f(n) = 2^(2n)Which of the following correctly represents the above function?Select one:O(2^n)Ω(2^n)Θ(2^n)None of these

Break Down the Problem

1. We are given the function $f(n) = 2^{2n}$.
2. We need to determine the growth rate of this function in relation to the options provided: $O(2^n)$, $\Omega(2^n)$, $\Theta(2^n)$, or "None of these".

Relevant Concepts

1. Big O Notation: It describes an upper bound on the time complexity.
2. Big Omega Notation: It provides a lower bound.
3. Big Theta Notation: It characterizes a tight bound, meaning both upper and lower bounds are equivalent.

Analysis and Detail

1. Function Transformation:

   • We can rewrite $f(n) = 2^{2n}$ as $f(n) = (2^n)^2$.
   • This indicates that $f(n)$ grows much faster than $2^n$.

2. Evaluating the Options:

   • Big O: $f(n) = 2^{2n}$ is not $O(2^n)$ because as $n \to \infty$, $f(n)$ grows faster than $2^n$.
   • Big Omega: $f(n)$ is $\Omega(2^n)$ since it serves as a lower bound, but we need to check if it is a tight bound.
   • Big Theta: $f(n)$ cannot be $\Theta(2^n)$ because it does not have both upper and lower bounds equal to $2^n$.

Verify and Summarize

Since $f(n) = 2^{2n}$ grows faster than $2^n$ and also satisfies that it grows lower than or equal to a function that grows similarly or faster (like $2^{2n}$), we deduce that:

• $f(n) = \Theta(2^{2n})$ but not $\Theta(2^n)$.
• Thus, $f(n)$ is $\Omega(2^n)$.

Final Answer

The correct representation of the function $f(n) = 2^{2n}$ is $\Omega(2^n)$.