1873_solutions

X n of points of e such that whenever m n we have x m

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Unformatted text preview: 2  3 n holds when 2/n  . Using the fact that the number 2/ is not an upper bound of the set Z of integers we choose an integer N such that N  2/ , in other words, 2. N Then, whenever n N we have 3 2  3 3  3  3 2 n N and so we have shown that x n is eventually in 3 , 3  . 3. Given that x n  1/n for each positive integer n and that x 0, prove that x is not a partial limit of x n . Solution: In the event that x  0, the interval Ý, 0 is a neighborhood of x and it is clear that x n fails to be frequently in this neighborhood. Therefore no negative number can be a partial limit of x n . Suppose now that x  0. The interval x/2, Ý is a neighborhood of x x 2 0 x and the condition x n x/2, Ý must fail to hold whenever n  2/x. Therefore the sequence x n cannot be frequently in the interval x/2, Ý and the number x cannot be a partial limit of x n . 4. Given that 1 n n 3 if n is a multiple of 3 xn  0 if n is one more than a multiple of 3 , 4 if n is two more than a multiple of 3 Prove that the partial limits of x n are Ý, Ý, 0 and 4. Since x n is unbounded both above and below, it follows from the discussion of infinite partial limits we saw earlier that both Ý and Ý are partial limits of x n . Since the equation x n  0 holds for infinitely many values of n we know that x n is frequently in every neighborhood of 0 and so 0 is a partial limit of x n . In the same way we can see that 4 is a partial limit of x n . Now we need to explain why any real number other than 0 and 4 must fail to be a partial limit of x n . Suppose that x R 0, 4 . In the event that 0  x  4, the fact that x is not a partial limit of x n follows from the fact that x n is not frequently (or ever) in the interval 0, 4 which is a neighborhood of x. Now suppose that x  0. x 1 x 0 In order to show that x is not a partial limit of x n we shall make the observation that x n is not frequently in the interval x 1, 0 which is a neighborhood of x. In fact, the inequality x 1  xn  0 can hold only when n is odd and x 1  n3 31 which is equivalent to saying that n x. Since there are only finitely many such positive integers n we conclude that x n is not frequently in the interval x 1, 0 134 Finally we must consider the case x  4. x 4 x1 In this case we observe that there can be only finitely many positive integers n for which 4  n3  x  1 and so, once again, x can’t be a partial limit of x n . 5. Give an example of a sequence of real numbers whose set of partial limits is the set 1 Þ 4, 5 . Hint: For each positive integer n, if n can be written in the form n  2m3k for some positive integers m and k and if 4 m k 5 then we define xn  m . k In all other cases we define x n  1. Observe that the range of the sequence x n is the set 1 Þ Q 4, 5 and then show that the set of partial limits of x n is 1 Þ 4, 5 . Since the equation x n  1 holds for infinitely many positive integers n the number 1 must be a partial limit of x n . To see that every number in the interval 4,...
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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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