   Chapter 11.2, Problem 23ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
1 views

# a. Use one of the methods of Example 11.2.4 to show that 1 5 n 2 − 42 n − 8 is Ω ( n 2 ) . b. Show that 1 5 n 2 − 42 n − 8 is O ( n 2 ) c. Justify the conclusion that 1 5 n 2 − 42 n − 8 is Θ ( n 2 ) .

To determine

(a)

Use one of the methods of Example 11.2.4 to show that 15n242n8 is Ω(n2)

Explanation

Given information:

15n242n8

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 ar such that,

Ag(n)f(n) for every integer na

Calculation:

Let m be a non-negative integer, let P(n) be a polynomial of degree m, and suppose the coefficient am of nm is positive.

To find big-omega for P(n): Let A=12am let’s find a as follows,

The coefficient of the highest power is 15

Sum of absolute values of coefficients is |42|+|8|. Thus,

A=12×15A=110 and a=215×(|42|+|8|)a=500

a=500, this is greater than 1

To determine

(b)

Show that 15n242n8 is O(n2)

To determine

(c)

Justify the conclusion that 15n242n8 is θ(n2)

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