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Unformatted text preview: EECS 203: Homework 13 Solutions Section 4.3 1. (E) 6cd c) This is not valid because, for example, f (2) can have many different values and still be consistent with the recursive definition. d) This is not valid for the opposite reason. There are no possible values of f (1) because according to the recursive definition, f (1) = 1 and yet f (1) = 2 f (0) = 0 . 2. (M) 14 Let the proposition P ( n ) be defined as follows: P ( n ) : f n +1 f n 1 f 2 n = ( 1) n (Note that f n is the Fibonacci sequence evaluated at n .) We show by induction that P ( n ) holds for all positive integers n . Basis Step: We show P (1) , which follows from the fact that f 1+1 f 1 1 f 2 1 = f 2 f f 2 1 = 1 · 1 2 = 1 = ( 1) 1 . Inductive Step: Given the induction hypothesis P ( k ) , we show P ( k + 1) as follows: f k +2 f k f 2 k +1 = ( f k +1 + f k ) f k f 2 k +1 expand f k +2 = f k +1 f k + f 2 k f 2 k +1 distribute over f k = f k +1 ( f k f k +1 ) + f 2 k factor out f k +1 = f k +1 ( f k ( f k + f k 1 )) + f 2 k expand f k +1 = f k +1 f k 1 + f 2 k simplify = ( f k +1 f k 1 f 2 k ) factor out 1 = ( 1) k apply the induction hypotheses P ( k ) = ( 1) k +1 simplify Therefore P ( k + 1) holds. This completes the proof by induction. 3. (E) 24 a) Basis Step: 1 ∈ S Recursive Step: if n ∈ S then n + 2 ∈ S b) Basis Step: 3 ∈ S Recursive Step: if n ∈ S then 3 n ∈ S c) Basis Step:...
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.
 Spring '07
 YaoyunShi

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