Question

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

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

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 \(O(n^3)\).

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 \(\Omega(n^3)\).

Key concepts

Big O Notation

Big O Notation is a mathematical way to describe the upper bound of an algorithm's running time or space requirements in terms of the size of the input, often denoted as \(n\). It gives us an idea of the worst-case scenario for an algorithm's performance.

When we say a function \(f(n)\) is in \(O(g(n))\), it implies that there exists a constant \(c\) and a threshold \(n_0\), such that for every \(n\) greater than \(n_0\), \(f(n)\) does not grow faster than \(g(n)\) when scaled by \(c\).

• The formula for Big O is: \(f(n) \leq c \cdot g(n)\) for all \(n \geq n_0\).
• Big O provides an upper limit on the growth of the function.

To apply Big O to the example \(6n^2 + 20n \in O(n^3)\), you choose a constant such as \(c = 26\) to show that \(6n^2 + 20n \leq 26n^3\) holds for \(n \geq 1\). Thus, proving this expression is in \(O(n^3)\). This highlights how polynomial terms at lower power don't affect Big O as they become insignificant compared to the higher power terms as \(n\) grows.

Big Omega Notation

Big Omega Notation is another mathematical concept used to describe the lower bound of an algorithm's running time or space requirements. It helps us to understand the best-case performance or the minimum amount of resources the algorithm needs.

For a function \(f(n)\) to be in \(\Omega(g(n))\), there must be a constant \(c'\) and a threshold \(n_0'\), so that \(f(n)\) does not shrink faster than \(g(n)\) when scaled down by \(c'\) for sufficiently large \(n\).

• The formula for Big Omega is: \(c' \cdot g(n) \leq f(n)\) for all \(n \geq n_0'\).
• Big Omega provides a lower limit on the growth of the function.

In the exercise, to show \(6n^2 + 20n otin \Omega(n^3)\), you note that no constant \(c'\) can satisfy \(c' \cdot n^3 \leq 6n^2 + 20n\). Even the smallest positive constant will cause \(c' \times n^3\) to surpass \(6n^2 + 20n\), especially as \(n\) becomes larger than 1. This implies the polynomial \(6n^2 + 20n\) does not grow quick enough to be bounded below by \(n^3\), thus it's not in \(\Omega(n^3)\).

Mathematical Proofs

Mathematical proofs are formal demonstrations showing the truth of a given statement. They are essential for verifying claims in mathematics, ensuring that assertions about algorithms and functions are logically sound.

Typically, a proof involves the following components:

• A starting point, based on assumptions or known truths.
• A series of logical steps, applying known definitions and theorems.
• A conclusion that directly follows from the logical steps.

In the context of Big O and Big Omega, proofs are used to affirm or refute these relationships.

To prove \(6n^2 + 20n \in O(n^3)\), you chose a constant \(c = 26\) and showed that \(6n^2 + 20n \leq 26n^3\) for all \(n \geq 1\). This required checking the inequality holds for increasing \(n\).

To show \(6n^2 + 20n otin \Omega(n^3)\), you attempted to find a \(c'\) that fits the opposite inequality but failed, as the inequality \(c' \cdot n^3 \leq 6n^2 + 20n\) could not be satisfied, proving the statement false using contradiction. Understanding these proof techniques helps solidify the grasp of algorithm performance and efficiency.