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Unformatted text preview: Modern Analysis, Homework 4 Possible solutions 1. [12] (5 points) Prove that every sequence of reals has a monotone (ie either decreas ing or increasing, though not necessarily strictly so) subsequence. Solution: Let { x n } be a sequence of reals. Suppose that lim sup n →∞ x n = + ∞ Then choose n 1 ∈ N so that x n 1 > 1. Having chosen n 1 , ··· ,n i ∈ N , choose n i < n i +1 ∈ N so that x n i +1 > x n i . Then { x n i } forms a monotone (increasing) subsequence. If instead lim sup n →∞ x n =∞ (for example x n = n ), then a similar argument shows that x n has a monotone decreasing subsequence. Now suppose that s * = lim sup n →∞ x n is finite. There are two cases to consider: (1) The set S > = { n ∈ N  x n ≥ s * } is infinite (2) The set S < = { n ∈ N  x n ≤ s * } is infinite (notice that these cases are not mutually exclusive, but at least one of them holds). First suppose that case 1 holds. Choose some n 1 ∈ S > (this gives us our x n 1 ). Having chosen n 1 , ··· ,n i ∈ S > , choose n i +1 ∈ S > such that n i +1 > n i and x n i +1 ≤ x n i (possible since we’re assuming that S > is infinite ). In this way, we construct a monotone decreasing subsequence { x n i } of { x n } . In a similar way, if case 2 holds, we can construct a monotone increasing subsequence. 2. [15] (10 points) Rudin, Ch 3, problem 16, parts (a) and (b). Fix a positive number α . Choose x 1 > √ α and define x 2 ,x 3 ,x 4 , ··· , by the recursion formula x n +1 = 1 2 x n + α x n . (a) Prove that { x n } decreases monotonically and that lim x n = √ α. Solution: As a first step, I claim that with the above definition, we have x n > √ α for all n ∈ N . Let’s prove this by induction: the statement for n = 1 is true by construction. Now suppose that x 1 , ··· ,x k > √ α for some k ∈ N . Then x k +1 √ α = x k 2 + α 2 x k √ α = 1 x k x 2 k 2 + α 2 x k √ α = 1 2 x k ( x 2 k 2 √ αx k + α ) = 1 2 x k ( x k √ α ) 2 > So by induction, we have x n > √ α for all n ∈ N . 1 2 Now with this in place, we can show that { x n } is decreasing: x n x n +1 = x n x n 2 + α 2 x n = x n 2 α 2 x n = x 2 n α 2 x n > 0 (since x n > √ α ) Since { x n } is a decreasing sequence which is bounded from below, it follows that it con verges to some limit, L . By the defining equation x n +1 = 1 2 x n + α x n , it follows that L = 1 2 ( L + α L ) . Solving for L gives L 2 = α . Now since x n > √ α > 0 for all n ∈ N , it follows that L too must be positive. Hence L = √ α . (b) Put ε n = x n √ α , and show that ε n +1 = ε 2 n 2 x n < ε 2 n 2 √ α so that, setting β = 2 √ α , ε n +1 < β ε 1 β 2 n for all n ∈ N ....
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This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
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