Chapter 10
Power Series
10.1
Approximating Functions with
Polynomials
Linear approximation near x = a
(1st order)
2nd order approximation near x = a
Slide 10 - 3
Example:
Approximation of sin(x) near x = a
(3rd order)
(1st order)
(5th order)
Slide 10 - 4
Slide 10 - 5
Brook Taylor was an
accomplished musician
and painter. He did
research in a variety of
areas, but is most
famous for his
development of ideas
regarding infinite series.
Brook Taylor
1685 - 1731
Slide 10 - 6
Greg Kelly, Hanford High School, Richland, Washington
Practice example:
Suppose we wanted to find a fourth degree polynomial of the
form:
P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
that approximates the behavior of
f  x   ln  x  1 at x  0
If we make P  0  f  0, and the first, second, third and fourth
derivatives the same, then we would have a pretty good
approximation.
Slide 10 - 7
P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
f  x   ln  x  1
f  x   ln  x  1
P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
f  0  ln 1  0
P  0   a0
P  x   a1  2a2 x  3a3 x2  4a4 x3
1

f  x 
1 x
1
f  0   1
1
f   x   
a0  0
P  0  a1
1
1  x 
1

f  0     1
1
2
a1  1
P  x   2a2  6a3 x  12a4 x2
P  0  2a2
1
a2  
2
Slide 10 - 8
P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
1  x 
1
f   0     1
1
f   x   2 
P  x   2a2  6a3 x  12a4 x2
1
f   x   
2
P  0  2a2
3
P  0  6a3
f   0  2
f
 4
f
 x   6
4
1
1  x 
 0  6
1
a2  
2
P  x   6a3  24a4 x
1
1  x 
f  x   ln  x  1
4
P
P
4
 4
2
a3 
6
 x   24a4
 0  24a4
6
a4  
24
Slide 10 - 9
P  x   a0  a1 x  a2 x2  a3 x3  a4 x4
f  x   ln  x  1
1 2 2 3 6 4
P  x   0  1x  x  x 
x
2
6
24
x 2 x3 x 4
P  x  0  x   
2 3 4
If we plot both functions, we see that
near zero the functions match very
well!
5
4
3
f  x
1
1
2
3
4
5
-2
-5
-1
-0.5
0
0.5
1
-0.5
-3
-4
1
0.5
2
-5 -4 -3 -2 -1 0
-1
f  x   ln  x  1
P  x
-1

Slide 10 - 10
Our polynomial:
0  1x 
1 2 2 3 6 4
x  x 
x
2
6
24
has the form:
f   0  2 f   0  3 f  4  0  4
f  0  f   0 x 
x 
x 
x
2
6
24
or:
f  0 f   0
f   0  2 f   0  3 f  4  0  4

x
x 
x 
x
0!
1!
2!
3!
4!
This pattern occurs no matter what the original function was!
Slide 10 - 11
Definition:
Taylor Series:
(generated by f at x  a )
P  x   f  a   f   a  x  a  
f   a 
2!
 x  a
2

f   a 
3!
 x  a
3
 
If we want to center the series (and it’s graph) at zero, we get
the Maclaurin Series:
Maclaurin Series:
(generated by f at x  0 )
P  x   f  0  f   0 x 
f   0 
2!
x 
2
f   0 
3!
x3  
Slide 10 - 12
Exercise 1: find the Taylor polynomial approximation at 0 (Maclaurin series)
for:
y  cos x
f  x   cos x
f   x    sin x
f  0  1
f   0  0
f   x   sin x f   0  0
f
 4
 x   cos x
f
 4
 0  1
f   x    cos x f   0   1
1x 2 0 x3 1x 4 0 x5 1x 6
P  x   1 0x 




 
2!
3!
4!
5!
6!
x 2 x 4 x 6 x8 x10
P  x  1    

2! 4! 6! 8! 10!

Slide 10 - 13
x 2 x 4 x 6 x8 x10
P  x  1    

2! 4! 6! 8! 10!
y  cos x
1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
The more terms we add, the better our approximation.
Slide 10 - 14
To find Factorial using the TI-83:
Slide 10 - 15
Exercise 2: find the Taylor polynomial approximation at 0 (Maclaurin
series) for:
y  cos  2 x 
Rather than start from scratch, we can use the function that we
already know:
2x   2x   2x   2x   2x 

P  x  1




2!
4!
6!
8!
10!
2
4
6
8
10


Slide 10 - 16
y  cos  x  at x 
Exercise 3: find the Taylor series for:
f  x   cos x
 
f  0
2

2
 


f  x   sin x f    1
2


f   x    sin x f     1
2
f

 4
f   x    cos x f     0
2
 x   cos x f
 4  

 0
2
 0  1 

P  x   0  1 x     x     x    
2  2! 
2  3! 
2

2

3
3

5


x

x 




2 
2

P  x    x    

 
2
3!
5!

Slide 10 - 17
When referring to Taylor polynomials, we can talk
about number of terms, order or degree.
x2 x4
cos x  1  
2! 4!
This is a polynomial in 3 terms.
It is a 4th order Taylor polynomial, because it was found
using the 4th derivative.
It is also a 4th degree polynomial, because x is raised to the
4th power.
The 3rd order polynomial for cos x
degree 2.
x2
is 1  , but it is
2!
The x3 term drops out when using the third derivative.
This is also the 2nd order polynomial.

Slide 10 - 18
.
Practice
1) Show that the Taylor series expansion of ex is:
2) Use the previous result to find the exact value of:
3) Use the fourth degree Taylor polynomial of
cos(2x) to find the exact value of
Slide 10 - 19
10.2 and 10.3
Properties of Power Series:
Convergence
Slide 10 - 21
Slide 10 - 22
Convergence of Power Series:
is

n
c
(
x

a
)
The Radius of Convergence for a power series  n
n0
cn
R  lim
n  c
n 1
is:
The center of the series is x = a. The series converges on the open
interval (a  R, a  R) and may converge at the endpoints.
You must test each series that results at the endpoints of the interval
separately for convergence.
( x  2) n
Examples: The series
2 is convergent on [-3,-1]
n 0 ( n  1)


but the series

n 0
(1) n ( x  3) n
5
n
n 1
is convergent on (-2,8].
Slide 10 - 23
Slide 10 - 24
Convergence of Taylor Series:
is
If f has a power series expansion centered at x = a, then the

power series is given by f ( x)  
n 0
f ( n ) (a)
( x  a) n
n!
And the series converges if and only if the Remainder satisfies:
lim Rn ( x)  0
n 
Where: Rn ( x) 
f ( n1) ( c )
( n 1)!
( x  a ) n 1
is the remainder at
x, (with c between x and a).
Slide 10 - 25
Common Taylor Series:
Slide 10 - 26