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Assume f(n) and g(n) are two functions such that f(n) = O (g(n)). Which of the following will always hold?

Let's analyze each statement given the condition \( f(n) = O(g(n)) \). 

### Given Condition:
- \( f(n) = O(g(n)) \) means that \( f(n) \) grows at most as fast as \( g(n) \), or that there exists a constant \( C > 0 \) and \( n_0 \) such that \( f(n) \leq C \cdot g(n) \) for all \( n \geq n_0 \).

### Statement 1: \( f(n) = o\left((f(n))^2\right) \)

- Small-o notation \( o\left((f(n))^2\right) \) means that \( f(n) \) grows strictly slower than \( (f(n))^2 \).
- \( f(n) \) cannot be strictly less than \( (f(n))^2 \) unless \( f(n) \) is a constant or tends to zero, which would make the left-hand side negligible compared to the right-hand side.
- In general, this relationship does not hold for arbitrary \( f(n) \) (e.g., if \( f(n) = n \), \( n = o(n^2) \) is true, but it is not universally true for all functions).

**Conclusion**: This statement **does not always hold**.

### Statement 2: \( f(n) = \Omega\left((f(n))^2\right) \)

- Big-Omega notation \( \Omega\left((f(n))^2\right) \) means that \( f(n) \) grows at least as fast as \( (f(n))^2 \).
- This cannot be true because \( (f(n))^2 \) grows faster than \( f(n) \) for sufficiently large \( n \), except in trivial cases like when \( f(n) \) is a constant.

**Conclusion**: This statement **does not always hold**.

### Statement 3: \( g(n) = o\left((f(n))^2\right) \)

- Small-o notation \( o\left((f(n))^2\right) \) means that \( g(n) \) grows strictly slower than \( (f(n))^2 \).
- Given that \( f(n) = O(g(n)) \), this implies \( (f(n))^2 \) is \( O\left((g(n))^2\right) \). However, \( g(n) = o\left((f(n))^2\right) \) suggests \( g(n) \) is strictly less than \( (f(n))^2 \).
- There is no guarantee that this condition will always hold. For example, if \( f(n) = g(n) \), then \( g(n) \) cannot grow slower than \( (f(n))^2 \).

**Conclusion**: This statement **does not always hold**.

### Statement 4: \( g(n) = \Omega(g(n)) \)

- Big-Omega notation \( \Omega(g(n)) \) means that \( g(n) \) grows at least as fast as \( g(n) \).
- This is trivially true because any function \( g(n) \) is trivially bounded below by itself.

**Conclusion**: This statement **always holds**.

### Final Answer:
- The statement that will **always** hold is **Statement 4**: \( g(n) = \Omega(g(n)) \).

Would you like more details or have any further questions?

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Here are some related questions that could deepen your understanding:

1. What conditions must be met for a function to be in \( O(g(n)) \)?
2. Can you provide an example where \( f(n) = o(g(n)) \)?
3. How does \( f(n) = \Theta(g(n)) \) differ from \( f(n) = O(g(n)) \)?
4. Why does statement 4 always hold regardless of \( f(n) \) or \( g(n) \)?
5. Can two different functions \( f(n) \) and \( g(n) \) both be \( O(n^2) \) and still differ in their growth rates?
6. What role do constants play in Big-O and Big-Omega notations?
7. How can we interpret \( f(n) = o(g(n)) \) in terms of algorithm efficiency?
8. What are the implications of \( g(n) = \Omega(g(n)) \) in algorithm analysis?

**Tip**: When evaluating asymptotic statements, consider the general properties of the functions involved and how the notation affects the relationships.

Explore the implications of f(n) = O(g(n)) in asymptotic notations. Learn why certain statements involving f(n) and g(n) hold or do not hold under this condition. Gain insights into mathematical relationships between functions in algorithm analysis.

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