find the interval of (a,b), f(x)=x to the power 4 + x to d power 3 - x to d power 2 - 2
Answers
Answer:
The power series is
∞
∑
n
=
0
(
−
1
)
n
2
n
+
1
n
+
1
x
n
+
2
, its radius of convergence is
R
=
1
2
.
Explanation:
Start with a basic power series:
1
1
−
x
=
∞
∑
n
=
0
x
n
Replace
x
with
−
2
x
:
1
1
−
(
−
2
x
)
=
∞
∑
n
=
0
(
−
2
x
)
n
1
1
+
2
x
=
∞
∑
n
=
0
(
−
1
)
n
2
n
x
n
Integrate both sides. In the series, we integrate just the part with
x
.
∫
d
x
1
+
2
x
=
∞
∑
n
=
0
(
−
1
)
n
2
n
∫
x
n
d
x
1
2
ln
(
1
+
2
x
)
=
C
+
∞
∑
n
=
0
(
−
1
)
n
2
n
x
n
+
1
n
+
1
Don't forget the
1
/
2
in the integral. Absolute value bars aren't needed for
ln
(
1
+
2
x
)
since we're only working in
x
≥
0
.
Add a constant of integration. Fortunately, letting
x
=
0
shows that
C
=
0
, so this isn't really an issue.
The next step to reaching the "goal function" of
x
ln
(
1
+
2
x
)
. To do this, multiply both sides by
2
x
.
x
ln
(
1
+
2
x
)
=
2
x
∞
∑
n
=
0
(
−
1
)
n
2
n
n
+
1
x
n
+
1
Bring
2
x
into the series:
x
ln
(
1
+
2
x
)
=
∞
∑
n
=
0
(
−
1
)
n
2
n
+
1
n
+
1
x
n
+
2
x
ln
(
1
+
2
x
)
=
2
x
2
−
2
x
3
+
8
3
x
4
−
4
x
5
+
...
For the radius of convergence of the series
∑
a
n
, find the ratio
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
. Here,
a
n
=
(
−
1
)
n
2
n
+
1
n
+
1
x
n
+
2
.
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
=
∣
∣
∣
∣
∣
(
−
1
)
n
+
1
2
n
+
2
n
+
2
x
n
+
3
(
−
1
)
n
2
n
+
1
n
+
1
x
n
+
2
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
(
−
1
)
n
+
1
(
−
1
)
n
2
n
+
2
2
n
+
1
x
n
+
3
x
n
+
2
n
+
1
n
+
2
∣
∣
∣
∣
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
=
∣
∣
∣
−
2
x
n
+
1
n
+
2
∣
∣
∣
Find its limit as
n
→
∞
:
lim
n
→
∞
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
=
lim
n
→
∞
∣
∣
∣
−
2
x
n
+
1
n
+
2
∣
∣
∣
The limit only concerns how
n
changes, so
|
2
x
|
can be extracted from the limit (we can also disregard the
−
1
as we're working within absolute values):
lim
n
→
∞
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
=
2
|
x
|
lim
n
→
∞
∣
∣
∣
n
+
1
n
+
2
∣
∣
∣
The limit is
1
:
lim
n
→
∞
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
=
2
|
x
|
The ratio test states that
∑
a
n
converges when
lim
n
→
∞
∣
∣
∣
a
n
+
1
a
n
∣
∣
∣
<
1
. So, our interval of convergence will occur when:
2
|
x
|
<
1
Or:
|
x
|
<
1
2
Thus, the radius of convergence is
R
=
1
2
.