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Unformatted text preview: Homework 3 Solutions Note: These solutions are not necessarily model answers. Rather, they are designed to be tutorial in nature, and sometimes contain a little more explanation than an ideal solution. Also, bear in mind that there may be more than one correct solution. 1. A formula for the Fibonacci Numbers Base cases : For n = 0 the formula evaluates to zero, which is the correct value of F (0) . check . For n = 1 , the value of the formula is  (1 ) 5 = 2  1 5 = 2 1+ 5 2 1 5 = 2 + 2 5 2 2 5 = 1 = F (1) . check So both base cases are verified. Inductive hypothesis : Assume that the formula is valid for k n , where n 1 is arbitrary. (Note that we are using strong induction.) Induction step : We want to show that the formula is valid for n + 1 . To do this, we first notice two facts that follow from the definition of and that will be useful to us in the algebra below: (a) 1 + = 2 ; (b) (1 ) 2 = 1 2 + 2 = 2 . Now, using the recursive definition of the Fibonacci sequence we have F ( n + 1) = F ( n ) + F ( n 1) = n (1 ) n 5 + n 1 (1 ) n 1 5 [by the inductive hypothesis] = 1 5 bracketleftbig n 1 ( + 1) (1 ) n 1 (1 + 1) bracketrightbig = 1 5 bracketleftbig n 1 ( + 1) (1 ) n 1 (2 ) bracketrightbig = 1 5 bracketleftbig n 1 2 (1 ) n 1 (1 ) 2 bracketrightbig [using (a) and (b) above] = 1 5 bracketleftbig n +1 (1 ) n +1 bracketrightbig . This completes the induction step, and thus we can conclude that the formula is valid for all n . 2. Strengthening the Induction Hypothesis First note that parenleftbigg a b c d parenrightbiggparenleftbigg 1 0 1 1 parenrightbigg = parenleftbigg a + b b c + d d parenrightbigg , so if we just try to use the inductive hypothesis that all the entries of the...
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This note was uploaded on 10/08/2008 for the course CMPSC 360 taught by Professor Haullgren during the Fall '08 term at Pennsylvania State University, University Park.
 Fall '08
 HAULLGREN

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