momin (rrm497) – Homework 14 – cheng – (58520)
1
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printout
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have
12
questions.
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before answering.
001
10.0 points
Suppose
T
4
(
x
) = 8

4(
x

2) + 8(
x

2)
2

6(
x

2)
3
+ 2(
x

2)
4
is the degree 4 Taylor polynomial centered at
x
= 2 for some function
f
.
What is the value of
f
(3)
(2)?
1.
f
(3)
(2) = 6
2.
f
(3)
(2) = 36
3.
f
(3)
(2) =

6
4.
f
(3)
(2) = 2
5.
f
(3)
(2) =

36
correct
6.
f
(3)
(2) =

2
Explanation:
Since
T
4
(
x
) =
f
(2) +
f
′
(2)(
x

2) +
f
′′
(2)
2!
(
x

2)
2
+
f
(3)
(2)
3!
(
x

2)
3
+
f
(4)
(2)
4!
(
x

2)
4
,
we see that
f
(3)
(2) =

3!
×
6 =

36
.
002
10.0 points
Find the Taylor series centered at
x
= 0 for
f
(
x
) = cos 5
x.
1.
∞
summationdisplay
n
=0
(

1)
n
5
2
n
n
!
x
2
n
2.
∞
summationdisplay
n
=0
(

1)
n
5
2
n
n
!
x
n
3.
∞
summationdisplay
n
=1
5
n
(2
n
)!
x
n
4.
∞
summationdisplay
n
=0
(

1)
n
5
2
n
(2
n
)!
x
2
n
correct
5.
∞
summationdisplay
n
=0
(

1)
n
5
2
n
(2
n
)!
x
n
6.
∞
summationdisplay
n
=1
5
n
(2
n
)!
x
2
n
Explanation:
The Taylor series centered at
x
= 0 for any
f
is
f
(0) +
f
′
(0)
x
+
1
2!
f
′′
(0)
x
2
+
...
=
∞
summationdisplay
n
=0
1
n
!
f
(
n
)
(0)
x
n
.
But when
f
(
x
) = cos(5
x
),
n
f
(
n
)
(
x
)
f
(
n
)
(0)
0
cos(5
x
)
1
1

5 sin(5
x
)
0
2

25 cos(5
x
)

25
3
125 sin(5
x
)
0
4
625 cos(5
x
)
625
.
.
.
.
.
.
in other words,
f
′
(0) =
f
′′′
(0) =
f
(5)
(0) =
...
= 0
,
while
f
(0) = 1
,
f
′′
(0) =

5
2
,
f
(4)
(0) = 5
4
.
Thus in general,
f
(2
n
+1)
(0) = 0
,
f
(2
n
)
(0) = (

1)
n
5
2
n
.
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momin (rrm497) – Homework 14 – cheng – (58520)
2
Consequently, the Taylor series of cos 5
x
cen
tered at
x
= 0 is
∞
summationdisplay
n
=0
(

1)
n
5
2
n
(2
n
)!
x
2
n
.
keywords:
003
10.0 points
Find the degree 2 Taylor polynomial of
f
centered at
x
= 2 when
f
(
x
) = 5
x
ln
x.
1.
10 + 5 ln 2(
x

2) +
5
2
(
x

2)
2
2.
10 ln 2 + 5(ln 2 + 1)(
x

2) +
5
4
(
x

2)
2
correct
3.
10 ln 2 + 5 ln 2(
x

2) +
5
4
(
x

2)
2
4.
10 + 5(ln 2 + 1)(
x

2) +
5
4
(
x

2)
2
5.
10 ln 2 + 5(ln 2 + 1)(
x

2) +
5
2
(
x

2)
2
6.
10 + 2 ln 5(
x

2) +
5
4
(
x

2)
2
Explanation:
The degree 2 Taylor polynomial of
f
cen
tered at
x
= 2 is given by
T
2
(
x
) =
f
(2) +
f
′
(2)(
x

2)
+
1
2!
f
′′
(2)(
x

2)
2
.
When
f
(
x
) = 5
x
ln
x
, therefore,
f
′
(
x
) = 5 ln
x
+ 5
,
f
′′
(
x
) =
5
x
.
But when
f
(2) = 10 ln 2,
f
′
(2) = 5(ln 2 + 1)
,
f
′′
(2) =
5
2
.
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 Fall '08
 RAdin
 Calculus, Power Series, Taylor Series, Natural logarithm

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