Exam_solutions_4_(2)

Exam_solutions_4_(2) - December 12, 2004 MATH 141 TEST 4...

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Unformatted text preview: December 12, 2004 MATH 141 TEST 4 (9.8 10.2 plus earlier) [Pilachowski] SOLUTIONS 1. a. (14 points) Find the radius of convergence of ( ) ( ) = 1 2 ! ! 2 n n x n n . (Show all your steps.) Ratio Test: ( ) [ ] ( ) [ ] ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( )( ) 4 1 2 2 2 1 lim 2 2 1 2 2 2 1 lim 2 1 2 1 lim 2 2 2 2 2 1 2 ! ! ! ! ! ! ! ! x n n x n x n n n n n x x n n n x n n x n n n n n n n n = + + + = + + + = + + + For convergence, 4 1 4 < < x x , so radius of convergence R = 4. b. (12 points) By making the appropriate adjustments to the power series expansion for x e , derive a power series expansion for ( ) 1 = x e x f . (Show all your steps, and write your answer in form.) ( ) ( ) ( ) ( ) = = = = = = + = = = = 1 1 ! ! ! ! ! 1 1 1 1 1 n n n x n n n n n n n n x n n x n x e n x n x n x e n x e 2. a. (12 points) Given ( ) = = + 1 1 1 n n n x x , derive a power series expansion for ( ) ( ) 2 1 1 x x f + = . (Show all your steps, and write your answer in...
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Exam_solutions_4_(2) - December 12, 2004 MATH 141 TEST 4...

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