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
6
Example 1 – First-Degree Polynomial Approximation of f(x) = ex
For the function f(x) = ex, find a first-degree polynomial
function
P1(x) = a0 + a1x
whose value and slope agree with the value and slope of f
at x = 0. In other words, find the tangent line approximation
to f(x) = ex at x = 0.
Solution:
Because f(x) = ex and f'(x) = ex, the value and the slope of f,
at x = 0, are given by
f(0) = e0 = 1
and
f'(0) = e0 = 1.
7
Example 1 – Solution
cont’d
Because P1(x) = a0 + a1x, you can use the condition that
P1(0) = f(0) to conclude that a0 = 1.
Moreover, because P1'(x) = a1, you can use the condition
that P1'(0) = f'(0) to conclude that a1 = 1.
Therefore, P1(x) = 1 + x.
Figure 9.11 shows the graphs
of P1(x) = 1 + x and f(x) = ex.
Figure 9.11
8
Example 1 – Solution
cont’d
P1(x) = 1 + x.
Can we do better?
Find P2(x), P3(x), and P4(x).
Approximate e0.1 using P4(x).
e0.1  P4  0.1  1.105170
Based on the observed pattern,
Find Pn(x) for f(x) = ex around x=0.
1 2 1 3
1 n
Pn  x   1  x  x  x  ...  x
2
3!
n!
9
Maclaurin Polynomials
cont’d
What would Pn(x) be for any f(x) around x=0.
f   0  2 f   0  3
f n  0 n
Pn  x   f  0   f   0  x 
x 
x  ... 
x
2!
3!
n!
10
Example 2
Find the Maclaurin polynomial of degree 6 for f(x) = cos x.
Use the 6th degree polynomial to approximate cos (0.1).
1 2 1 4 1 6
P6  x   1  x  x  x
2!
4!
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.
1st derivatives – slopes agree at the point
2nd derivatives – curvature agrees at the point, etc.
The polynomial approximation at x=c:
Tangent line approximation: y  f  c   f   c  x  c 
At x=c, y=f (c) and At x=c, y  f   c 
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), 1st derivatives agree, and 2nd derivatives agree.
y f
0
 c  x  c 
0
f   c 
2
 f   c  x  c  
 x  c
2
1
Ex: Find the quadratic approximation for f  x   sin x at x 
1

P2  x   1   x  
2
2

2
.
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
15
Maclaurin (1698–1746).
Example 3
Find the Taylor polynomial P4(x) for f(x) = ln x
centered at c=1. Use it to estimate ln(1.2).
P4  x    x  1 
1
1
1
2
3
4
 x  1   x  1   x  1
2
3
4
16
Day 2
Remainder of a Taylor Polynomial
18
Taylor Polynomials
Taylor Polynomial approximation at x=c:
f   c 
f   c 
f n c
2
3
n
Pn  x   f  c   f   c  x  c  
x

c

x

c

...

x

c






2!
3!
n!
Kth Taylor Polynomial of f(x) centered at x=c:
k

n 0
f
n
c
n!
 x  c
n
19
Example 4
Find the 3rd Taylor polynomial for f(x) = sin x,
expanded about c   6.
1
3

1 

3 

P3  x   
x


x


x







2 2 
6  2  2! 
6  2  3! 
6
2
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 Pn(x), you can use the concept
of a remainder Rn(x), defined as follows.
21
Remainder of a Taylor Polynomial
So, Rn(x) = f(x) – Pn(x). The absolute value of Rn(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
n 1
xc
Rn  x  
 max f n 1  z 
 n  1!
where f n 1  z  is the maximum value of f n 1  z  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  x   f  c   f   z  x  c 
f  x  f c
or f   z  
xc
24
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 P3(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
cont’d
Therefore,
Because f (4)(z) = sin z, it follows that the error |R3(0.1)| can
be bounded as follows.
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:
12 13 14 15
e  11



2! 3! 4! 5!
1 2 1 3
1 n
Solution: f  x   e , Pn  x   1  x  x  x  ...  x
2!
3!
n!
x
f 6  z   ez  max on 0,1  e1.
e1 6
e
R5  x   1 
 0.00378
6!
720
28
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