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12p8a9spr2006

# 12p8a9spr2006 - 12.8 Power Series Find the radius of...

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12.8. Power Series Find the radius of convergence and interval of convergence of the series. 10. X n =1 x n n 3 n solution: If a n = x n n 3 n , then lim n →∞ | a n +1 a n | = lim n →∞ | x n +1 ( n + 1)3 n +1 n 3 n x n | = lim n →∞ | xn ( n + 1)3 | = | x | 3 lim n →∞ n n + 1 = | x | 3 . By the Ratio test, the series converges when | x | 3 < 1. That is, | x | < 3. So, R = 3. When x = - 3, the series is the alternating harmonic series, which converges by the Alter- nating Series test. When x = 3, it is the harmonic series, which diverges. Thus, the interval of convergence is [ - 3 , 3). 14. X n =0 ( - 1) n x 2 n (2 n )! . solution: a n = ( - 1) n x 2 n (2 n )! , so lim n →∞ | a n +1 a n | = lim n →∞ | x | 2 n +2 (2 n + 2)! · (2 n )! | x | 2 n = lim n →∞ | x | 2 (2 n + 1)(2 n + 2) = 0 Thus, by the Ratio test, the series converges for all real x and we have R = and the interval of convergence ( -∞ , ). 12.9. Representation of Functions as Power Series 8. Find a power series representation for the function f ( x ) = x 4 x + 1 and determine the interval of convergence. solution: f ( x ) = x 4 x + 1 = x · 1 1 - ( - 4 x ) = x X n =0 ( - 4 x ) n = X n =0 ( - 1) n 4 n x n +1 . 1

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2 The series converges when | - 4 x | < 1 ( that is, | x | < 1 4 ) since n =0 ( - 4 x ) n is a geometric series. It is easy to see that the series diverges at the end points. So, the interval of conver- gence is ( - 1 4 , 1 4 ). 14. (a) Find a power series representation for
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