1873_solutions

# X n is convergent and discuss its limit assuming an

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Unformatted text preview: ten times, starting at the number 0. Evaluate the column of numbers that you have obtained accurately to ten decimal places and, in this way, show the first ten members of the sequence x n . fx  b. Show that x n 2  2  x n 5  2x n for every integer n 1, and then show that the sequence x 2n 1 is increasing and that the sequence x 2n is decreasing and that these two sequences have the same limit 2 1. Solution: For every positive number x we define g x  2x . 5  2x Whenever 0  t  x we see that 2t  xt 0 g t  2x 5  2t 5  2x 5  2x 5  2t and so the function g is strictly increasing. Since x 1  x 3 we have g x 1  g x 3 from which we deduce that x 3  x 5 . Continuing in this way we see that the sequence x 2n 1 is increasing. Since x 2  x 4 we have g x 2  g x 4 from which deduce that x 4  x 5 . Continuing in this way we see that the sequence x 2n is decreasing. Therefore the sequences x 2n 1 and x 2n are convergent. If we write x  n Ý x 2n 1 lim gx then it follows from the identity x 2 n 1  2  x 2 n 1 5  2x 2n 1 that x  2x 5  2x and we can see that the only positive solution of this equation is 2 lim x n Ý 2n 1  2 1. Thus 1 and we can see similarly that lim x n Ý 2n c. Deduce that x n 2 1 as n  2 1. Ý. Exercises on the Cantor Intersection Theorem 149 1. Suppose that H n is a sequence (not necessarily contracting) of closed bounded sets and that for every positive integer n we have n  Hi . i1 Prove that Ý  Hi . i1 For each n we define n Kn   Hi i1 Since the sequence K n is a contracting sequence of nonempty closed bounded sets, we know from the Cantor intersection theorem that Ý  Kn . n1 We observe finally that, since Ý Ý n1 n1  Hn   Kn the intersection of all of the sets H n must also be nonempty. 2. Suppose that H is a closed bounded set of real numbers and that U n is an expanding sequence of open sets. a. Explain why the sequence of sets H U n is a contracting sequence of closed bounded sets. For each n, it follows at once from the inclusion U n U n1 that H U n 1 H Un. b. Use the Cantor intersection theorem to deduce that if H for every n then Un Ý H Un . n1 Since H is closed and bounded and since H Un  H R Un for each n, the sets H U n must be closed and bounded. The desired result therefore follows at once from the Cantor intersection theorem. c. Prove that if Ý H  Un n1 then there exists an integer n such that H We suppose that Un. Ý H  Un. n1 Since the set Ý  H n1 Un  H Ý  Un n1 is empty we deduce from part b that there is a value of n for which H integer n we have H U n . Un  and for any such 3. Suppose that U n is a sequence of open sets (not necessarily expanding) and that H is a closed bounded set 150 and that Ý  Un. H n1 Prove that there exists a positive integer N such that N  Un. H n1 For each n we define n Vn   Ui. i1 The sequence V n is expanding and, since Ý H  Un  n1 Ý  Vn n1 it follows from Exercise 2 that there exists a positive integer n for which...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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