Question

Using the Properties of Order in Section \(1.4 .2,\) show that \\[ 5 n^{5}+4 n^{4}+6 n^{3}+2 n^{2}+n+7 \in \Theta\left(n^{5}\right) \\]

Short answer

The function \(5n^{5} + 4n^{4} + 6n^{3} + 2n^{2} + n + 7\) is indeed in the complexity class Theta(n^5) as it was shown to be both \(O(n^{5})\) and \(Omega(n^{5})\).

Step-by-step solution

Prove Big O(n^5)

We start by proving the function is in Big O of \(n^{5}\), i.e., f(n) is less than or equal to \(c \cdot n^{5}\) for all \(n \geq n_0\) where c > 0 and \(n_0 \geq 1\) are constants. To show this, consider the function \(f(n) = 5n^{5} + 4n^{4} + 6n^{3} + 2n^{2} + n + 7\). One can note that: For \(n \geq 1\), this function is less than or equal to \(5n^{5} + 4n^{5} + 6n^{5} + 2n^{5} + n^{5} + 7n^{5}\), which equals to \(25n^{5}\). Therefore we can say that \(f(n) = O(n^{5})\)

Prove Big Omega(n^5)

We also need to show that the function is in Big Omega of \(n^{5}\), i.e. f(n) is more than or equal to \(c \cdot n^{5}\) for all \(n geq n_0\) where c > 0 and \(n_0 \geq 1\) are constants. Consider the function \(f(n) = 5n^{5} + 4n^{4} + 6n^{3} + 2n^{2} + n + 7\). One can note that: For \(n \geq 1\), this function is more than or equal to \(5n^{5}\) which is indeed bigger than or equal to \(c \cdot n^{5}\) when \(c = 5\). Therefore we can say \(f(n) = \Omega(n^{5})\)

Conclusion

Since the function \(f(n) = 5n^{5} + 4n^{4} + 6n^{3} + 2n^{2} + n + 7\) is both \(O(n^{5})\) and \(Omega(n^{5})\), we can conclude that \(f(n) \in Theta(n^{5})\), meaning that the function grows at the same rate as \(n^{5}\) in the limit as n approaches infinity.