Transcript 9.7 Taylor Polynomials
9
Infinite Series
Copyright © Cengage Learning. All rights reserved.
9.7
Taylor Polynomials and Approximations
Copyright © Cengage Learning. All rights reserved.
Objectives
Find polynomial approximations of elementary functions and compare them with the elementary functions.
Find Taylor and Maclaurin polynomial approximations of elementary functions.
Use the remainder of a Taylor polynomial.
3
Polynomial Approximations of Elementary Functions
4
Polynomial Approximations of Elementary Functions To find a polynomial function
P
that approximates another function
f
, begin by choosing a number
c
in the domain of
f
at which
f
and
P
have the same value. That is, The approximating polynomial is said to be
expanded about c
or
centered at c.
Geometrically, the requirement that
P
(
c
) =
f
(
c
) means that the graph of
P
passes through the point (
c
,
f
(
c
)). Of course, there are many polynomials whose graphs pass through the point (
c
,
f
(
c
)).
5
Polynomial Approximations of Elementary Functions The goal is to find a polynomial whose graph resembles the graph of
f
near this point. One way to do this is to impose the additional requirement that the slope of the polynomial function be the same as the slope of the graph of
f
at the point (
c
,
f
(
c
)). With these two requirements, you can obtain a simple linear approximation of
f
, as shown in Figure 9.10.
Figure 9.10
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Example 1 –
First
-
Degree Polynomial Approximation of f
(
x
)
= e x
For the function
f
(
x
)
= e x ,
find a first-degree polynomial function
P
1 (
x
) =
a
0 +
a
1
x
whose value and slope agree with the value and slope of
f
at to
f x
(
x
= 0. In other words, find the tangent line approximation )
= e x
at x = 0.
Solution: Because
f
(
x
)
= e x
and
f'
(
x
)
= e x
, the value and the slope of
f,
at
x
= 0, are given by
f
( 0 )
= e
0 = 1 and
f'
( 0 )
= e
0 = 1.
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Example 1 –
Solution
Because
P
1 (
x
) =
a
0 +
a
1
x
, you can use the condition that
P
1 (0) =
f
(0) to conclude that
a
0 = 1. Moreover, because
P
1
'
(
x
) =
a
1 , you can use the condition that
P
1
'
(0) =
f'
(0) to conclude that
a
1 = 1. cont’d Therefore,
P
1 (
x
) = 1 +
x.
Figure 9.11 shows the graphs of
P
1 (
x
) = 1 +
x
and
f
(
x
)
= e x .
Figure 9.11
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Example 1 –
Solution
P
1 (
x
) = 1 +
x.
Can we do better?
Find P
2 (
x
),
P
3 (
x
), and
P
4 (
x
).
Approximate
e 0.1
u
sing P
4 (
x
).
e
0.1
P
4 1.105170
Based on the observed pattern, Find P
n (
x
) for
f
(
x
)
= e x
around x=0.
n
x
1 2
x
2 1 3!
x
3 1
n
!
x n
cont’d 9
Maclaurin Polynomials
cont’d
What would P
n (
x
) be for any
f
(
x
) around x=0.
n
f
f
x
f
2!
x
2
f
3!
x
3
f n n
!
x n
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Example 2 Find the Maclaurin polynomial of degree 6 for
f
(
x
)
= cos x.
Use the 6 th degree polynomial to approximate cos (0.1).
6 1 2!
x
2 1 4!
x
4 1 6!
x
6 11
Taylor and Maclaurin Polynomials
12
Taylor and Maclaurin Polynomials
When approximating a function with a polynomial, we need all the derivative to agree. 1 st derivatives – slopes agree at the point 2 nd derivatives – curvature agrees at the point, etc.
The polynomial approximation at x=c: Tangent line approximation:
y
f
At x=c, y=f (c)
and
At x=c,
y
f
How do we build a polynomial so that every derivative agrees? 13
Taylor and Maclaurin Polynomials
If we want to build a quadratic approximation to y=f(x) at x=c, we need the following: y=f(c), 1 st derivatives agree, and 2 nd derivatives agree.
y
f
0
x
c
0
f
x
c
1
f
2
2 Ex: Find the quadratic approximation for 2 1 2
x
2 2 sin
x at x
2 .
14
Taylor and Maclaurin Polynomials
Taylor polynomials
are named after the English mathematician Brook Taylor (1685 –1731) and
Maclaurin polynomials
are named after the English mathematician Colin Maclaurin (1698 –1746).
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Example 3 Find the Taylor polynomial
P
4 (
x
) for
f
(
x
)
= ln x centered at c=1. Use it to estimate ln(1.2).
x
1 2
x
1 2 1 3
x
1 3 1 4
x
1 4 16
Day 2 Remainder of a Taylor Polynomial
18
Taylor Polynomials
n
Taylor Polynomial approximation at x=c:
f
f
2!
c
2
f
3!
3
f n n
!
n
K th Taylor Polynomial of f(x) centered at x=c:
n k
0
f n n
!
c n
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Example 4 Find the 3 rd Taylor polynomial for
f
(
x
)
= sin x,
3 2 2 3
x
6 1
x
6 2 3
x
6 3 20
Remainder of a Taylor Polynomial
An approximation technique is of little value without some idea of its accuracy.
To measure the accuracy of approximating a function value
f
(
x
) by the Taylor polynomial
P n
(
x
), you can use the concept of a remainder R
n
(
x
), defined as follows.
21
Remainder of a Taylor Polynomial
So,
R n
(
x
) =
f
(
x
) –
P n
(
x
). The absolute value of
R n
(
x
) is called the
error
associated with the approximation. That is, The next theorem gives a general procedure for estimating the remainder associated with a Taylor polynomial. This important theorem is called
Taylor’s Theorem,
and the remainder given in the theorem is called the
Lagrange form of the remainder.
22
Remainder of a Taylor Polynomial
23
Remainder of a Taylor Polynomial
One useful consequence of Taylor’s Theorem is that
R n x
n
n
1 max
f n
1 where
f n
1 is the maximum value of
f n
1 between x and c.
For n=0, Taylors Theorem states that if f is differentiable in an interval containing c, then, for each x in the interval, there exists z between x and c such that:
f
or f
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Example 8 –
Determining the Accuracy of an Approximation
The third Maclaurin polynomial for sin
x
is given by Use Taylor’s Theorem to approximate sin(0.1) by
P
3 (0.1) and determine the accuracy of the approximation.
Solution: Using Taylor’s Theorem, you have where 0 <
z
< 0.1.
25
Example 8 –
Solution
Therefore, Because
f
(4) (
z
) = sin
z
, it follows that the error |
R
3 (0.1)| can be bounded as follows.
cont’d This implies that the error is between 0 and 0.000004
26
From Harvard Series Booklet 27
Example 9 –
Finding the upper bound for the error of an approximation
Use Taylor’s Theorem to obtain an upper bound for the error of the approximation:
e
1 2 2!
1 3 3!
1 4 4!
1 5 5!
Solution:
e x
,
n
x
1 2!
x
2 1 3!
x
3 1
n
!
x n f
6
e z
e
1 .
5
e
1 6!
6
e
720 0.00378
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Homework
Day 1: pg.656 1-4 all, 13-29 odd, 41,43 Day 2: pg.657 45-51 odd, MMM pg.188-189 29