Explanation Given information: 5 n 3 + 65 n + 30. Concept used: Let f and g be real-valued functions defined on the same set of nonnegative integers, with g ( n ) ≥ 0 for every integer n ≥ r , where r is a positive real number. Then f is of order g, written f ( n ) is Θ ( g ( n )) ( f of n is big-Theta of g of n ), if and only if, there exist positive real numbers A, B and k ≥ r such that f is of order at least g, written f (n) is Ω ( g (n)) ( f of n is big-Omega of g of n ), if and only if, there exist positive real numbers A and a ≥ r such that A g ( n ) ≤ f ( n ) for every integer n ≥ a . f is of order at most g, written f (n) is O ( g ( n )) ( f of n is big- O of g of n ), if and only if, there exist positive real numbers B and b ≥ r such that 0 ≤ f ( n ) ≤ B g ( n ) for every integer n ≥ b . Proof: Firstly, to show that 5 n 3 + 65 n + 30 is Ω ( n 3 ) , we need to show that 5 n 3 + 65 n + 30 is greater than or equal to a positive multiple of n 3 for all values of n that are sufficiently large. Now 5 n 3 ≤ 5 n 3 + 65 n + 30 for every integer n ≥ 1 Because when n ≥ 1 , 65 n + 30 is positive. Thus we can let A = 5 and a = 1 to obtain A n 3 ≤ 5 n 3 + 65 n + 30 for every integer n ≥ a , Concluding by definition of Ω -notation, that 5 n 3 + 65 n + 30 is Ω ( n 3 ) . Secondly, to show that 5 n 3 + 65 n + 30 is O ( n 3 ) , we need to show that 5 n 3 + 65 n + 30 is greater than or equal to 0 and less than or equal to some positive multiple of n 3 for all values of n that are sufficiently large. First note that because all terms of 5 n 3 + 65 n + 30 are positive, 0 ≤ 5 n 3 + 65 n + 30 for every integer n ≥ 1

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ISBN: 9780357540244

Author: EPP

Publisher: CENGAGE L

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Chapter 11.2, Problem 13ES

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To show that 5n3+65n+30 is Θ(n3).

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Chapter 11.2, Problem 12ES

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To show that 5 n 3 + 65 n + 30 is Θ ( n 3 ) .

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