Your Federal Quarterly Tax Payments are due April 15th

_ _ ... ... f x c cx cx cx = + + + + + _ _ 1 ... ... f x x x x x = + + + + + = ∑ by jizhen1947

VIEWS: 3 PAGES: 5

• pg 1
```									                                                                                            6/21/2011

Math 104 – Rimmer
12.8 Power Series

A power series is a series of the form
∞

∑c x
n =0
n
n
= c0 + c1 x + c2 x 2 + c3 x3 + ...

where:
a ) x is a variable
b) The cn 's are constants called the coefficients of the series.

For each fixed x, the series above is a series of constants
that we can test for convergence or divergence.

A power series may converge for some values of x
and diverge for other values of x.

Math 104 – Rimmer
12.8 Power Series

The sum of the series is a function
f ( x) = c0 + c1 x + c2 x 2 + ... + cn x n + ...
whose domain is the set of all x for which the series converges.
f ( x) is reminiscent of a polynomial but it has infinitely many terms

If all cn 's = 1, we have
∞
f ( x) = 1 + x + x + ... + x + ... = ∑ x n
2                 n

n=0
This is the geometric series with r = x.
The power series will converge for x < 1 and diverge for all other x.
a    1            1                                      ∞
a = 1, r = x ⇒ s =        =                  = 1 + x + x 2 + ... + x n + ... = ∑ x n
1− r 1− x         1− x                                   n=0

1
6/21/2011

Math 104 – Rimmer
12.8 Power Series

In general, a series of the form
∞

∑ c ( x − a)
n =0
n
n
= c0 + c1 ( x − a ) + c2 ( x − a ) 2 + ...
is called a power series centered at a or a power series about a

We use the Ratio Test (or the Root Test) to find for what values of x the series converges.
an +1
lim             = L < 1 for convergence
n →∞       an
solve for x − a to get x − a < R                                                       of convergence ( R.O.C.) .

⇒ −R < x − a < R
⇒a−R< x<a+R
use square brackets [ or ]

This is called the interval                                                  Plug in the endpoints to check for convergence
of convergence ( I.O.C.) .                                                   or divergence at the endpoints.
use parentheses ( or )

Math 104 – Rimmer
Find the radius of convergence and the interval of convergence.                                                                                   12.8 Power Series

n
∞
( −1)                    n2 xn
∑n =1                     2n
1
n +1       2                        n                  2

lim
an +1
= lim
( −1) ⋅ ( n + 1) ⋅ 2n ⋅ x n+1 = lim ( −1) ⋅ ( −1) ⋅ ( n + 1) ⋅ 2n ⋅ x n ⋅ x = − x
n                                     n
n →∞ a
n
n →∞
( −1)      n2    2n +1 x n     n →∞
( −1)            n2    2n ⋅ 2 x n       2
For convergence, this limit
needs to be less than 1
−x     1
<1 ⇒ x <1 ⇒ x < 2                                                  so, − 2 < x < 2
2      2
Now we need to solve                                   This is the radius                             Plug in x = 2 and x = −2 to see if there
this inequality for x .                                of convergence.                                is conv. or div. at the endpoints.

x=2                                                       x = −2                                        2n
n
∞
( −1)
n    2
n 2   n         ∞                    ∞
( −1)
n    2
n ( −2 )
n    ∞
=∑
(( −1) ⋅ ( −2 ) )       n2     ∞
= ∑ n2
∑n =1       2n
= ∑ ( −1) n 2
n
∑
n =1               2n              n =1              2   n
n =1
n =1

Diverges by the Test for Divergence                                Diverges by the Test for Divergence
n
since lim ( −1) n 2 does not exist.                                since lim n 2 = ∞.
n →∞                                                                   n →∞
Radius of convergence: R = 2
Interval of convergence:        ( −2, 2 )

2
6/21/2011

Math 104 – Rimmer
Find the radius of convergence and the interval of convergence.                                                                                            12.8 Power Series

n
∞
3n ( x + 4 )
∑n =1                            n
1
n +1                         n
an+1        3n+1    n ( x + 4)              n
3 ⋅3   n ( x + 4) ( x + 4)
lim       = lim       ⋅      ⋅         n  = lim n ⋅       ⋅            n   = 3( x + 4)
n →∞ a
n
n →∞
3n     n + 1 ( x + 4)      n →∞ 3   n +1     ( x + 4)
For convergence, this limit
needs to be less than 1
1                             1         1
so, − < x + 4 <
3( x + 4) < 1 ⇒ 3 x + 4 < 1 ⇒ x + 4 <                                                                                        3         3
3
Now we need to solve                                                                                                        1         1
− −4< x < −4
this inequality for x + 4 .                                                          This is the radius                     3         3               −13           −11
of convergence.                        −13       −11 Plug in x = 3 and x = 3
<x<
3        3 to see if there is conv. or div.
−13                                                                              x=      −11                                                  at the endpoints.
x=       3                                                                                        3
n                           n                  n                                      n
n
(   −13
+ 4)                   n
( )
−1
( −1)                       n                               n        1 n
∞
3                             ∞
3                ∞                          ∞
3    ( −311 + 4 )         ∞
3   ( )
∑                3

n
=∑              3

n
=∑
n                ∑                              =∑                3

n =1                                 n =1                    n =1                       n =1             n              n =1         n                       1
Converges by the Alt. Series Test                                                                                         ∞
R.O.C.: R =
1                        3
1                          1                                                                                       =∑
bn =    is decreasing and lim      = 0.                                                                                   n =1        n                    −13 −11 
n                    n →∞  n                                                                                                               I.O.C. :      ,
1                                            
Divergent p − series with p = .
2                                   3     3 

Math 104 – Rimmer
Find the radius of convergence and the interval of convergence.                                                                                            12.8 Power Series
n
∞
( 4 x + 1)
∑n =1                    n2
n +1                             1                             n
a
lim n+1 = lim
n2
⋅
( 4 x + 1)                                                = lim
( 4 x + 1) ⋅ ( 4 x + 1)
n2
⋅
2              n                                                                       2         n                          = 4x + 1
n →∞ a
n
n →∞
( n + 1) ( 4 x + 1)                                                       n →∞
( n + 1)       ( 4 x + 1)                                For convergence, this limit
needs to be less than 1

4x + 1 < 1                 For this one, the value a isn't very obvious, so we will proceed as follows:
−1        +1                                         −1
4        4                       it turns out a =
4( x + 1 )
4                <1                                                                            4
4 x + 1 <1                                                                                     solve 4x + 1 = 0
4
−12
−14      0
x+ 1 <
4
1
4
−1                               −1
so in this case with a =    and the interval going from      to 0,
4                                2
1
the radius of convergence is R =
4
Check endpoints:
−1                                                     x=0                                                                                    1
x=         2                                                                                                                               R.O.C.: R =
n                                                      n                                                                    4
∞
( 4 ( ) + 1)
−1                       ∞
( −1)
n            ∞
(1)          =∑
∞
1
∑
2

n2
=∑
n   2               ∑
n =1       n   2
n =1 n2
 −1 
I.O.C. :  , 0 
n =1                                  n =1

convergent Alt. series                                               convergent p − series
2 

3
6/21/2011

Math 104 – Rimmer
12.8 Power Series

Sometimes the Root Test can be used just as the Ratio Test.
n
When an can be written as ( bn ) , then the Root Test should be used.
n                                            n
∞
3n ( x − 5 )               ∞
 3 ( x − 5) 
∑             n   n          = ∑
n

No value of x will
make this limit > 1
n =1                           n =1                                                                   to give divergence

n                    3 ( x − 5)                         We get convergence
lim n an = lim n
n →∞                    n →∞
(   3( x −5)
n      )         = lim
n →∞            n
= 0 <1        no matter what x is

R.O.C. = ∞
I.O.C. = ( −∞, ∞ )

an +1
lim            = 0 ⇒ R.O.C. = ∞ ⇒ I .O.C. ( −∞, ∞ )                                                             ( or lim
n →∞
n   an = 0   )
n →∞      an
the power series only converges for all x

Math 104 – Rimmer
n
n !( x − 7 )
12.8 Power Series
∞

∑n =1         2n
n +1                                            n

lim
an+1
= lim
( n + 1)! ⋅ 2n ⋅ ( x − 7 )                = lim
( n + 1) n ! ⋅    2n ( x − 7 ) ⋅ ( x − 7 )
⋅
n →∞    an    n →∞     n ! 2n+1 ( x − 7 )n                     n →∞          n!         2n ⋅ 2   ( x − 7)
n

= lim
n →∞
1
2   ( n + 1)( x − 7 ) = ∞     >1
R.O.C. = 0                                         No value of x will               We get divergence
I.O.C. = {7}                                       make this limit < 1              for all values of x
to give convergence              except at x = a
at x = a, each term of the series is 0

an +1
lim            = ∞ ⇒ R.O.C. = 0 ⇒ I .O.C. {a}                                                       ( or lim
n →∞
n   an = ∞       )
n →∞      an
the power series only converges at the point x = a

4
6/21/2011

Find the radius of convergence.                                                                      Math 104 – Rimmer
12.8 Power Series
n            2       2n
∞
( −1) ( n !)                     x
∑         ( 2n ) !                                         2
n =1                                  ( n + 1) n !
             
n +1                2

lim
an+1
= lim
( −1) ⋅ ( n + 1)! ⋅ ( 2n )! ⋅ x 2( n+1)
           
n →∞      an          n →∞
( −1)
n
( n !)
2
 2 ( n + 1)  ! x 2 n
                                                           1
n                     2     2
4
= lim
( −1) ( −1) ⋅ ( n + 1) ( n !) ⋅           ( 2n ) !         x2n x2
⋅ 2n                             2  n 2 + 2n + 1
= lim ( −1) x ⋅ 2
n                2
n →∞
( −1)           ( n !)       ( 2n + 2 )( 2n + 1)( 2n )! x                    n →∞         4n + 6n + 2

=
( −1) x 2           For convergence, this limit       ( −1) x 2   <1 ⇒
1 2
x <1 ⇒ x < 4 ⇒ x < 2
2

4            needs to be less than 1              4               4