Assignment 11 Solutions

Assignment 11 Solutions - MATH 138 Calculus 2, Solutions to...

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1: (a) Find the radius and interval of convergence of the power series X n =1 (2 - 3 x ) n n . Solution: Let a n = (2 - 3 x ) n n . Then ± ± ± ± a n +1 a n ± ± ± ± = | 2 - 3 x | n +1 n + 1 n | 2 - 3 x | n = n + 1 n | 2 - 3 x | -→ | 2 - 3 x | = 3 ± ± x - 2 3 ± ± . By the R.T, the power series converges when 3 ± ± x - 2 3 ± ± < 1, that is when ± ± x - 2 3 ± ± < 1 3 , and diverges when ± ± x - 2 3 ± ± > 1 3 , so the radius of convergence is R = 1 3 . The endpoints of the interval of convergence are the points where ± ± x - 2 3 ± ± = 1 3 , that is x = 1 3 and x = 1. When x = 1 3 , we have a n = 1 n and so X a n diverges. When x = 1 we have a n = ( - 1) n n and so X a n converges by the A.S.T. Thus the interval of convergence is I = ( 1 3 , 1 ² . (b) Find the set of all values of x such that the series X n =1 ( x 2 + x - 1) n n converges. Solution: Let a n = ( x 2 + x - 1) n n . Then ± ± ± ± a n +1 a n ± ± ± ± = | x 2 + x - 1 | n +1 n + 1 · n 5 | x 2 + x - 1 | = n + 1 n | x 2 + x - 1 | and so lim n →∞ ± ± ± ± a n +1 a n ± ± ± ± = | x 2 + x - 1 | . By the R.T, the series converges (absolutely) when | x 2 + x - 1 | < 1 and diverges when | x 2 + x - 1 | > 1. From the graph of y = x 2 + x - 1 we see that | x 2 + x - 1 | < 1 when x ( - 2 , - 1) (0 , 1) and | x 2 + x - 1 | > 1 when x ( -∞ , - 2) ( - 1 , 0) (1 , ). We check the endpoints of these intervals. When x = - 2 , 1 we have a n = 1 n so a n converges, and when x = - 1 , 0 a n = ( - 1) n n so a n converges. Thus the series converges when
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

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Assignment 11 Solutions - MATH 138 Calculus 2, Solutions to...

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