Chapter 5

Chapter 5 - 5 Series Solutions of Linear Equations Exercises 5.1 1 lim n an 1 2n 1 xn 1(n 1 2n = lim = lim |x| = 2|x| n n n 1 an 2n xn/n-1)n

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5 Series Solutions of Linear Equations Exercises 5.1 1. lim n →∞ ¯ ¯ ¯ ¯ a n +1 a n ¯ ¯ ¯ ¯ = lim n →∞ ¯ ¯ ¯ ¯ 2 n +1 x n +1 / ( n +1) 2 n x n /n ¯ ¯ ¯ ¯ = lim n →∞ 2 n n +1 | x | =2 | x | The series is absolutely convergent for 2 | x | < 1or | x | < 1 / 2. At x = 1 / 2, the series X n =1 ( 1) n n converges by the alternating series test. At x =1 / 2, the series X n =1 1 n is the harmonic series which diverges. Thus, the given series converges on [ 1 / 2 , 1 / 2). 2. lim n →∞ ¯ ¯ ¯ ¯ a n +1 a n ¯ ¯ ¯ ¯ = lim n →∞ ¯ ¯ ¯ ¯ 100 n +1 ( x +7) n +1 / ( n + 1)! 100 n ( x n /n ! ¯ ¯ ¯ ¯ = lim n →∞ 100 n | x +7 | =0 The series is absolutely convergent on ( −∞ , ). 3. lim k →∞ ¯ ¯ ¯ ¯ a k +1 a k ¯ ¯ ¯ ¯ = lim k →∞ ¯ ¯ ¯ ¯ ( x 5) k +1 / 10 k +1 ( x 5) k / 10 k ¯ ¯ ¯ ¯ = lim k →∞ 1 10 | x 5 | = 1 10 | x 5 | The series is absolutely convergent for 1 10 | x 5 | < 1, | x 5 | < 10, or on ( 5 , 15). At x = 5, the series X k =1 ( 1) k ( 10) k 10 k = X k =1 1 diverges by the k -th term test. At x = 15, the series X k =1 ( 1) k 10 k 10 k = X k =1 ( 1) k diverges by the k -th term test. Thus, the series converges on ( 5 , 15). 4. lim k →∞ ¯ ¯ ¯ ¯ a k +1 a k ¯ ¯ ¯ ¯ = lim k →∞ ¯ ¯ ¯ ¯ ( k + 1)!( x 1) k +1 k !( x 1) k ¯ ¯ ¯ ¯ = lim k →∞ ( k | x 1 | = ,x 6 The radius of convergence is 0 and the series converges only for x =1. 5. sin x cos x = µ x x 3 6 + x 5 120 x 7 5040 + ··· ¶µ 1 x 2 2 + x 4 24 x 6 720 + = x 2 x 3 3 + 2 x 5 15 4 x 7 315 + 6. e x cos x = µ 1 x + x 2 2 x 3 6 + x 4 24 −··· 1 x 2 2 + x 4 24 x + x 3 3 x 4 6 + 7. 1 cos x = 1 1 x 2 2 + x 4 4! x 6 6! + =1+ x 2 2 + 5 x 4 4! + 61 x 6 6! + Since cos( π/ 2) = cos( 2) = 0, the series converges on ( 2 ,π/ 2). 8. 1 x 2+ x = 1 2 3 4 x + 3 8 x 2 3 16 x 3 + Since the function is undeFned at x = 2, the series converges on ( 2 , 2). 212
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Exercises 5.1 9. X n =1 2 nc n x n 1 + X n =0 6 c n x n +1 =2 · 1 · c 1 x 0 + X n =2 2 nc n x n 1 | {z } k = n 1 + X n =0 6 c n x n +1 | {z } k = n +1 c 1 + X k =1 2( k +1) c k +1 x k + X k =1 6 c k 1 x k c 1 + X k =1 [2( k c k +1 +6 c k 1 ] x k 10. X n =2 n ( n 1) c n x n +2 X n =2 n ( n 1) c n x n 2 +3 X n =1 nc n x n · 2 · 1 c 2 x 0 · 3 · 2 c 3 x 0 · 1 · c 1 x 0 + X n =2 n ( n 1) c n x n | {z } k = n +2 X n =4 n ( n 1) c n x n 2 | {z } k = n 2 +3 X n =2 nc n x n | {z } k = n =4 c 2 + (12 c 3 + (12 c 3 c 1 ) x + X n =2 k ( k 1) c k x k X n =2 ( k + 2)( k c k +2 x k X n =2 kc k x k c 2 +(3 c 1 +12 c 3 ) x + X n =2 ( [ k ( k 1) + 3 k ] c k +2( k + 2)( k c k +2 ) x k c 2 c 1 c 3 ) x + X n =2 £ k ( k +2) c k k + 1)( k c k +2 ± x k 11. y 0 = X n =1 ( 1) n +1 x n 1 ,y 0 = X n =2 ( 1) n +1 ( n 1) x n 2 ( x y 0 + y 0 =( x X n =2 ( 1) n +1 ( n 1) x n 2 + X n =1 ( 1) n +1 x n 1 = X n =2 ( 1) n +1 ( n 1) x n 1 + X n =2 ( 1) n +1 ( n 1) x n 2 + X n =1 ( 1) n +1 x n 1 = x 0 + x 0 + X n =2 ( 1) n +1 ( n 1) x n 1 | {z } k = n 1 + X n =3 ( 1) n +1 ( n 1) x n 2 | {z } k = n 2 + X n =2 ( 1) n +1 x n 1 | {z } k = n 1 = X k =1 ( 1) k +2 kx k + X k =1 ( 1) k +3 ( k x k + X k =1 ( 1) k +2 x k = X k =1 £ ( 1) k +2 k (
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This note was uploaded on 02/13/2008 for the course ENG 342 taught by Professor Delahanty during the Fall '07 term at TCNJ.

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Chapter 5 - 5 Series Solutions of Linear Equations Exercises 5.1 1 lim n an 1 2n 1 xn 1(n 1 2n = lim = lim |x| = 2|x| n n n 1 an 2n xn/n-1)n

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