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Unformatted text preview: 2.4.2. De&ne the sequence ( x n ) by x 1 = 3 and x n +1 = 1 4 & x n (a) Prove that the sequence ( x n ) converges. The &rst three terms of the sequence are 3 , 1 , 1 = 3 . We want to show that this sequence is decreasing (it will be bounded below by because x n ¡ 3 ). So we want to show that 1 4 & x n < x n ¡ 3 hold for all natural numbers n . The inequality x n ¡ 3 makes sure that certain quantities, like 4 & x n , stay positive. Of course we do this by induction. We know the inequalities hold for n = 1 , because 1 < 3 ¡ 3 . So suppose the inequalities hold for for n = k . That is 1 4 & x k < x k ¡ 3 Then, because x k +1 = 1 4 & x k , we see trivially that x k +1 ¡ 3 . It remains to show that 1 4 & x k +1 < x k +1 that is, that 1 4 & 1 4 & x k < 1 4 & x k Multiplying the numerator and denominator of the left hand side by 4 & x k gives 4 & x k 4 (4 & x k ) & 1 Note that the denominator of this fraction is positive because x k ¡ 3 , so 4 & x k ¢ 1 . So, since both denominators in the desired inequality are positive, we can verify the inequality 4 & x k 4 (4 & x k ) & 1 < 1 4 & x k 1 by cross multiplying and showing that (4 & x k ) 2 < 4 (4 & x k ) & 1 But the right hand side of this inquality can be written as (4 & x k ) 2 + x k (4 & x k ) & 1 , so the inequality becomes 1 < x k (4 & x k ) which is simply the induction hypothesis...
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 Spring '10
 MetCalfe
 Natural number, Don, Mathematics in medieval Islam

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