T4Solutions - At x = 5 ; 1 X n =1 1 n ¢ 2 n 2 n = 1 X n =1...

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MATH 1005A Test 4 Solutions March 20, 2009 Questions 1-5 are multiple choice. Circle the correct answer. Only the answer will be marked. [Marks] 1. The series 1 X n =1 ( ¡ 1) n ¡ 1 p n +1 [3] (a) Converges absolutely (b) Converges conditionally (c) Diverges Solution: (b) 2. The series 1 X n =0 ( ¡ 1) n n 3 3 n [3] (a) Converges absolutely (b) Converges conditionally (c) Diverges Solution: (a) 3. The series 1 X n =1 ( ¡ 1) n (2 n ) n ( n +1) n [3] (a) Converges absolutely (b) Converges conditionally (c) Diverges Solution: (c) 4. The radius of convergence of the series 1 X n =0 n 2 n (3 x ¡ 1) n is [3] (a) 2 3 (b) 3 2 (c) 2 (d) 1 2 (e) 1 Solution: (a) 5. The coe±cient c 3 of ( x ¡ 2) 3 in the Taylor series of f ( x )=ln( x )about a =2is [3] (a) 1 (b) 3 (c) 1 4 (d) 1 24 (e) None of these Solution: (d) 6. Determine the radius and interval of convergence of the series 1 X n =1 1 n ¢ 2 n ( x ¡ 3) n . [5] Solution: R =l im n !1 ¯ ¯ ¯ ¯ c n c n +1 ¯ ¯ ¯ ¯ = lim n !1 ( n +1)2 n +1 n ¢ 2 n =2 : j x ¡ 3 j < 2 2 <x ¡ 3 < 2 ) 1 <x<
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2 At x =1 ; 1 X n =1 1 n ¢ 2 n ( ¡ 2) n = 1 X n =1 ( ¡ 1) n n converges (conditionally).
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Unformatted text preview: At x = 5 ; 1 X n =1 1 n ¢ 2 n 2 n = 1 X n =1 1 n diverges. Thus, I = [1 ; 5). 7. Find the Maclaurin series (Taylor series about a = 0) of f ( x ) = x 1 + x 3 . [5] Solution: 1 1 ¡ t = 1 X n =0 t n ) 1 1 + x 3 = 1 X n =0 ( ¡ x 3 ) n = 1 X n =0 ( ¡ 1) n x 3 n ) x 1 + x 3 = 1 X n =0 ( ¡ 1) n x 3 n +1 . 8. Find the Maclaurin series (Taylor series about a = 0) of f ( x ) = 1 (1 ¡ x ) 3 . [5] Solution: 1 1 ¡ x = 1 X n =0 x n ) 1 (1 ¡ x ) 2 = d dx 1 1 ¡ x = d dx 1 X n =0 x n = 1 X n =0 nx n ¡ 1 , and 2 (1 ¡ x ) 3 = d dx 1 (1 ¡ x ) 2 = d dx 1 X n =0 nx n ¡ 1 = 1 X n =0 n ( n ¡ 1) x n ¡ 2 ) 1 (1 ¡ x ) 3 = 1 X n =0 n ( n ¡ 1) 2 x n ¡ 2 = 1 X n =2 n ( n ¡ 1) 2 x n ¡ 2 = 1 X n =0 ( n + 2)( n + 1) 2 x n ....
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This note was uploaded on 10/16/2009 for the course MATH math1005 taught by Professor Someone during the Winter '08 term at Carleton CA.

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T4Solutions - At x = 5 ; 1 X n =1 1 n ¢ 2 n 2 n = 1 X n =1...

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