Exam_solutions_4_(2)

# Exam_solutions_4_(2) - MATH 141 – TEST 4(9.8 – 10.2...

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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 Σ 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) - MATH 141 – TEST 4(9.8 – 10.2...

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