Math 301 Problem Set 3 Solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online