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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 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 ) 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 This is the radius of convergence. Radius of convergence: R = 2 5

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