hw04soln - ECE 310, Spring 2005, HW 4 solutions Problem...

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Unformatted text preview: ECE 310, Spring 2005, HW 4 solutions Problem E3.5 a) Using Definition 3.4.1 , (1- 1 n ) r n- 1 ( A ) + 1 n I A ( n ) = n- 1 n 1 n- 1 n- 1 X i =1 I A ( i ) + 1 n I A ( n ) = 1 n n X i =1 I A ( i ) = r n ( A ) / b) From part a), r n ( A )- r n- 1 ( A ) = 1 n ( I A ( n )- r n- 1 ( A )). | r n ( A )- r n- 1 ( A ) | = 1 n | I A ( n )- r n- 1 ( A ) | 6 1 n c) No. Consider the following construction of a binary sequence as a counter- example: Let x 1 = 1 and r i = 1 i j = i j =1 x j For i > 2 if r i > 2 / 3, then x i +1 = 0 else if r i 6 1 / 3, then x i +1 = 1 else if 1 / 3 < r i < 2 / 3 and x i = 1, then x i +1 = 1 else if 1 / 3 < r i < 2 / 3 and x i = 0, then x i +1 = 0 The sequence of relative frequencies will oscillate between 1/3 and 2/3. Problem E3.8 The polynomial has the following roots: p 1 = 2, p 2 = 2 - 1, p 3 =- 2 - 1, p 4 =- . 5, and p 5 = . 5 . The fifth root, p 5 = 0 . 5, is the only that belongs to the interval [0 , 1], therefore p = 0 . 5 is the answer.5 is the answer....
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hw04soln - ECE 310, Spring 2005, HW 4 solutions Problem...

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