Explanation Given information: 1 5 n 2 − 42 n − 8 Concept used: f is of order at least g , written f ( n ) is Ω ( g ( n ) ) , if and only if there exists positive real numbers A and a ≥ r such that, A g ( n ) ≤ f ( n ) for every integer n ≥ a Calculation: Let m be a non-negative integer, let P ( n ) be a polynomial of degree m , and suppose the coefficient a m of n m is positive. To find big-omega for P ( n ) : Let A = 1 2 a m let’s find a as follows, The coefficient of the highest power is 1 5 Sum of absolute values of coefficients is | − 42 | + | − 8 | . Thus, A = 1 2 × 1 5 A = 1 10 and a = 2 1 5 × ( | − 42 | + | − 8 | ) a = 500 a = 500 , this is greater than 1 To determine (b) Show that 1 5 n 2 − 42 n − 8 is O ( n 2 ) To determine (c) Justify the conclusion that 1 5 n 2 − 42 n − 8 is θ ( n 2 )

5th Edition

ISBN: 9780357035283

Author: EPP

Publisher: Cengage

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Question

Chapter 11.2, Problem 23ES

To determine

(a)

Use one of the methods of Example 11.2.4 to show that 15n2−42n−8 is Ω (n2)

To determine

(b)

Show that 15n2−42n−8 is O (n2)

To determine

(c)

Justify the conclusion that 15n2−42n−8 is θ (n2)

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Chapter 11.2, Problem 22ES

Chapter 11.2, Problem 24ES

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