Applied Calculus - 7e - c 11.pdf
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11-W3979 11/2/06 1:51 PM Page 697
11
Taylor Polynomials
and Infinite Series
W
hat percentage of the nonfarm
workforce will be in the service
industries one decade from now?
In Example 4, page 703, you
will see how a Taylor polynomial
can be used to help answer this
question.
I
N THIS CHAPTER we show how certain functions can be represented by a
power series
. A power series involves infinitely many terms, but when
truncated it is just a polynomial. By approximating a function with a
Taylor
polynomial
, we are often able to obtain approximate solutions to problems
that we cannot otherwise solve.
We also look at
Newton’s method
for finding the zeros of a function. For
example, Newton’s method can be used to find the critical points of a func-
tion, which, as you may recall, are candidates for the solution of the opti-
mization problems considered in Chapter 4.
697
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698
11
TAYLOR POLYNOMIALS AND INFINITE SERIES
11.1
Taylor Polynomials
As we saw earlier, obtaining an exact solution to a problem is not always possible; in
such cases we have to settle for an approximate solution. In this section we show
how a function may be approximated near a given point by a polynomial. Poly-
nomials, as we have seen time and again, are easy to work with; for example, they
are easy to evaluate, differentiate, and integrate. Thus, by using polynomials rather
than working with the original function itself, we can often obtain approximate solu-
tions to a problem that we might otherwise not be able to solve.
Taylor Polynomials
Suppose we are given a differentiable function
f
and a number
a
in the domain of
f
.
Then the polynomial of degree 0 that best approximates
f near x
a
is the constant
polynomial
P
0
(
x
)
f
(
a
)
which coincides with
f
at
x
a
(Figure 1a).
Now, unless
f
itself is a constant function, it is possible in many cases to obtain
a better approximation of
f near x
a
by using a polynomial function of degree 1.
Recall that the linear function
L
(
x
)
f
(
a
)
f
(
a
)(
x
a
)
is just an equation of the tangent line to the graph of the function
f
at the point
(
a, f
(
a
)) (Figure 1b). As such, the value of the function
L
coincides with the value of
f
at
x
a
, and its slope coincides with the slope of
f
at
x
a;
that is,
L
(
a
)
f
(
a
)
and
L
(
a
)
f
(
a
)
Let’s write
L
(
x
)
P
1
(
x
)
y
y
y
=
P
1
(
x
)
=
f
(
a
)
+
f'
(
a
) (
x
–
a
)
y
=
f
(
x
)
y
=
f
(
x
)
(
a, f
(
a
))
(
a, f
(
a
))
y
=
P
0
(
x
)
=
f
(
a
)
x
x
a
a
(a)
P
0
(
x
)
f
(
a
) is a zero-degree poly-
(b)
P
1
(
x
)
f
(
a
)
f
(
a
)(
x
a
) is a first-
nomial that approximates
f
near
x
a
.
degree polynomial that approximates
f
near
FIGURE
1
x
a
.
11-W3979 11/2/06 1:51 PM Page 699
11.1
699
TAYLOR POLYNOMIALS
Then
P
1
(
x
)
f
(
a
)
f
(
a
)(
x
a
)
is the required polynomial approximation (of degree 1) of
f
near
x
a
.
This discussion suggests that yet a better approximation to
f
at
x
a
may be
found by using a polynomial of degree 2,
P
2
(
x
), and requiring that its value, slope,
and concavity coincide with those of
f
at
x
a
. In other words,
P
2
(
x
) should satisfy
the three conditions
2
(
a
)
2
(
a
)
P
2
(
a
)
f
(
a
)
P
f
(
a
)
P
f
(
a
)
The third condition ensures that the graph of the polynomial bends in the right way,
at least near
x
a
. Pursuing this line of reasoning, we are led to the search for a
polynomial of degree
n
in
x
a
,
1
2
1
2
1
2
2
P
n
x
a
0
a
1
x
a
a
2
x
a
p
1
2
3
1
2
n
a
3
x
a
a
n
x
a
(where
a
0
,
a
1
, . . . ,
a
n
are constants), that satisfies the conditions
1
2
1
2
,
P
n
1
2
f
¿
1
2
,
P
n
1
2
f
–
1
2
p
,
P
1
n
2
1
2
1
n
2
1
2
P
n
a
f
a
a
a
a
a
,
a
f
a
(1)
n
To determine the required polynomial, we compute
p
2
n
1
P
n
1
x
2
a
1
2
a
2
1
x
a
2
3
a
3
1
x
a
2
na
n
1
x
a
2
3
#
2
a
3
p
n
2
P
n
1
x
2
2
a
2
1
x
a
2
n
1
n
1
2
a
n
1
x
a
2
3
#
2
a
3
4
#
3
#
2
a
4
p
n
3
P
n
1
x
2
1
x
a
2
n
1
n
1
21
n
2
2
a
n
1
x
a
2
o
p
1
n
2
1
P
n
x
2
n
1
n
1
21
n
2
2
1
1
2
a
n
n
(
x
),
P
n
(
x
), . . . ,
P
(
n
)
(
x
) in suc-
Setting
x
a
in each of the expressions for
P
n
(
x
),
P
cession and using the conditions in (1), we find
1
2
1
2
P
n
a
a
0
f
a
P
n
1
2
f
¿
1
2
a
a
1
a
P
n
1
2
f
–
1
2
a
2
a
2
a
3
#
2
a
3
P
n
1
2
f
‡
1
2
a
a
o
p
1
n
2
1
2
1
21
2
1
2
1
n
2
1
2
P
a
n
n
1
n
2
1
a
n
f
a
n
from which we deduce that
1
2
f
–
1
3
#
2
f
‡
1
2
f
¿
1
2
1
2
1
2
p
,
a
0
f
a
,
a
1
a
,
a
2
a
,
a
3
a
,
1
1
n
2
1
2
a
n
2
f
a
p
1
21
2
1
n
n
1
n
2
1
Let’s introduce the expression
n
! (read “
n
factorial”), defined by
p
3
#
2
#
1
1
21
21
2
1
2
n
!
n
n
1
n
2
n
3
for
n
1
0!
1
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700
11
TAYLOR POLYNOMIALS AND INFINITE SERIES
Thus,
#
#
#
1!
1
4!
4321
24
#
#
#
#
#
2!
21
2
5!
54321
120
#
#
3!
321
6
and so on. Using this notation, we may write the coefficients of
P
n
(
x
) as
1
2!
f
–
1
3!
f
‡
1
n
!
f
1
2
f
¿
1
2
1
2
1
2
p
,
a
n
1
n
2
1
2
a
0
f
a
,
a
1
a
,
a
2
a
,
a
3
a
,
a
so that the required polynomial is
f
–
1
2
1
n
2
1
2
a
f
a
p
2
n
P
n
1
x
2
f
1
a
2
f
¿
1
a
21
x
a
2
1
x
a
2
1
x
a
2
2!
n
!
The
n
th Taylor Polynomial
Suppose that the function
f
and its first
n
derivatives are defined at
x
a
. Then
the
n
th Taylor polynomial of
f
at
x
a
is the polynomial
1
2
1
2
f
¿
1
21
2
P
n
x
f
a
a
x
a
f
–
1
2
1
n
2
1
2
a
f
a
p
2
n
1
x
a
2
1
x
a
2
(2)
2!
n
!
which coincides with
f
(
x
),
f
(
x
), . . . ,
f
(
n
)
(
x
) at
x
a
; that is,
1
n
2
1
n
2
1
P
n
1
a
2
f
1
a
2
,
P
n
1
a
2
f
¿
1
a
2
,
p
,
P
1
a
2
f
a
2
n
Using a Taylor Polynomial to Approximate a Function
In many instances the Taylor polynomial
P
n
(
x
) provides us with a good approxima-
tion of
f
(
x
) near
x
a
.
EXAMPLE 1
e
x
0
and sketch the graph of each polynomial superimposed upon the graph of
f
(
x
)
Find the first four Taylor polynomials of
f
(
x
)
at
x
e
x
.
Solution
Here
a
0 and, since
f
(4)
(
x
)
e
x
f
(
x
)
f
(
x
)
f
(
x
)
f
(
x
)
we find
f
(4)
(0)
f
(0)
f
(0)
f
(0)
f
(0)
1
Using Formula (2) with
n
1, 2, 3, and 4 in succession, we find that the first four
Taylor polynomials are
11-W3979 11/2/06 1:51 PM Page 701
11.1
701
TAYLOR POLYNOMIALS
1
2
1
2
f
¿
1
21
2
P
1
x
f
0
0
x
0
1
x
f
–
1
2
0
1
2
x
2
2
P
2
1
x
2
f
1
0
2
f
¿
1
0
21
x
0
2
1
x
0
2
1
x
2!
f
–
1
0
2
f
‡
1
0
2
1
2
1
2
f
¿
1
21
2
1
2
2
1
2
3
P
3
x
f
0
0
x
0
x
0
x
0
2!
3!
1
2
x
2
1
6
x
3
1
x
f
–
1
2
f
‡
1
2
0
0
2
3
P
4
1
x
2
f
1
0
2
f
¿
1
0
21
x
0
2
1
x
0
2
1
x
0
2
2!
3!
1
4
2
1
f
0
2
1
2
x
2
1
6
x
3
1
24
x
4
1
2
4
x
0
1
x
4!
The graphs of these polynomials are shown in Figure 2. Observe that the ap-
proximation of
f
(
x
) near
x
0 improves as the degree of the approximating
Taylor polynomial increases.
y
y
y
=
e
x
y
=
e
x
y
= 1 +
x
+
1
x
2
2
y
= 1 +
x
x
x
y
y
y
=
e
x
y
=
e
x
y
= 1 +
x
+
1
x
2
2
+
1
x
3
+
1
x
4
6
24
FIGURE
2
The graphs of the first four Taylor poly-
nomials of
f
(
x
)
e
x
at
x
0 superim-
posed upon the graph of
f
(
x
)
e
x
.
x
x
y
= 1 +
x
+
1
x
2
+
1
x
3
2
6
EXPLORING WITH TECHNOLOGY
Let
f
(
x
)
xe
x
.
1.
Find the first four Taylor polynomials of
f
at
x
0.
2.
Use a graphing utility to plot the graphs of
f
,
P
1
,
P
2
,
P
3
, and
P
4
on the same set of
axes in the viewing window [
0.5, 1]
[
0.5, 0.5].
3.
Comment on the approximation of
f
by the polynomial
P
n
near
x
0 for
n
1, 2, 3,
and 4. What happens to the approximation if
x
is “far” from the origin?
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