Math 301 Problem Set 3 Solutions

# Math 301 Problem Set 3 Solutions - Math 301/ENAS 513...

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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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Math 301 Problem Set 3 Solutions - Math 301/ENAS 513...

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