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Unformatted text preview: Section 3.3 #4) Clearly, the sequence is bounded below by 0. First, x 1 = 1 < 2. If x n < 2, then 2 + x n < 4 and x n +1 = √ 2 + x n < 2 as well. So the sequence is bounded above by 2. Now notice that x 2 n +1 = 2 + x n > x n + x n = 2 x n > x 2 n . Hence, for all n , x n +1 > x n , so the sequence is monotone increasing. So the sequence has a limit. Let x = lim( x n ). Taking the limit of both sides of the defining relation of the sequence, we see that x = lim( √ 2 + x n ) = √ 2 + x . So x 2 x 2 = 0, and thus, x = 1 or x = 2. Clearly, the limit must be greater than 0, so lim( x n ) = 2. #5) Clearly, the sequence is bounded below by 0. First we prove the hint: y 1 = √ p < 1 + 2 √ p . If y n < 1 + 2 √ p , then y n +1 = √ p + y n < p p + 2 √ p + 1 = √ p + 1 < 1 + 2 √ p . So 1 + 2 √ p is an upper bound for this sequence. We will now show that the sequence is increasing. First note that y 2 = p p + √ p > √ p = y 1 . Now we procede by induction: If y n > y n 1 for some n ≥ 2, then y n +1 = √ p + y n > p p + √ y n 1 = y n as desired. Hence the sequence is bounded and monotone, and so converges to some y = lim( y n ). We must have y = √ p + y , and solving this yields the two solutions...
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 Spring '08
 JUNGE
 Tn, Order theory, xk k

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