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psAnalysis1Ans - Fall 2007 Problem Set#01 First Analysis...

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Fall 2007 ARE211 Problem Set #01 First Analysis Problem Set Due date: Sep 11 Problem 1 Please use the Pythagorean metric in the following problem a) Consider the sequence x n = 2 + ( - 1) n n defined on R . Prove (i) that the sequence is a convergent sequence using the definition of a convergent sequence and show (ii) that the sequence is a Cauchy sequence using the definition of a Cauchy sequence . b) Now consider the sequence x n = 2 + ( - 1) n n defined on S = R \ { 2 } . Using your proof from part a) argue that it is still a Cauchy sequence in S . Prove that it is not a convergent sequence in S . (Note: The set A \ B is defined as: A \ B = { x | x A , x / B } ). Ans: Recall the following definitions for convergence: (C) Convergent Sequence: A sequence x n in X converges to an element x X in the metric d if epsilon1 > 0 N N such that n > N we have d ( x n , x ) < epsilon1 . (CA) Cauchy Sequence: A sequence x n in X is a Cauchy sequence under the metric d if epsilon1 > 0 N N such that n, m > N we have d ( x n , x m ) < epsilon1 . Now let’s apply those definitions to the sequence x n = 2 + ( - 1) n n using the Pythagorian metric. (a) (i) The sequence converges to 2 X as epsilon1 > 0 pick N = ceilingleft 1 epsilon1 ceilingright ∈ N and n > N d ( x n , x ) = | 2 + ( - 1) n n - 2 | = | ( - 1) n n | = 1 n (The absolute value of ( - 1) n is 1) < 1 N (Given n > N ) 1 1 epsilon1 = epsilon1 (Given N = ceilingleft 1 epsilon1 ceilingright ) (ii) The sequence is a Cauchy sequence as epsilon1 > 0 pick N = ceilingleft 2 epsilon1 ceilingright ∈ N and n, m > N d ( x n , x m ) = | 2 + ( - 1) n n - 2 - ( - 1) m m | = | ( - 1) n n + - ( - 1) m m | ≤ | ( - 1) n n | + | - ( - 1) m m | (The triangle inequelity of the absolute value) = 1 n + 1 m (The absolute value of ( - 1) n and - ( - 1) m is 1) < 1 N + 1 N (Given n, m > N ) = 2 N 2 2 epsilon1 = epsilon1 (Given N = ceilingleft 2 epsilon1 ceilingright ) 1
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2 (b) (i) Convergent sequence: One easy way out might be to assume it converges to a point other than 2 and then argue that this is a violation of problem 6 on problem set 2 (i.e. that a sequence converges to at most one point). However, be careful as you are dealing with two different universes here. It might be a good exercise to show this part using the definition (C). A sequence does not converge to a point x X in the metric d if epsilon1 > 0 s.t. N N n > N with d ( x n , x ) epsilon1 . Every point x in the universe X = R \ { 2 } can be written as x = 2 + δ, δ negationslash = 0 We know have to consider two cases: case 1) Let δ > 0 . We now show that this can’t be a limit: Let epsilon1 = δ and N N n = 2 N + 1 > N with d ( x n , x ) = | 2 + ( - 1) n n - 2 - δ | = | ( - 1) 2 N +1 2 N +1 - δ | = | - 1 2 N +1 - δ | (As 2 N + 1 is an odd number) = 1 2 N +1 + δ (As both - 1 2 N +1 and - δ are negative) > δ (As 1 2 N +1 is positive) = epsilon1 case 2) Let δ < 0 . We now show that this can’t be a limit: Let epsilon1 = - δ and N N n = 2 N > N with d ( x n , x ) = | 2 + ( - 1) n n - 2 - δ | = | ( - 1) 2 N 2 N - δ | = | 1 2 N - δ | (As 2 N is an even number) = 1 2 N - δ (As both 1 2 N and - δ are positive) > - δ (As 1 2 N is positive) = epsilon1 (ii) Cauchy sequence: The proof that x n is a Cauchy sequence in part (a) does not depend on whether the limit x is in the set or not. Clearly all points of the sequence { x n } lie in S = R \ { 2 } and the same proof thus holds.
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