Problem 13 - Discrete Mathematics with Applications

Question

Verify each. $$2^{n}=\mathbf{O}(n !)$$

Step-by-Step Solution

Answer

We have verified that \(2^n = \mathbf{O}(n!)\) by mathematical induction. By choosing the constant \(c = 1\) and \(n_0 = 4\), we demonstrated that the inequality \(2^n \leq c * n!\) holds for all \(n \geq n_0\).

Step 1: Find inequality to be proved

The goal is to find values for constant c and \(n_0\) such that the inequality \(2^n \leq c * n!\) is true for all n greater than or equal to \(n_0\).

Step 2: Check examples for values of n greater than or equal to some threshold

It is useful to look at some examples to get a feel for the inequality: For \(n = 1\), \(2^1 = 2\) and \(1! = 1\). Since \(2 > 1\), we need to find a constant c that satisfies the inequality. For \(n = 2\), \(2^2 = 4\) and \(2! = 2\). Since \(4 > 2\), we still need a constant c that satisfies the inequality. For \(n = 3\), \(2^3 = 8\) and \(3! = 6\). The inequality is not satisfied here, either. For \(n = 4\), \(2^4 = 16\) and \(4! = 24\). The inequality is now satisfied without using any constant c. To make it more clear, let's examine values for n greater than 5. For \(n = 5\), \(2^5 = 32\) and \(5! = 120\). The inequality is satisfied. For \(n = 6\), \(2^6 = 64\) and \(6! = 720\). The inequality is satisfied. From the provided examples, it seems like the inequality holds for \(n\geq 4\). Let's choose the constant c to be 1 and now we need to prove that the inequality holds for \(n \geq 4\).

Step 3: Prove the inequality for all n greater than or equal to 4

We have chosen the constants that we found in the examples, \(c=1\) and \(n_0 = 4\). Now we need to prove that the inequality \(2^n \leq n!\) holds for all \(n \geq n_0\). We will now show this using mathematical induction. Claim: \(2^n \leq n!\) is true for all integers \(n \geq 4\) Base Case: \(n = 4\). We have \(2^4 = 16\) and \(4! = 24\). Since \(16 \leq 24\), the claim holds for \(n = 4\). Inductive Step: Let's assume the claim is true for \(n=k\), where \(k \geq 4\), i.e., \(2^k \leq k!\). Now we need to show that the claim is true for \(n=k+1\). \(2^{k+1} = 2 * 2^k\) Using the inductive hypothesis, we know that \(2^k \leq k!\). So, \(2 * 2^k \leq 2 * k!\) We need to show that \(2 * k! \leq (k+1)!\). Since \(k\geq 4\),giving us \(k+1 \geq 5 \Rightarrow 2\leq \frac{k+1}{2}\).So, \(2 * k! \leq (k+1) * k!\) This proves that our claim also holds for \(n=k+1\). Therefore, by mathematical induction, the inequality \(2^n \leq n!\) holds for all \(n \geq 4\).

Step 4: Conclusion

We proved that for \(c=1\) and \(n_0 = 4\), the inequality \(2^n \leq c * n!\) holds for all \(n \geq n_0\). Hence, the given statement \(2^n=\mathbf{O}(n!)\) is verified.

Key Concepts

Big O Notation

Big O notation is a way to describe the upper bound or the worst-case scenario of an algorithm's complexity. It gives us insight into how the runtime of a program increases with the size of inputs. In our given exercise, we are asked to verify that \(2^n = \mathbf{O}(n!)\). This means we need to prove that the function \(2^n\) grows at most as fast as a constant multiple of \(n!\) as \(n\) becomes very large.

The reason big O notation is used is that it allows mathematicians and computer scientists to make generalized statements about growth rates without getting bogged down by smaller-detail computations. It's a tool to evaluate which algorithms might perform better in the context of large inputs.

• We search for constants \(c\) and \(n_0\) such that for all \(n \geq n_0\), the inequality \(f(n) \leq c \, g(n)\) holds.
• In our exercise, we found that \(c = 1\) and \(n_0 = 4\) satisfied \(2^n \leq n!\) for all \(n \geq 4\).

This provides the assurance that \(2^n\) is indeed \(\mathbf{O}(n!)\).

Factorial Growth

Factorial growth refers to a function that grows very rapidly, much faster than polynomial or exponential growth. The factorial of a number \(n!\) is the product of all positive integers up to \(n\), and it appears in many areas such as permutations and combinations.

In our exercise, \(n!\) was compared to \(2^n\). The exercise shows that as \(n\) becomes large enough, the factorial growth \(n!\) exceeds the exponential growth \(2^n\) for \(n \geq 4\). It's important to understand when working with algorithm complexity that while exponential growth is fast, factorial growth is even faster:

• For \(n = 5\), \(5! = 120\) while \(2^5 = 32\). Factorial dominance continues for higher \(n\).
• Understanding how quickly \(n!\) grows can help identify which algorithms are practically feasible for large input sizes.

Overall, factorial growth is significantly faster and illustrates the importance of considering it while analyzing algorithm complexity.

Mathematical Induction

Mathematical induction is a proof technique used to prove statements about integers. It works by proving a base case and then showing that if a statement holds for an integer \(k\), it must also hold for \(k+1\). This method is useful for demonstrating the truth of an infinite number of cases.

In the exercise, we used mathematical induction to prove that \(2^n \leq n!\) for all \(n \geq 4\). The process was as follows:

The base case was proved for \(n = 4\), demonstrating that the statement holds true. Then the inductive step assumed that \(2^k \leq k!\) is true and showed this implies \(2^{k+1} \leq (k+1)!\).

• The assumption, called the inductive hypothesis, helps in moving the proof from \(k\) to \(k+1\).
• Verification for a general case \(k+1\), ensures the statement holds for all integers starting from the base case.

This systematic approach reinforces not only the statement being proved but also the strength and utility of mathematical induction as a proof method.