33. continued
(c)
The coefficient is
}
f
(
k
k
)
!
(0)
}
5
(d)
f
(0)
5
1,
f
9
(0)
5
m
, and we’re done by part (c).
34.
Because
f
(
x
)
5
(1
1
x
)
m
is a polynomial of degree m.
Alternately, observe that
f
(
k
)
(0)
5
0 for
k
$
m
1
1.
■
Section 9.3
Taylor’s Theorem
(pp. 480–487)
Exploration 1
Your Turn
1.
We need to consider what happens to
R
n
(
x
) as
n
→
‘
.
By Taylor’s Theorem,
R
n
(
x
)
5
}
f
(
n
(
n
1
1
1)
1
(
c
)!
)
}
(
x
2
0)
n
, where
f
(
n
1
1)
(
c
) is the (
n
1
1)st derivative of cos
x
evaluated at
some
c
between
x
and 0. As with sin
x
, we can say that
f
(
n
1
1)
(
c
) lies between
2
1 and 1 inclusive. Therefore, no
matter what
x
is, we have
)
R
n
(
x
)
)
5
)
}
f
(
n
(
n
1
1
1)
1
(
c
)!
)
}
(
x
2
0)
n
)
#
)
}
(
n
1
1
1)!
}
x
n
)
5
}
(
n
1
)
x
)
n
1)!
}
.
The factorial growth in the denominator, as noted in
Example 3, eventually outstrips the power growth in the
numerator, and we have
}
(
n
1
)
x
)
n
1)!
}
→
0 for all
x
. This means
that
R
n
(
x
)
→
0 for all
x
, which completes the proof.
Exploration 2
Euler’s Formula
1.
e
ix
5
1
1
ix
1
}
(
i
2
x
!
)
2
}
1
}
(
i
3
x
!
)
3
}
1
…
1
}
(
i
n
x
!
)
n
}
1
…
5
1
1
ix
2
}
x
2
2
!
}
2
i
}
x
3
3
!
}
1
}
x
4
4
!
}
1
i
}
x
5
5
!
}
2
…
1
(
i
)
n
}
x
n
n
!
}
1
…
2.
If we isolate the terms in the series that have
i
as a factor,
we get:
e
ix
5
1
1
ix
2
}
x
2
2
!
}
2
i
}
x
3
3
!
}
1
}
x
4
4
!
}
1
i
}
x
5
5
!
}
2
…
1
(
i
)
n
}
x
n
n
!
}
1
…
5
1
2
}
x
2
2
!
}
1
}
x
4
4
!
}
2
}
x
6
6
!
}
1
…
1
(
2
1)
n
}
(2
x
n
2
n
)!
}
1
…
1
i
1
x
2
}
x
3
3
!
}
1
}
x
5
5
!
}
2
}
x
7
7
!
}
1
…
1
(
2
1)
n
}
(2
x
n
2
1
n
1
1
1)!
}
1
…
2
5
cos
x
1
i
sin
x
.
(We are assuming here that we can rearrange the terms of a
convergent series without affecting the sum. It happens to
be true in this case, but we will see in Section 9.5 that it is
not always true.)
3.
e
i
p
5
cos
p
1
i
sin
p
5 2
1
1
0
5 2
1
Thus,
e
i
p
1
1
5
0
Quick Review 9.3
1.
Since
)
f
(
x
)
)
5
)
2 cos (3
x
)
)
#
2 on [
2
2
p
, 2
p
] and
f
(0)
5
2,
M
5
2.
2.
Since
f
(
x
) is increasing and positive on [1, 2],
M
5
f
(2)
5
7.
3.
Since
f
(
x
) is increasing and positive on [
2
3, 0],
M
5
f
(0)
5
1.
4.
Since the minimum value of
f
(
x
) is
f
(
2
1)
5 2
}
1
2
}
and the
maximum value of
f
(
x
) is
f
(1)
5
}
1
2
}
,
M
5
}
1
2
}
.
5.
On [
2
3, 1], the minimum value of
f
(
x
) is
f
(
2
3)
5 2
7 and
the maximum value of
f
(
x
) is
f
(0)
5
2. On (1, 3],
f
is
increasing and positive, so the maximum value of
f
is
f
(3)
5
5. Thus
)
f
(
x
)
)
#
7 on [
2
3, 3] and
M
5
7.
6.
Yes, since each expression for an
n
th derivative given by
the Quotient Rule will be a rational function whose
denominator is a power of
x
1
1.
7.
No, since the function
f
(
x
)
5
)
x
2
2
4
)
has a corner at
x
5
2.
8.
Yes, since the derivatives of all orders for sin
x
and cos
x
are defined for all values of
x
.
9.
Yes, since the function
f
(
x
)
5
e
2
x
has derivatives of the
form
f
(
n
)
(
x
)
5 2
e
2
x
for odd values of
n
and
f
(
n
)
(
x
)
5
e
2
x
for even values of
n
, and both of these expressions are
defined for all values of
x
.
10.
No, since
f
(
x
)
5
x
3/2
, we have
f
9
(
x
)
5
}
3
2
}
x
1/2
and
f
0
(
x
)
5
}
3
4
}
x
2
1/2
, so
f
0
(0) is undefined.
Section 9.3 Exercises
1.
f
(0)
5
e
2
2
x
)
x
5
0
5
1
f
9
(0)
5 2
2
e
2
2
x
)
x
5
0
5 2
2
f
0
(0)
5
4
e
2
2
x
)
x
5
0
5
4, so
}
f
0
2
(
!