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Unformatted text preview: Math 301/ENAS 513: Homework 3 Solution Triet Le October 3, 2007 1. Let { z n } be a complex sequence. Show that lim n z n = z lim n Re ( z n ) = Re ( z ) and lim n Im ( z n ) = Im ( z ) . Solution 1 . Let { z n } be a complex sequence. ( ) : Suppose lim n z n = z . Then for each > 0, there exists N = N ( ) such that  z n z  < , for all n N. But  Re ( z n ) Re ( z )  =  Re ( z n z )   z n z  . Thus,  Re ( z n ) Re ( z )  < , for all n N . Similarly,  Im ( z n ) Im ( z )  < for all n N . By the definition of convergence, Re ( z n ) Re ( z ) and Im ( z n ) Im ( z ). ( ): Suppose lim n Re ( z n ) = Re ( z ) and lim n Im ( z n ) = Im ( z ). Pick any > 0, there exist N 1 = N 1 ( / 2) and N 2 = N 2 ( / 2) such that  Re ( z n ) Re ( z )  < / 2 , n N 1 , and  Im ( z n ) Im ( z )  < / 2 , n N 2 . Let N = max ( N 1 , N 2 ), then  z n z   Re ( z n z )  +  Im ( z n z )  =  Re ( z n ) Re ( z )  +  Im ( z n ) Im ( z )  < / 2 + / 2 = . Thus, by the definition of convergence, z n z . 2. Page 32: 1. Suppose that { x n } is a sequence of real numbers that converges to x , and suppose that a x n b for all n . Prove that a x b . Solution 2 . Suppose a > x , and let = ( a x ) > 0. Then there exists N = N ( ) such that  x n x  < , for all n N . Thus x n < x + = x + ( a x ) = a, n N. This contradicts the assumption that x n a , for all n . Thus a x . Similarly, x b . 1 3. Page 32: 9. Solution 3 . (Skip) 4. Page 32: 12. Suppose that x 1 = 1 and x n +1 = 2 + x n , n = 1 , 2 , ... . Prove that { x n } converges and find its limit. Solution 4 . We will show that { x n } is bounded above by 2 and increasing (thus it converges). Note x 1 = 1 2, x 2 = 3 2. Suppose that x n 2, then x n +1 = 2 + x n 2 + 2 = 2. Thus, by induction, x n 2, for all n . Since x n 1 > 0, for all n , one way to show { x n } is increasing is to show x n +1 x n 1. But x n +1 x n 1 x 2 n +1 x 2 n 1. x 2 n +1 x 2 n = 2 + x n x 2 n x n + x n x 2 n = 2 x n x 2 n x 2 n x 2 n = 1 . Thus there x n x for some x R such that 1 x 2. We have x = lim n x n +1 = lim n 2 + x n = 2 + x....
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 Fall '07
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