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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
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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