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The recurrence relation given is:

$T(n) = 2T\left(\frac{n}{2}\right) + n^3$

To solve this recurrence and determine whether $T(n)$ is $O(n^3)$, we will apply the Master Theorem for divide-and-conquer recurrences. The general form of such recurrences is:

$T(n) = aT\left(\frac{n}{b}\right) + f(n)$

In your case, comparing it with the standard form, we have:

• $a = 2$
• $b = 2$
• $f(n) = n^3$

Now, we check the form of $f(n)$ relative to $n^{\log_b{a}}$, where $b = 2$ and $a = 2$.

First, calculate $\log_b a$: $\log_b a = \log_2 2 = 1$

So, $n^{\log_b a} = n^1$.

Now we compare $f(n) = n^3$ to $n^{\log_b a} = n^1$:

• Since $f(n) = n^3$ grows faster than $n^1$, we are in Case 3 of the Master Theorem, where $f(n)$ is polynomially larger than $n^{\log_b a}$.

According to Case 3 of the Master Theorem, if $f(n) = \Omega(n^{\log_b a + \epsilon})$ for some constant $\epsilon > 0$, then $T(n) = O(f(n))$.

In this case, $f(n) = n^3$ satisfies the condition $f(n) = \Omega(n^{1 + \epsilon})$ for any $\epsilon$ where $\epsilon \leq 2$. Therefore, we conclude:

$T(n) = O(n^3)$

Summary:

The recurrence $T(n) = 2T(n/2) + n^3$ solves to $T(n) = O(n^3)$ according to the Master Theorem, because $f(n) = n^3$ grows faster than $n^1$, satisfying Case 3 of the Master Theorem.

Would you like a deeper explanation or have further questions on the topic?

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Here are 5 questions to help expand on this topic:

1. How do you apply the Master Theorem to different forms of recurrence relations?
2. What happens if $f(n)$ is smaller than $n^{\log_b a}$?
3. Can you derive the recurrence for the merge sort algorithm using the Master Theorem?
4. How do you deal with recurrences that do not fit the form of the Master Theorem?
5. What is the significance of $\epsilon$ in Case 3 of the Master Theorem?

Tip: When applying the Master Theorem, always make sure the recurrence is in the correct form, and ensure that the conditions for each case are met before concluding the time complexity.