HW8_sol - COT 3100 Applications of Discrete Structures...

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Unformatted text preview: COT 3100: Applications of Discrete Structures March 31, 2010 Solution to Homework 8 Lecturer: Prof. My T. Thai TA: Thang N. Dinh 1 Section 4.3 Problem 3 Solution: In each case we compute the subsequent terms by plugging into the recursive formula, using the previously given or computed values. a. f (2) = f (1)+3 f (0) = 2+3(- 1) =- 1; f (3) = f (2)+3 f (1) =- 1+3 . 2 = 5; f (4) = f (3)+3 f (2) = 5 + 3(- 1) = 2; f (5) = f (4) + 3 f (3) = 2 + 3 · 5 = 17 b. f (2) = f (1) 2 f (0) = 2 2 · (- 1) =- 4; f (3) = f (2) 2 f (1) = (- 4) 2 · 2 = 32; f (4) = f (3) 2 f (2) = 32 2 · (- 4) =- 4096; f (5) = f (4) 2 f (3) = (- 4096) 2 · 32 = 536 , 870 , 912 c. f (2) = 3 f (1) 2- 4 f (0) 2 = 3 · 2 2- 4 · (- 1) 2 = 8; f (3) = 3 f (20 2- 4 f (1) 2 = 3 · 8 2- 4 · 2 2 = 176; f (4) = 3 f (3) 2- 4 f (2) 2 = 3 · 176 2- 4 · 8 2 = 92 , 672; f (5) = 3 f (4) 2- 4 f (3) 2 = 3 · 92672 2- 4 · 176 2 = 25 , 764 , 174 , 174 , 848 d. f (20 = f (0) /f (1) = (- 1) / 2 =- 1 / 2; f (3) = f (10 /f (0) = 2 / (- 1 2 ) =- 4; f (4) = f (2) /f (3) = (- 1 / 2) / (- 4) = 1 / 8; f (5) = f (3) /f (4) = (- 4) / (1 / 8) =- 32 Problem 5 Solution: a. This is not valid, since letting n = 1 we would have f (1) = 2 f (- 1), but f (- 1) is not defined. b. This is valid. The basis step tells us what f (0) is, and the recursive step tells us how each subsequent value is determined from the one before. It is not hard to look at the pattern and conjecture that f ( n ) = 1- n . We prove this by induction. The basis step is f (0) = 1 = 1- 0; and if f ( k ) = 1- k , then f ( k + 1) = f ( k )- 1 = 1- k- 1 = 1- ( k + 1). c. The basic conditions specify f (0) and f (1), and the recursive step gives f ( n ) in terms of f ( n- 1) for n ≥ 2, so this is a valid definition. If we compute the first several values, we conjecture that f ( n ) = 4- n if n > 0, but f (0) = 2. That is our “formula.” To prove it correct by induction we need two basis steps: f (0) = 2, and f (1) = 3 = 4- 1. For the inductive step (with k ≥ 1), f ( k + 1) = f ( k )- 1 = ( f- k )- 1) = 4- ( k + 1). Page 1-1 d. The basis conditions specify f (0) and f (1), and the recursive step gives f ( n ) in terms of f ( n- 2) for n ≥ 2, so this is a valid definition. The sequence of function values is 1 , 2 , 2 , 4 , 4 , 8 , 8 ,..., and we can fit a formula to this if we use the floor function: f ( n ) = 2 b ( n +1) / 2 c . For a proof, we check the base case f (0) = 1 = 2 b (0+1) / 2 c and f (1) = 2 = 2 b (1+2) / 2 c . For the inductive step: f ( k + 1) = 2 f ( k- 1) = 2 · 2 b k/ 2 c = 2 b (( k +1)+1) / 2 c e. The definition tells us explicitly what f (0) is. The recursive step specifies f (1) ,f (3) ,... in terms of f (0) ,f (2) ,... ; and it also gives...
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HW8_sol - COT 3100 Applications of Discrete Structures...

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