quizsol11 - > , let N = 7 9 . Then for any n...

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Math 104, Solution to Quiz 11 Instructor: Guoliang Wu August 4, 2009 1. (10pts) Find the radius of convergence of the following power series X n =0 (2 n )! n 2 n x n Solution: β = lim sup ± ± ± ± (2 n )! n 2 n ± ± ± ± 1 /n = lim sup (2 n + 2)! ( n + 1) 2 n +2 · n 2 n (2 n )! = lim sup (2 n + 2)(2 n + 1) ( n + 1) 2 · 1 ( 1 + 1 n ) 2 n = 4 e 2 . So the radius of convergence R = 1 β = e 2 4 . 2. Let f n ( x ) = n + sin x 3 n + x 2 for n N ,x [0 , 2] . (a) (5pts) Find the (pointwise) limit function f of the sequence ( f n ) . Solution: For any x [0 , 2] , - 1 n sin x n 1 n and 0 x 2 n 4 n , by squeeze theorem, lim n →∞ sin x n = lim n →∞ x 2 n = 0 . Hence f n ( x ) = 1 + sin x n 3 + x 2 n 1 3 . 1
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So f = 1 3 . (b) (10pts) Show that f n converges uniformly to the above f . Solution: For any
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Unformatted text preview: > , let N = 7 9 . Then for any n > N and x [0 , 2] , | f n ( x )-f ( x ) | = n + sin x 3 n + x 2-1 3 = | n + sin x-n-x 2 3 | 3 n + x 2 | sin x | + x 2 3 3 n + x 2 1 + 4 3 3 n < 7 9 N = . By denition, f n converges uniformly to f = 1 3 . (c) (5pts) Calculate lim n Z 2 f n ( x ) dx Solution: Since f n converges uniformly to 1 3 , lim n Z 2 f n ( x ) dx = Z 2 1 3 dx = 2 3 . 2...
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quizsol11 - > , let N = 7 9 . Then for any n...

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