Final_sln - Prof Bjorn Poonen(solutions by J Steever...

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Prof. Bjorn Poonen (solutions by J. Steever) December 6, 2001 MATH 55 PRACTICE FINAL SOLUTIONS (1) For each of (a)-(d) below: If the statement is true (always), write TRUE. Otherwise write FALSE. (Please do not use the abbreviations T and F, since in handwriting they are sometimes indistiguishable.) No explanations are required in this problem. (a) If f : S T is a function and A B S , then f ( B - A ) = f ( B ) - f ( A ). FALSE. Consider the constant function f : { 1 , 2 } → { 1 } (i.e. f (1) = f (2) = 1). If B = { 1 , 2 } and A = { 1 } then f ( B - A ) = f ( { 2 } ) = 1, however f ( B ) - f ( A ) = { 1 } - { 1 } = . (b) If f and g are independent random variables defined on the same sample space, then the standard deviation σ ( f + g ) of f + g equals ( σ ( f )) 2 + ( σ ( g )) 2 . TRUE. σ ( f + g ) = V ( f + g ) = V ( f ) + V ( g ) = ( σ ( f )) 2 + ( σ ( g )) 2 (c) There exists an injection from the set of rational numbers to the set of integers. TRUE. The set of rationals is countable, therefore there is a bijection, and hence an injection, from the set of rationals into the set of integers. (d) The program procedure fibonacci(n : nonnegative integer) if n = 0 then fibonacci (0) := 0 else if n = 1 then fibonacci (1) := 1 else fibonacci ( n ) := fibonacci ( n - 1) + fibonacci ( n - 2) performs a total of Θ( n ) additions to compute the n th Fibonacci number. FALSE. Let a n denote the number of additions required by the algorithm to produce the n th Fibonacci number. Then a simple analysis of the algorithm yield the relation a n = a n - 1 + a n - 2 + 1, with initial conditions a 0 = 0 , a 1 = 0. Solving this relation, we get the formula a n = 5 - 5 10 · ( 1 + 5 2 ) n + 5 + 5 10 · ( 1 - 5 2 ) n - 1 which is not Θ( n ). (2) How many elements of { 1 , 2 , . . . , 10 } × { 1 , 2 , . . . , 10 } must be taken in order to guarantee that there are two elements, say ( i, j ) and ( i , j ), such that i + j = i + j ? 1
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2 A total of 20 elements must be picked. Over all ( i, j ) in the set, i + j ranges over the 19 natural numbers 2 thru 20. Thus, by the Pigeonhole Principle, any twenty pairs must contain at least two whose coordinate sums are equal.
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