Math 152 Section 10.6 Functions as Power Series
J. Lewis

We have already seen that the geometric series
n
x 
n0
1
on (  1,1). From this series we
1 x
can represent other functions as power series on this same interval.
Example1:
Example 2:


x
n
 x  x   x n 1 on (  1,1).
1 x
n0
n0
1
1 x
2

 x

n0

2 n


x
2n
on (  1,1).
n0


1
n
  ( 2 x )   ( 1) n 2 n x n
Example 3:
1  2 x n0
n0
1 1
the interval of convergence is ( , ) .
2 2
This series converges only for | 2 x | 1 so
The following theorem says we can differentiate and integrate term by term.
Theorem: If
f ( x) 

 cn x
n
has radius of convergence R then
n0
i)
f ' ( x) 

 cn nx
n 1
with radius of convergence R and
n 1

ii)
c
n 1
n
 f ( x ) dx   n  1x  C
n0
with radius of convergence R.
Note the index in the derivative, i) , starts at n=1 because the derivative of the constant term is 0.
Note the new series i) might not have convergence at either endpoint where the original series does
converge and ii) might have convergence at one or both endpoints where the original does not
converge.
Example4: From the series

1
  (  1) n x n
1  x n0
with radius of convergence 1, we can say
i)
1
(1  x )
2

convergence is

n
n 1
 (  1) nx 
n 1
 (  1)
m 1
( m  1) x m where m  n  1 .
The interval of
m 0
( 1,1).
x n 1
ln(1  x )   (  1)
C
n 1
n0

ii)

n
and this series converges on
( 1,1]. By substituting x=0 into
x n 1
ln(1  x )   (  1)
.
n 1
n0

both sides of the equation, we see the constant of integration is 0 so
We can re-index to get
ln(1  x ) 

 (  1)
m 1
m 1
xm
m
n
.
Note: All can be done for a power series about any a by substituting x-a for x in the above.

Example 6: What is the function represented by
xn

n  0 n!
that the derivative is the same as the original. f(0)=1.)
? (Take the derivative term by term and see
Examples: Find a power series for each function and give the radius and interval of convergence.
1
x2
1.
f ( x) 
2.
f ( x )  arctan x
3.
f ( x )  ln( 4  x )
4.
f ( x )  ln( 4  x 2 )
x2
5.
f ( x) 
6.
f ( x )  ln(1  x ) about a  1
(1  x ) 2