MATH 152 - Recitation Quiz 9 - Spring 2015
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1. Find a power series representation for the following function and determine the interval of convergence
for that series:
f (x) =
x
x−3
Solution 1:
First note that
− x3
x
−x
−x
=
.
f (x) =
=
=
x−3
3−x
3 1 − x3
1 − x3
Therefore, the power series for
f
is
f (x) =
Note that this converges so long as
is
∞ ∞
X
x x n X x n+1
−
=
.
−
3
3
3
n=0
n=0
x
< 1 =⇒ |x| < 3.
3
Therefore, the interval of convergence for the series
(−3, 3) .
Solution 2:
First note that
f (x) =
Therefore, the power series for
f
x
x
1
=
3 =
x−3
1−
x 1− x
is
f (x) =
∞ n
X
3
n=0
Note that this converges so long as
is
(−∞, −3) ∪ (3, ∞) .
x
g (x) = ln (5 − x)
.
.
3
< 1 =⇒ 3 < |x| . Therefore,
x
2. Find a power series representation for
3
x
the interval of convergence for the series
and determine the radius of convergence.
First note that
ˆ
g (x) = ln (5 − x) = −
1
dx = −
5−x
ˆ
1
1
dx = −
·
5 1 − x5
∴ g (x) = C −
∞ X
n=0
ˆ X
∞ n
1
x
1
n+1
∴ g (0) = ln 5 = C =⇒ g (x) = ln 5 −
n=0
Since the series converges if and only if
5
1
5
x n+1
.
5
∞ X
n=0
x
< 1,
5
dx = C−
1
n+1
x n+1
.
5
the radius of convergence is 5.
∞ ˆ n
X
1
x
dx.
5
5
n=0
3. Find the Taylor series for
h
centered at 2 in addition to its interval of convergence if
n
h(n) (2) =
The Taylor series for
h
centered at 2 is given by:
h (x) =
Note now that
(−1) n!
.
5n (n + 5)
∞
∞
n
n
X
X
h(n) (2)
(−1) (x − 2)
n
(x − 2) =
.
n!
5n (n + 5)
n=0
n=0
(x − 2)n+1 5n (n + 5) 1
lim |x − 2| < 1 =⇒ |x − 2| < 5.
n =
n→∞ 5n+1 (n + 6) (x − 2) 5
Therefore, the radius of convergence is 5 which means that if
converges. We now have to check the endpoints. If
h (x) =
lim
n→∞
P
1
n
P 1
p-test,
n+5
x = −3. If x = 7, then
diverges by the
Taylor series diverges when
x ∈ (2 − 5, 2 + 5) = (−3, 7) ,
then the series
then
∞
∞
n
n
X
X
(−1) (−5)
1
=
.
n
5 (n + 5)
n+5
n=0
n=0
Note that
Therefore, since
x = −3,
h (x) =
1
n+5
1
n
= 1.
diverges by the Limit Comparison Test. Therefore, the
∞
∞
n
n
X
X
(−1)
(−1) · 5n
=
,
5n (n + 5) n=0 n + 5
n=0
which converges by the alternating series test. Therefore, the interval of convergence for the Taylor series is
(−3, 7] .
2