Finding and Using Taylor
Series
Finding Taylor Series
Taylor Series and Limits
Approximate Integration
Comparison of Functions
Elusive Limit
Other Types of Limits
Index
FAQ
Summary Formulae
Taylor
Polynomials at
x=a
f´(a)
f´´(a)
P( x )  f(a) 
( x  a) 
( x  a)2 
1!
2!
f ( n ) (a )

( x  a)n
n!
Maclaurin series = Taylor series at x = 0. Basic Maclaurin series:
1
1
1
cos( x )  1  x 2  x 4 
2!
4!
1 3 1 5
x  x 
3!
5!
2
sin( x )  x 
3
1
1
e  1 x 2  x 3 
2!
3!
x
Index
( 1)k 2k

x
k 0 (2k )!



2 k 1
( 1) x
.
k 0 (2k  1)!

k
Formulae 1 – 3
can be used
for all x.
1 k
x
k 0 k !

Mika Seppälä: Taylor Series
FAQ
Finding Taylor Series (1)
Find the Maclaurin series of the function sin  x2 .
Problem

Solution
Start with the Taylor series sin  z   
k 0
 1
k
 2k  1!
z 2k 1.
Substitute z  x 2 to get
 

sin x 2  
Index
k 0
 1
k
x 

 2k  1!
2
2 k 1


k 0
Mika Seppälä: Taylor Series
 1
k
 2k  1!
x 4k 2 .
FAQ
Finding Taylor Series (2)
Problem
Find the Maclaurin series of the function
sin  x 
x

Solution
Start with the Maclaurin series sin  x   
k 0
.
 1
k
 2k  1!
x 2k 1.
Divide all terms by x to get
sin  x 
x
Index


 1
k

 1
k
1
x 2k 1  
x 2k .

x k 0  2k  1!
k 0  2k  1 !
Mika Seppälä: Taylor Series
FAQ
Finding Taylor Series (3)
Find the first three non-zero terms for
Problem
Maclaurin series of the function f  x   cos  sin  x   .
We have sin  x   x 
Solution
1 3 1 5
x  x 
3!
5!
and cos(u )  1 
1 2 1 4
u  u 
2!
4!
Use the shown beginnings of the Taylor series and substitute
ux
1 3 1 5
x  x in the Taylor polynomial for cos u  .
3!
5!
One gets
2
4
1
1
1 
1
1
1 
cos  sin  x    1   x  x 3  x 5    x  x 3  x 5   higher order terms
2! 
3!
5!  4! 
3!
5! 
1
1 2 1 4
 1 x 2  
  x  higher order terms
2!
2!
3!
4! 

1
5
 1  x 2  x 4  higher order terms
2!
4!
Index
Mika Seppälä: Taylor Series
FAQ
.
Error Estimates
For alternating Taylor or Maclaurin series, use the error estimates for
alternating series.
Assume that there is a constant L such that for all positive integers k
and for all t between 0 and x :
f ( k ) (t )  L.
This number L usually
depends on x.
Error when Approximating the Function f with its Taylor polynomial of degree m
f´(0)
f´´(0) 2
Em ( x )  f(x)  f(0) 
x
x 
1!
2!
Error Estimate
Index
Em  x   L
x
m
f ( m ) (0) m
f ( n ) (0) n

x  f( x )  
x
m!
n
!
n 0
m 1
(m  1)!
Mika Seppälä: Taylor Series
FAQ
Taylor Series and Limits (1)
Problem
The functions f, g and h satisfy the following:
f(3)  g(3)  h(3)  0, f´(3)  h´(3)  0, g´(3)  10 and f´´(3)  5,
f( x )
f( x )
g´´(3)  7, h´´(3)  10. Determine the limits lim
and lim
.
x 3 g( x )
x 3 h( x )
Solution
Hence
The properties of the functions f, g and h imply that their Taylor
polynomials of degree 2 at x  3 are
Pf ( x ) 
5
7
10
( x  3)2 , Pg ( x )  10( x  3)  ( x  3)2, and Ph ( x ) 
( x  3)2.
2!
2!
2!


5
( x  3)2  terms divisible by (x  3)3
f( x )
2!

g( x ) 10( x  3)  7 ( x  3)2  terms divisible by (x  3)3
2!
5
( x  3)   terms containing powers of (x  3) 
 2!

 0.
x 3
7
10  ( x  3)   terms containing powers of (x  3) 
2!

Index
Mika Seppälä: Taylor Series

FAQ
Taylor Series and Limits (2)
Problem
The functions f, g and h satisfy the following:
f(3)  g(3)  h(3)  0, f´(3)  h´(3)  0, g´(3)  10 and f´´(3)  5,
f( x )
f( x )
g´´(3)  7, h´´(3)  10. Determine the limits lim
and lim
.
x 3 g( x )
x 3 h( x )
Solution (part b)
Using the Taylor series expansions at x  3 one gets
5
( x  3)2   terms divisible by (x  3)3 
f( x ) 2!

h( x ) 10 ( x  3)2  terms divisible by (x  3)3


2!
5
  terms containing powers of (x  3)
1
2!



.
x 3
10
2
  terms containing powers of (x  3)
2!
Index
Mika Seppälä: Taylor Series
FAQ
Comparison of Functions (1)
Problem
Let f( x )  sin( x ), g( x )  e x  1, and h( x ) 
1
 1.
1 x
Decide which of the above functions takes the smallest values and which
the largest values for small positive values of x.
2
Solution
We solve the problem by comparing the Taylor series at x = 0 of
the above functions. The smallest power terms of the series
determine the behavior of the function near the origin.
1
The Taylor expansion for the function
starts
1 z
 1  3 
  2   2 
1
1
1


 z2  .
 1  z  2  1  z  
2
2!
1 z
Substituting z   x 2 one gets
1
1
3
 1 x 2  x 4  .
Solution continues
2
8
1 x 2
Index
Mika Seppälä: Taylor Series
FAQ
Comparison of Functions (2)
Solution continues
1
We have
1- x
2
1
1 2 3 4
x  x 
2
8
.
Taylor series for the other functions are basic Taylor series:
1 3
x 
and
3!
1
1
1
1


ex  1  1  x  x 2  x 3    1  x  x 2  x 3  .
2!
3!
2!
3!


For values of x near 0 it suffices to look at the Taylor polynomials of degree 2.
sin( x )  x 
We have sin( x )  x, e x  1  x 
Since for small positive values of x,
that for small positive values of x,
Index
1 2
x and
2!
1
1-x 2
1
1 2
x
2
1 2
1
x  x  x  x 2 , we deduce
2
2!
1
 1  sin( x )  e x  1
1 x 2
Mika Seppälä: Taylor Series
FAQ
Index
Mika Seppälä: Taylor Series
FAQ