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9.8
Maclaurin and Taylor Series;
Power Series
Liberty Bell, Philadelphia, PA
Photo by Vickie Kelly, 2003
There are some Maclaurin series that occur often
enough that they should be memorized. They are
on your formula sheet, but today we are going to
look at where they come from.
Maclaurin Series:
(generated by f at x  0 )
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
1
1
 1  x 
1 x
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
f  n  x 
1  x 
1
1  x 
2
2 1  x 
3
6 1  x 
4
24 1  x 
5
List the function and its
derivatives.
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
1
1
 1  x 
1 x
f  n  x 
1  x 
1
1  x 
2
2 1  x 
3
6 1  x 
4
24 1  x 
5
f  n  0
List the Evaluate
function and
column
its one
1 derivatives.
for x = 0.
1
2 2 3! 3 4! 4
 1  1x  x  x  x  
1 x
2!
3!
4!
1
2
6  3!
24  4!
1
 1  x  x 2  x3  x 4  
1 x
This is a geometric series
with a = 1 and r = x.
1
We could generate this same series for
1 x
polynomial long division:
1  x  x  x  
2
1  x 
3
1
1 x
x 2
xx
2
x
x 2  x3
x
3
with
We wouldn’t expect to use the previous two series to
evaluate the functions, since we can evaluate the
functions directly.
They do help to explain where the formula for the
sum of an infinite geometric comes from.
We will find other uses for these series, as well.
A more impressive use of Taylor series is to
evaluate transcendental functions.
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
cos  x 
f  n  x 
f  n  0
cos  x 
1
 sin  x 
0
 cos  x 
1
sin  x 
0
cos  x 
1
1 2 0 3 1 4
cos  x   1  0 x  x  x  x  
2!
3!
4!
x2
x4 x6
cos  x   1  
 
2!
4! 6!
Both sides are even functions.
cos (0) = 1 for both sides.
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
sin  x 
f  n  x 
f  n  0
sin  x 
0
cos  x 
1
 sin  x 
0
 cos  x 
1
sin  x 
0
0 2 1 3 0 4
sin  x   0  1x  x  x  x  
2!
3!
4!
x3
x5 x 7
sin  x   x  
 
3!
5! 7!
Both sides are odd functions.
sin (0) = 0 for both sides.
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
ln 1  x 
f  n  x 
f  n  0
ln 1  x 
0
1  x 
1
1
 1  x 
2 1  x 
2
3
6 1  x 
4
1
2
6  3!
1 2 2 3 3! 4
ln 1  x   0  1x  x  x 
x  
2!
3!
4!
x 2 x3 x 4
ln 1  x   x     
2 3 4
We have saved the best for last!
e
f   0  2 f   0  3
P  x   f  0  f   0 x 
x 
x  
2!
3!
x
f  n  x 
f  n  0
ex
1
1 2 1 3 1 4
e  1  1x  x  x  x  
2!
3!
4!
x
ex
1
ex
1
ex
1
ex
1
x 2 x3 x 4
e  1  x     
2! 3! 4!
x
There are three possibilities for power series convergence.
1
The series converges over some finite interval:
(the interval of convergence).
There is a positive number R such that the series
diverges for x  a  R but converges for x  a  R .
The series may or may not converge at the
endpoints of the interval.
2
The series converges for every x. ( R   )
3
The series converges at x  a and diverges
everywhere else. ( R  0 )
(As in the next example.)
The number R is the radius of convergence.
Example:
n
n
!
x
 n 0

If x  1 then
n ! x n grows without bound.
n!
n
If 0  x  1 then lim n ! x  lim n  
n 
n  1
 
 
 x
As n  , eventually n is
larger than
1
x
therefore the
numerator grows faster than
the denominator.
The series diverges.
(except when x=0)
Geometric series have a constant ratio between
terms. Other series have ratios that are not
constant. If the absolute value of the limit of the
ratio between consecutive terms is less than one,
then the series will converge.

For
t
n 1
n
tn 1
, if L  lim
n  t
n
then:
if
L 1
the series converges.
if
L 1
the series diverges.
if
L 1
the series may or may not converge.
Example:
2
3
If we replace x with
x – 1, we get:
4
x
x x
   
2 3 4
1
1
1
2
3
4
ln  x    x  1   x  1   x  1   x  1   
2
3
4
ln 1  x   x 
1  x  1

L  lim
n2
n 
n 1
n 1
x  1  x  1

 lim

n
n 
n 1
x  1 n

 lim
n 
n 1

n
 1  x  1
n 1
n

  1
n 1
n 1
1
n
  x  1
n
an 1
1
 an 1 
an
an
n
 x  1
 x 1
n
If the limit of the ratio
between consecutive terms
is less than one, then the
series will converge.
If the limit of the ratio between consecutive
terms is less than one, then the series will
converge.
x 1  1
1  x 1  1
0 x2
The interval of convergence is (0,2).
The radius of convergence is 1.

n
n
x  5

n 
n 1 3
Example:
1
2
3
2
3
 x  5   x  5   x  5  
3
9
27
n  1 x  5

L  lim
n 
n 1

n 1
3
3n
n  x  5
n
n

1
x

5
x

5

3
    
n
L  lim
n 
3  3  n  x  5
n
n  1 x  5

L  lim
n 
3n
n
n

Example:
n
n
x  5

n 
n 1 3
n  1 x  5

L  lim
n 
3n
n 1
L  x  5 lim
n  3n
1
L  x 5 
3
1
x 5 1
3
x 5  3
3  x  5  3
2 x8
The interval of convergence is (2,8).
82
The radius of convergence is
3.
2

Example:
n!
n
x  3

4 
n 1 n
1
2
3
2
3
4
 x  3   x  3   x  3   x  3  
8
27
32
n  1! x  3

L  lim
4
n 
 n  1
n 1

n4
n ! x  3
n ! n  1 x  3  x  3
n
L  lim
n 
 n  1
4
1
 n 
L  x  3 lim  n  1 

n 
n

1


4

n
n4
n ! x  3
n

Example:
n!
n
x  3

4 
n 1 n
1
 n 
L  x  3 lim  n  1 

n 
 n 1 
L
At
for all x
 3.
4
Radius of convergence = 0.
x  3, the series is 0  0  0   , which converges to zero.
Note: If R is infinite, then the series converges for all values
of x.

x2n
Example: Prove that 
2


n 0 n !
converges for all real x.
There are no negative terms: 


n 0
 x 2 n
n!
x
2n
 n !
2

x 
2 n
n!
larger denominator
is the Taylor series for
e
x2
, which converges.
 The original series converges.
The direct comparison test only works when the
terms are non-negative.

Example:

 sin x 
n
n!
n 0
We test for absolute convergence:
sin x
n!
n
1

n!
2
3
n
x
x
x
Since e x  1  x 
   
2! 3!
n!

1
converges to

n 0 n !


n 0
sin x
e e
1
n
converges by the direct comparison test.
n!

Since
,

n 0
 sin x 
n!
n
converges absolutely, it converges.