hw9solutions - x k ≤ x k-1 , it follows that-1 x k ≤ -1...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: February 10, 2010 Assignment 9 1. Suppose that { x n } is a convergent sequence and { y n } is such that for any ε > 0 there exists M such that | x n - y n | < ε for all n M . Does it follow that { y n } is convergent? Yes. Say that x n x . This means that for any ε > 0 there is an N N so that | x n - x | < ε whenever n N . For a given ε > 0, let K = max ( N,M ) where N and M are given above for the corresponding ε . Then, | y n - x | ≤ | y n - x n | + | x n - x | < ε + ε = 2 ε, n K. Thus, y n x . 2. Let x 1 > 1 and x n +1 := 2 - 1 /x n for n N . Show that { x n } is bounded and monotone. Find the limit. We first show that the sequence is decreasing. We shall do this using induction. Since x 1 > 1, we have ( x 1 - 1) 2 > 0 = x 2 1 - 2 x 1 + 1 > 0 = x 2 1 > 2 x 1 - 1 = x 1 > 2 - 1 x 1 = x 2 . This establishes the base case. We now assume that x k x k - 1 and use this to show the k +1 case. Since
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Unformatted text preview: x k ≤ x k-1 , it follows that-1 x k ≤ -1 x k-1 , and x k +1 = 2-1 x k ≤ 2-1 x k-1 = x k . Thus the sequence is monotonically decreasing. We next show that it is bounded below. Indeed, we show that x n > 1. We, again, use induction. The base case is given by hypothesis. Assuming that x k > 1, we examine x k +1 . Indeed, x k > 1 gives us that-1 x k >-1 and x k +1 = 2-1 x k > 2-1 = 1 which establishes the bound from below. Since the sequence is monotonically decreasing and bounded below, it must converge. Set L = lim n →∞ x n . Then, we must have that L = 2-1 L 1 or L 2 = 2 L-1 . Thus, L = 1. 2...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

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hw9solutions - x k ≤ x k-1 , it follows that-1 x k ≤ -1...

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