Lecture 29 – Power Series
Def: The power series centered at x = a:

 c ( x  a)
n 0
n
n
 c0  c1 ( x  a)  c2 ( x  a)  
2
x is the variable and the c’s are constants (coefficients)

2
3
k
1
x
x
x
x
1
n
(
x

0
)
 1     
cn 
,a  0  
1 2! 3!
k!
n!
n 0 n !
cn
n

 1

,a  2 
2 n 1


n 0
 1n ( x  2) n  1  x  2  ( x  2) 2    (1) k ( x  2) k
2
n 1
2
4
8
2
k 1

1
For any power series, exactly one of the following is true:
1.
converges for all x
2.
converges for only x  a
3.
converges( absolutely ) for all x in ( R, R)
& diverges for all x outside ( R, R)
& at x  R or x   R, may converge or diverge
2
Example 1 – Radius and Interval of Convergence

xn
x x 2 x3
xk
 1     

1 2! 3!
k!
n 0 n !
Ratio Test:
x n 1 n !
lim
 n 
n  ( n  1)! x
Series converges for
3
Example 2 – Radius and Interval of Convergence

2
k


n
!
x

10

(
x

10
)

2
!
(
x

10
)



k
!
(
x

10
)


n
n 1
Ratio Test:
(n  1) !( x  10) n 1
lim

n
n 
n !( x  10)
Series converges for
4
Example 3 – Radius and Interval of Convergence

2 x n
n 1
n

(2 x) 2 (2 x) 3
(2 x) k
 2x 

 

2
3
k
Ratio Test:
(2 x) n 1
n
lim


n
n  n  1
( 2 x)
5
Example 3 – continued (testing endpoints)


2 x n
n 1
x
n
(2 x) 2 (2 x) 3
(2 x) k
 2x 

 

2
3
k
R
Interval :


,
1
:
2
1
x :
2
6
Example 4 – Radius and Interval of Convergence

x  1
n 0
3n

n
x  1 ( x  1) 2
( x  1) k
 x 1
 

 

  1
k
3
9
3
n 0  3 
n

Root Test:
x 1

3
n
lim
n 
n
7
Example 4 – continued (testing endpoints)

x  1
n 0
3n

n
 x 1
 

n 0  3 

n
R
Interval :

,

x  4 :
x  2:
8
Example 5 – Radius and Interval of Convergence


n 0
 1n x  2n
2 n 1
k
1 x  2 ( x  2) 2
(
x

1
)
 

   (1) k

k 1
2
4
8
2
Geometric Series:


n 0
 1n x  2n
2n 1
1  1 x  2
 
2n
n 0 2

n
n
9
Example 5 – continued – what is the converging value?


n 0
 1n x  2n
k
1 x  2 ( x  2) 2
(
x

1
)
 

   (1) k

k 1
2
4
8
2
2 n 1
Geometric Series:
1  x2



2 
n 0 2 

n
10
Lecture 30 – More Power Series
The geometric series:
a
a  ar  ar   
when |r|  1.
1 r
2
As a power series with a = 1, r = x and cn = 1 for all n:
1
1 x  x  
when |x|  1.
1 x
2
In other words, the function f(x) can be written as a power series.

1
f ( x) 
  x n with
1  x n 0
R 1
Interval : 1, 1
11
Create new power series for other functions through:
sum, difference, multiplication, division, composition
and
differentiation and integration

1
  x n with R  1 Interval : 1,1
Example 1 f ( x) 
1  x n 0
g ( x)  f (3x) 
12

1
  x n with R  1 Interval : 1,1
Example 2 f ( x) 
1  x n 0
x
h( x ) 
2
1 x
13
Consider the graphs:
x
3
5
7
h( x ) 

x

x

x

x
  Interval : 1,1
2
1 x
y1  x
1
h(x)
y3  x  x
3
y5  x  x  x
5
3
1
1
y7  x  x 3  x 5  x 7
14
Example 3

1
  xn
1  x n 0
with R  1 Interval :  1,1
f ( x)  tan 1 x
15
tan 1 x
Need to solve for C. Set x = 0 to get:
2 n 1
x
tan 1 x   (1) n
2n  1
n 0

x  1:
Test endpoints???
x  1 :
16
Example 4

1
  xn
1  x n 0
with R  1 Interval :  1,1
f ( x)  ln( 1  x)
17
ln( 1  x)
Need to solve for C. Set x = 0 to get:
n 1
x
ln( 1  x)   (1) n
n 1
n 0

x  1:
Test endpoints???
x  1 :
18