This preview shows page 1. Sign up to view the full content.
Unformatted text preview: he series n=0 ∞ ∞ cn z n x (−1)n t2n dt = = 2, n=0 0 n=0 is
∞ =
n=0 (−1)n
2n + 1 x2n+1 . 1
(i) convergent when z  < , and
2
1
(ii) divergent when z  > .
2 Consequently, on (−1, 1)
∞ f ( x) =
n=0 008 (−1)n 2n+5
x
.
2n + 1 On the other hand, since
lim 10.0 points Compare the radius of convergence, R1 , of
the series n→∞ (n + 1)cn+1
ncn = lim n→∞ cn+1
,
cn the Ratio Test ensures also that the series ∞ ∞ cn z n n cn z n−1 n=0 with the radius of convergence, R2 , of the
series n=1 is ∞ n cn z n−1
n=1 1
(i) convergent when z  < , and
2 stiurca (mas7745) – 11.9: Representations as Series – meth – (91825)
1
(ii) divergent when z  > .
2 Thus
∞ f ( x) = x3 Consequently, 009 n=0 1
.
2 R1 = R2 = x3 ∞ n=2
∞ 2. f (x) =
n=3
∞ 3. f (x) =
n=3
∞ 4. f (x) =
n=3
∞ 5. f (x) =
n=2 (6 − x)2 ∞ . n−1 n
x
6n
1
6n−3 xn nn
x
6n
n−2 n
x correct
6n−1
1
6n−1 xn Explanation:
By the known result for geometric series,
1
1
=
x
6−x
6 1−
6
= 1
6 ∞ x
6 n=0 n ∞ 1 =
n=0 6n+1 xn . This series converges on (−6, 6).
On the other hand,
1
1
d
,
=
2
(6 − x)
dx 6 − x
and so on (−6, 6),
d
1
=
(6 − x)2
dx
∞ =
n=1 n
6n+1 ∞ n=0 xn
6n+1
∞ x n−1 =
n=0 ∞ n=0 n + 1 n+3
x
.
6n+2 Consequently, Find a power series representation centered
at the origin for the function 1. f (x) = n+1 n
x=
6n+2 10.0 points f ( x) = 6 n+1 n
x.
6n+2 f ( x) =
n=3 n−2 n
x.
6n−1...
View
Full
Document
This document was uploaded on 09/17/2013.
 Summer '13

Click to edit the document details