )
1 - x 2 " 1 + 2x
= (1 + 2:c)- 1
-- 2 [by (11.2.3) with r = x 2 ] --+ -
1
-- 2 as n--+ oo by (11.2.4), when lx l < 1.
- X - X
I
2n 1 +2x. 2n 0" I I 1Th &:
A so
1+2x. .
S2n = S2n-1 + x --+ _ x 1 2 smce x --+ •Or x < . ere,ore, Sn --+ _ x 1 2
smce S2n and S2n-1 both
1 + 2x . . 1 + 2x
approach --- 2 as n --+ oo. Thus, the mterval of convergence IS ( - 1, 1) and f ( x) = - - 2 .
. 1-x 1 - x
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