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Chapter 1: Problem 16

Using the definitions of \(O\) and \(\Omega\), show that $$6 n^{2}+20 n \in O\left(n^{3}\right) \quad \text { but } \quad 6 n^{2}+20 n \notin \Omega\left(n^{3}\right)$$

Short Answer

Expert verified
The function \(6n^2 + 20n\) does, indeed, belong to \(O(n^3)\) but it does not belong to \(\Omega(n^3)\), as per the provided definitions and analysis.

Step by step solution

01

Prove \(6n^2 + 20n \in O(n^3)\)

According to the definition of Big O notation, a function \(f(n)\) is in \(O(g(n))\) if there exists a positive constant \(c\) and an integer \(n_0\) such that \( 0 \leq f(n) \leq c*g(n) \) for all \(n \geq n_0\). Here, \(f(n)\) is \(6n^2 + 20n\) and \(g(n)\) is \(n^3\). If there are constants \(c\) and \(n_0\) such that the inequality holds, then \(6n^2 + 20n \in O(n^3)\). For \(6n^2 + 20n \leq c*n^3\), taking \(c = 26\) and \(n_0 = 1\), it can be seen that the inequality will hold. So, \(6n^2 + 20n \) is indeed in \(O(n^3)\).
02

Show \(6n^2 + 20n \notin \Omega(n^3)\)

According to the definition of Big Omega notation, a function \(f(n)\) is in \(\Omega(g(n))\) if there exists a positive constant \(c\')\) and an integer \(n_0\'\) such that \( 0 \leq c\'*g(n) \leq f(n) \) for all \(n \geq n_0\'\). If there are constants \(c'\) and \(n_0'\), such that the inequality holds, then \(6n^2 + 20n \in \Omega(n^3)\). However, it can be observed that no such constants can be found since \(n^3\) will always be greater than \(6n^2 + 20n\) for \(n \geq 1\). Therefore \(6n^2 + 20n\) is not in \(\Omega(n^3)\).

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Most popular questions from this chapter

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