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

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_ _ ... ... f x c cx cx cx = + + + + + _ _ 1 ... ... f x x x x x = + + + + + = ∑ Powered By Docstoc
					                                                                                            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




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                                                                                                                                                  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
                                                                                           R is called the radius
    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 )




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                                                                                                                                                                                        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 




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                                                                                                                                               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
                                                                                                          This is the radius
                                                                                                          of convergence.

                                                                                   Radius of convergence: R = 2




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