CIS
lab2soln

• Notes
• 3

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Your proof should follow just from the definition of Big-Oh. (As usual in problems with Big-Oh, we assume that the functions map nonnegative reals to strictly positive reals.) Prove Theorem 5 on slide 30 of Lecture Notes 2, that is, prove that for any(all) f, g, h, k if f ( n ) is O ( h ( n )) and g ( n ) is O ( k ( n )) then f ( n ) + g ( n ) is O ( h ( n ) + k ( n )). Answer Assume f ( n ) is O ( h ( n )) therefore N 1 , c 1 > 0 such that n N 1 we have f ( n ) c 1 h ( n ). Assume also that g ( n ) is O ( k ( n )) therefore N 2 , c 2 > 0 such that n N 2 we have g ( n ) c 2 k ( n ). We want to show that these assumptions imply that N 3 , c 3 > 0 such that n N 3 we have f ( n ) + g ( n ) c 3 ( h ( n ) + k ( n )). From f ( n ) c 1 h ( n ) and g ( n ) c 2 k ( n ), we can add the sides of these two inequalities obtaining f ( n ) + g ( n ) c 1 h ( n ) + c 2 k ( n ). Now take c 3 = max( c 1 , c 2 ). Therefore c 1 h ( n ) + c 2 k ( n ) c 3 h ( n ) + c 3 k ( n ) = c 3 ( h ( n ) + k ( n )) so f ( n ) + g ( n ) c 3 ( h ( n ) + k ( n )). 1

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Moreover, to be sure that both f ( n ) c 1 h ( n ) and g ( n ) c 2 k ( n ) we have to have both n N 1 n N 2 . A simple way to ensure this is to have n max( N 1 , N 2 ). To recap, we have shown that there exist c 3 = max( c 1 , c 2 ) and N 3 = max( N 1 , N 2 ) such that for n N 3 we have f ( n ) g ( n ) c 3 h ( n ) k ( n ).
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