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.

Key concepts

Big O Notation

Big O Notation, often symbolized as \(O(g(n))\), describes an upper bound on the growth rate of a function. This notation is essential for understanding how algorithms perform as their input size increases.

### Basics of Big O- **Purpose**: To capture the worst-case scenario for algorithm growth.- **Usage**: We say a function \(f(n)\), such as our example \(5n^{5} + 4n^{4} + 6n^{3} + 2n^{2} + n + 7\), is \(O(n^5)\) if it doesn't grow faster than \(kn^5\) for some constant \(k\) and sufficiently large \(n\). In simple terms, it shows that, beyond a certain point, \(f(n)\) will not exceed a constant multiple of \(n^5\) in terms of its growth.

### Example DemonstrationTo verify this, consider the given polynomial. We hypothesized that it is dominated by the term \(5n^5\) because for \(n \geq 1\), adding the lesser terms still results in \(25n^5\) being an upper bound. Thus, \(f(n) = O(n^5)\) as shown in Step 1 of the solution.

Big Omega Notation

Big Omega Notation, represented as \(\Omega(g(n))\), provides a lower bound. This concept tells us at the very least, how fast a function will grow.

### Understanding Big Omega- **Purpose**: To show the best-case scenario for a function's growth rate.- **Usage**: For any function \(f(n)\), it is said to be \(\Omega(g(n))\) if there exists a positive constant \(c\) and a value \(n_0\) such that \(f(n) \geq c\cdot g(n)\) for all \(n \geq n_0\).

### Application on ExampleFor our function \(f(n) = 5n^{5} + 4n^{4} + 6n^{3} + 2n^{2} + n + 7\), it clearly cannot grow slower than \(5n^5\) for \(n \geq 1\). Even after accounting for lower degree terms, the leading term ensures that the growth is at least as quick as \(c \cdot n^5\), making \(f(n) = \Omega(n^5)\) as shown in Step 2 of the solution.

Theta Notation

Theta Notation, denoted as \(\Theta(g(n))\), represents functions that are tightly bound both from above and below by \(g(n)\). This means the function grows at exactly the rate of \(g(n)\).

### Conceptual Overview- **Purpose**: To indicate precise asymptotic behavior.- **Usage**: A function \(f(n)\) is in \(\Theta(g(n))\) if and only if it is both \(O(g(n))\) and \(\Omega(g(n))\). In essence, there exist constants \(c_1\), \(c_2\), and \(n_0\) such that \(c_1\cdot g(n) \leq f(n) \leq c_2\cdot g(n)\) for all \(n \geq n_0\).

### Conclusion from ExerciseOur polynomial \(f(n) = 5n^{5} + 4n^{4} + 6n^{3} + 2n^{2} + n + 7\) met the criteria for both Big O and Big Omega for \(n^5\). Thus, it qualifies for \(\Theta(n^5)\) implying it strikes a balance between upper and lower bounds, as confirmed in the conclusion of the solution.