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hw4-sol - Modern Analysis Homework 4 Possible solutions...

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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 > 0 So by induction, we have x n > α for all n N . 1
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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 . Solution: In part (a), we already computed ε n +1 = x n +1 - α = ( x n - α ) 2 2 x n , so ε n +1 = ε 2 n 2 x n . Since we know that x n > α for all n N , it follows that ε n +1 < ε 2 n 2 α = (2 α ) ε n 2 α 2 . Setting β = 2 α and applying the previous formula recursively in n gives ε n +1 < β ε 1 β 2 n .
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