This preview shows pages 1–2. 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: 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....
View Full
Document
 Spring '05
 HAAS

Click to edit the document details