To solve this asymptotic notation problem, we need to prove the following statement:

Statement:
Let $f(n) = n^4 + 6n^2 + 5$. Then, for all integers $k$, $f(n) = O(n^k)$ if and only if $k \geq 4$.

This is an "if and only if" statement, which means we need to prove it in two parts:

1. (If): Show that $f(n) = O(n^k)$ when $k \geq 4$.
2. (Only if): Show that $f(n) \neq O(n^k)$ when $k < 4$.

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Step 1: Prove $f(n) = O(n^k)$ when $k \geq 4$

We start by analyzing the function $f(n) = n^4 + 6n^2 + 5$.

Case 1: $k = 4$

To show that $f(n) = O(n^4)$, we need to find constants $c > 0$ and $n_0 \geq 1$ such that for all $n \geq n_0$: $f(n) \leq c n^4$

We have: $f(n) = n^4 + 6n^2 + 5$ For large $n$, the dominant term in $f(n)$ is $n^4$, while the other terms $6n^2$ and 5 are smaller in comparison.

For sufficiently large $n$, we know that: $n^4 + 6n^2 + 5 \leq n^4 + n^4 + n^4 = 3n^4$ Thus, for large $n$, we can choose $c = 3$ and $n_0 = 1$, so: $f(n) \leq 3n^4 \quad \text{for all } n \geq 1$ Therefore, $f(n) = O(n^4)$.

Case 2: $k > 4$

When $k > 4$, we need to show that $f(n) = O(n^k)$. Since $n^k$ grows faster than $n^4$ for large $n$, we know that: $f(n) = n^4 + 6n^2 + 5 \leq n^k \quad \text{for sufficiently large } n$ For large $n$, $n^4$ is much smaller than $n^k$. So, for sufficiently large $n$, the inequality holds. Thus, $f(n) = O(n^k)$ when $k > 4$.

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Step 2: Prove $f(n) \neq O(n^k)$ when $k < 4$

Now, we need to show that $f(n) \neq O(n^k)$ when $k < 4$. The main idea here is that the highest degree term in $f(n)$ is $n^4$, and if we try to bound it by $n^k$ for $k < 4$, the bound will not hold for large $n$.

Case 1: $k = 3$

Suppose $f(n) = O(n^3)$. This would mean there exists a constant $c > 0$ and $n_0 \geq 1$ such that: $f(n) \leq c n^3 \quad \text{for all } n \geq n_0$ However, for large $n$, the dominant term in $f(n)$ is $n^4$, and we have: $f(n) = n^4 + 6n^2 + 5$ Since $n^4$ grows faster than $n^3$, it is impossible to find such a constant $c$ such that $n^4 \leq c n^3$ for large $n$. Therefore, $f(n) \neq O(n^3)$.

Case 2: $k < 3$

If $k < 3$, the argument is even stronger. The term $n^4$ will dominate even more significantly, and no constant $c$ will satisfy $f(n) \leq c n^k$ for large $n$. Hence, $f(n) \neq O(n^k)$ when $k < 4$.

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Conclusion

• We have shown that $f(n) = O(n^k)$ for $k \geq 4$.
• We have also shown that $f(n) \neq O(n^k)$ for $k < 4$.

Thus, $f(n) = O(n^k)$ if and only if $k \geq 4$, completing the proof.

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Do you want further clarifications on any step or details? Here are five related questions you might consider:

1. How does the behavior of asymptotic notation change when $f(n)$ has logarithmic terms?
2. How can we use limits to formally define Big-O and prove the bounds?
3. How would the proof change if $f(n)$ had higher-degree terms like $n^5$?
4. Can we provide a visual comparison of $f(n)$ with $n^k$ for different $k$?
5. What are the common misunderstandings with asymptotic notation in algorithm analysis?

Tip: When using Big-O notation, always focus on the term with the highest growth rate, as it dominates the behavior of the function for large $n$.