Keywords 020 100points find a power series

keywords:
020 10.0points
Find a power series representation for the
function
f ( z ) =
1
5 + 16 z 2
.
1. f ( z ) =
∞ summationdisplay
n =0
( - 1)
n
4
2 n
z
2 n
5
n
2. f ( z ) =
∞ summationdisplay
n =0
4
n
z
n
5
n +1

choice (hac762) – HW Quest Week 8 – cepparo – (53850) 10
3. f ( z ) =
∞
summationdisplay
n =0
( - 1)
n
4
2 n
z
2 n
5 n +1
correct
4. f ( z ) =
∞
summationdisplay
n =0
4
2 n
z
2 n
5 n +1
5. f ( z ) =
∞
summationdisplay
n =0
( - 1)
n
4
n
z
n
5 n +1
Explanation:
After simplification,
f ( z ) =
1
5 + 16 z
2
=
1
5
parenleftbigg
1
1 +
16
5
z
2
parenrightbigg
=
1
5
parenleftbigg
1
1 - ( -
16
5
z 2 )
parenrightbigg
.
On the other hand,
1
5
∞
summationdisplay
n =0
parenleftbigg
-
16
5
z
2
parenrightbigg
n
=
∞
summationdisplay
n =0
( - 1)
n
4
2 n
z
2 n
5
n +1
keywords:
021 10.0points
Find a power series representation centered
at the origin for the function
f ( y ) =
y
3
(6 - y )
2
.
1. f ( y ) =
∞ summationdisplay
n =3
n - 2
6
n - 1
y
n
correct
2. f ( y ) =
∞ summationdisplay
n =3
n
6
n
y
n
3. f ( y ) =
∞ summationdisplay
n =2
1
6
n - 1
y
n
4. f ( y ) =
∞ summationdisplay
n =3
1
6
n - 3
y
n
5. f ( y ) =
∞ summationdisplay
n =2
n - 1
6
n
y
n
Explanation:
By the known result for geometric series,
1
6 - y
=
1
6
parenleftBig
1 -
y
6
parenrightBig
=
1
6
∞
summationdisplay
n =0
parenleftBig
y
6
parenrightBig
n
=
∞
summationdisplay
n =0
1
6 n +1
y
n
.
This series converges on ( - 6 , 6).
On the other hand,
1
(6 - y )
2
=
d
dy
parenleftBig
1
6 - y
parenrightBig
,
and so on ( - 6 , 6),
1
(6 - y ) 2
=
d
dy
parenleftBigg
∞
summationdisplay
n =0
y
n
6 n +1
parenrightBigg
=
∞ summationdisplay
n =1
n
6
n +1
y
n - 1
=
∞ summationdisplay
n =0
n + 1
6
n +2
y
n
.
Thus
f ( y ) = y
3
∞
summationdisplay
n =0
n + 1
6 n +2
y
n
=
∞
summationdisplay
n =0
n + 1
6 n +2
y
n +3
.
Consequently,
f ( y ) =
∞ summationdisplay
n =3
n - 2
6
n - 1
y
n
.
022 10.0points
Find a power series representation for the
function
f ( x ) = tan
- 1
(2 x ) .
1. f ( x ) =
∞ summationdisplay
n =0
1
2 n + 1
x
2 n +1
2. f ( x ) =
∞ summationdisplay
n =0
2
2 n +1
2 n + 1
x
2 n +1

choice (hac762) – HW Quest Week 8 – cepparo – (53850) 11
3. f ( x ) =
∞
summationdisplay
n =0
( - 1)
n
2
2 n +1
2 n + 1
x
2 n +1
correct
4. f ( x ) =
∞
summationdisplay
n =0
( - 1)
n
2
2 n
2 n + 1
x
2 n
5. f ( x ) =
∞ summationdisplay
n =0
( - 1)
n
1
2 n + 1
x
2 n +1
6. f ( x ) =
∞ summationdisplay
n =0
1
2 n + 1
x
2 n
Explanation:
We know that
tan
- 1
( x ) =
integraldisplay
x
0
1
1 + t
2
dt .
On the other hand,
1
1 - x
= 1 + x + x
2
+ . . .
on the interval ( - 1 , 1). Replacing x by - x
2
we thus see that
1
1 + x
2
= 1 - x
2
+ x
4
- x
6
+ . . .
=
∞ summationdisplay
n =0
( - 1)
n
x
2 n
on ( - 1 , 1). But then
f ( x ) =
integraldisplay
x
0
parenleftBig
∞
summationdisplay
n =0
( - 1)
n
t
2 n
parenrightBig
dt
=
∞
summationdisplay
n =0
parenleftBig
( - 1)
n
integraldisplay
x
0
t
2 n
dt
parenrightBig
=
∞ summationdisplay
n =0
( - 1)
n
2 n + 1
x
2 n +1
on ( - 1 , 1). Consequently,
tan
- 1
(2 x ) =
∞
summationdisplay
n =0
( - 1)
n
2
2 n +1
2 n + 1
x
2 n +1
is a power series representation for tan
- 1
(2 x )
on
parenleftbigg
-
1
2
,
1
2
parenrightbigg
.
023 10.0points
Evaluate the integral
f ( t ) =
integraldisplay
t
0
s
1 - s 4
ds .
as a power series.
1. f ( t ) =
∞
summationdisplay
n =0
t
4 n +2
4 n + 2
correct
2. f ( t ) =
∞
summationdisplay
n =4
t
4 n
4 n + 2
3. f ( t ) =
∞
summationdisplay
n =0
t
4 n
4 n
4. f ( t ) =
∞ summationdisplay
n =0
( - 1)
n
t
4 n +2
4 n + 2
5. f ( t ) =
∞ summationdisplay
n =0
( - 1)
n
t
4 n
4 n
Explanation:
By the geometric series representation,
1
1 - s
=
∞ summationdisplay
n =0
s
n
,
and so
s
1 - s 4
=
∞
summationdisplay
n =0
s
4 n +1
.
But then
f ( t ) =
integraldisplay
t
0
parenleftBig
∞ summationdisplay
n =0
s
4 n +1
parenrightBig
ds
=
∞
summationdisplay
n =0
parenleftBig
integraldisplay
t
0
s
4 n +1
ds
parenrightBig
.
Consequently,
f ( t ) =
∞
summationdisplay
n =0
t
4 n +2
4 n + 2
.