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# a3sol - MATH 242 Assignment 3 Solutions 1(i It is easy to...

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MATH 242 Assignment 3 Solutions 1. (i) It is easy to verify that x n +1 - x n = - x 3 n - a 3 x 2 n . (1) We have x 3 n +1 - a = (2 x 3 n + a ) 3 - a (3 x 2 n ) 3 (3 x 2 n ) 3 = 8 x 9 n + 12 ax 6 n + 6 a 2 x 3 n + a 3 - 27 ax 6 n 27 x 6 n = 8 x 9 n - 15 ax 6 n + 6 a 2 x 3 n + a 3 27 x 6 n On the other hand (8 x 3 n + a )( x 3 n - a ) 2 27 x 6 n = (8 x 3 n + a )( x 6 n - 2 ax 3 n + a 2 ) 27 x 6 n = 8 x 9 n - 15 ax 6 n + 6 a 2 x 3 n + a 3 27 x 6 n = x 3 n +1 - a (ii) A simple induction now shows that x 3 n > a for all n N . It then follows from (1) that x n +1 - x n < 0 for all n N . We already know that x n > 0 for all n N . (iii) So since ( x n ) is a decreasing sequence bounded below, it must converge to a limit x and we have x 0 since x n 0 for all n N . Next mutliply up the relation x n +1 = 2 x 3 n + a 3 x 2 n to get 3 x 2 n x n +1 = 2 x 3 n + a . Passing to the limit on both sides (and using the uniqueness of limits) gives 3 x 3 = 2 x 3 + a . Hence x 3 = a as desired. 2. This may look plausible, but in fact it’s FALSE! Let z be any irrational number. Let q j for j = 1 , 2 , 3 , . . . be an enumeration of the rational numbers. Then set a j = q j -| q j - z | , b j = q j + | q j - z | and I j =] a j , b j [ . Then z / I j for every j . It’s then obvious that both Q j =1 I j and z / j =1 I j . 3. By hypothesis, the subsequence x 2 n converges, say to x even , the subsequence x 2 n +1 converges, say to x odd and the subsequence x 3 n converges, say to ξ . We claim that x even = x odd . Suppose not. Then choose = 1 5 | x even - x odd | in the definition of the convergence of all three subsequences. Then for all sufficiently large n we have | x even - x 2 n | < , | x odd - x 2 n +1 | < and | ξ - x 3 n | < . Then, for n chosen sufficiently large, we have | ξ - x even | ≤ | ξ - x 6 n | + | x 6 n - x even | < 2 and | ξ - x odd | ≤ | ξ - x 6 n +3 | + | x 6 n +3 - x odd | < 2 leading to the contradiction 5 = | x even - x odd | < 4 . The claim is proved.

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