This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 11 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...
View
Full Document
 Spring '08
 Staff
 Mathematical Induction, Natural number, basis step, recursive step, Thang N. Dinh

Click to edit the document details