This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 > 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 > N N such that n,m > N we have d ( x n ,x m ) < epsilon1 . Now lets 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 > 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 > 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 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 > 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 + , negationslash = 0 We know have to consider two cases: case 1) Let > . We now show that this cant be a limit: Let epsilon1 = and N N n = 2 N + 1 > N with d ( x n ,x ) =  2 + ( 1) n n 2  =  ( 1) 2 N +1 2 N +1  =  1 2 N +1  (As 2 N + 1 is an odd number) = 1 2 N +1 + (As both 1 2 N +1 and are negative) > (As 1 2 N +1 is positive) = epsilon1 case 2) Let < . We now show that this cant be a limit: Let....
View Full
Document
 Fall '07
 Simon

Click to edit the document details