To show that $n^2 + 100n + 4 \in O(n^3)$, we need to prove that there exist constants $c > 0$ and $n_0 \geq 0$ such that for all $n \geq n_0$, the inequality

$n^2 + 100n + 4 \leq c \cdot n^3$

holds true. This is the formal definition of Big-O notation, which indicates that the function $n^2 + 100n + 4$ grows asymptotically no faster than $n^3$.

Step-by-Step Proof

1. Consider the expression:

   $n^2 + 100n + 4$

2. Find an upper bound for $n^2 + 100n + 4$ in terms of $n^3$:

   Notice that $n^2, 100n,$ and $4$ are all smaller than $n^3$ for sufficiently large $n$. Thus, we can express the sum of these terms in terms of $n^3$:

   $n^2 \leq n^3 \quad \text{for } n \geq 1$ $100n \leq 100n^2 \leq 100n^3 \quad \text{for } n \geq 1$ $4 \leq 4n^3 \quad \text{for } n \geq 1$

3. Combine these inequalities:

   $n^2 + 100n + 4 \leq n^3 + 100n^3 + 4n^3 = (1 + 100 + 4)n^3 = 105n^3$

4. Choose constants $c$ and $n_0$:

   Here, we can choose $c = 105$ and $n_0 = 1$.

5. Verify the inequality:

   For all $n \geq 1$:

   $n^2 + 100n + 4 \leq 105n^3$

Thus, we have shown that for $c = 105$ and $n_0 = 1$, the function $n^2 + 100n + 4 \leq c \cdot n^3$ for all $n \geq n_0$.

Conclusion

Since we found constants $c = 105$ and $n_0 = 1$ such that the inequality $n^2 + 100n + 4 \leq 105n^3$ holds for all $n \geq 1$, we have demonstrated that $n^2 + 100n + 4 \in O(n^3)$.

Would you like more details on this proof or have any questions?

Here are 5 related questions that expand upon this information:

1. How would you prove that $n^2 + 100n + 4 \in O(n^2)$?
2. Can we find a smaller constant $c$ that works for $n^2 + 100n + 4 \in O(n^3)$?
3. What is the difference between Big-O, Big-Theta, and Big-Omega notations?
4. How can we prove $n^2 + 100n + 4 \notin O(n)$?
5. How would the proof change if we needed to show $n^2 + 100n + 4 \in O(n^2 \log n)$?

Tip: When proving Big-O notation, always look for the dominant term in the expression, as it often determines the simplest upper bound.