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Unformatted text preview: MA 2930, March 16, 2011 Worksheet 8 Solutions 1. Find the interval of convergence of the following power series: ( i ) ∞ X n =1 (2 x + 1) n ( n + 1) 2 , ( ii ) ∞ X n =0 ( x 1) n 3 n , ( iii ) ∞ X n =2 n ! x n n n In each case we just need to apply the ratio test. (i) R = lim n →∞ a n +1 a n = lim n →∞ (2 x +1) n +1 ( n +2) 2 (2 x +1) n ( n +1) 2 = (2 x + 1) lim n →∞ ( n + 1) 2 ( n + 2) 2 = (2 x + 1) The series converges whenever  R  =  2 x +1  < 1, i.e., when 1 < 2 x +1 < 1, i.e., when 1 < x < 0. Note that the radius of convergence is 1/2. (ii) R = lim n →∞ a n +1 a n = lim n →∞ ( x 1) n +1 3 n +1 ( x 1) n 3 n = lim n →∞ x 1 3 = x 1 3 The series converges whenever  R  =  x 1  / 3 < 1, i.e., when 3 < x 1 < 3, i.e., when 2 < x < 4. The radius of convergence is 3. (iii) R = lim n →∞ a n +1 a n = lim n →∞ ( n +1)! x n +1 ( n +1) n +1 n ! x n n n = x lim n →∞ ( n + 1) n n ( n + 1) n +1 = x lim n →∞ 1 (1 + 1 n ) n = x/e The series converges whenever  R  =  x/e  < 1, i.e., when e < x < e . The radius of convergence is e. 2. In the following series find the coefficient of x n by rewriting the series with x n as the generic term: ( i ) ∞ X n =1 a n x n +2 + ∞ X n =2 a n +1 nx n 1 , ( ii ) x ∞ X n =1 na n x n 1 +(1 x 2 ) ∞ X n =2 n ( n 1) a n x n 2 The yoga of index shifting is simply this: if you raise n by a certain amount inside the sum, you should lower the starting value of n by the same amount. (i) ∞ X n =1 a n x n +2 + ∞ X n =2 a n +1 nx n 1 = ∞ X n =3 a n 2 x n + ∞ X n =1 a n +2 ( n + 1) x n = ∞ X n =3 a n 2 x n + 2 a 3 x + 3 a 4 x 2 + ∞ X n =3...
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This note was uploaded on 06/10/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 TERRELL,R
 Differential Equations, Equations, Power Series

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