The initial approach has a few issues, so let's work through the problem carefully to verify and correct it.

Given: $T(n) = 3T\left(\frac{n}{2}\right) + n \log n$ with $T(1) = 1$.

We will use the recurrence relation and the Master Theorem to solve this.

Step 1: Verify Applicability of the Master Theorem

For the recurrence relation $T(n) = a \, T\left(\frac{n}{b}\right) + f(n)$, we have:

• $a = 3$,
• $b = 2$, and
• $f(n) = n \log n$.

1. Calculate $n^{\log_b a}$: $\log_b a = \log_2 3.$ Thus, $n^{\log_b a} = n^{\log_2 3} \approx n^{1.585}$.

2. Compare $f(n) = n \log n$ with $n^{\log_b a}$:

   • $n \log n$ grows more slowly than $n^{\log_2 3}$ because $\log_2 3 > 1$.
   • Therefore, $f(n) = O(n^{\log_2 3 - \epsilon})$ for some $\epsilon > 0$.

Since $f(n)$ grows more slowly than $n^{\log_b a}$, by the Master Theorem, the solution to this recurrence relation is: $T(n) = \Theta(n^{\log_2 3}).$

Final Answer

$T(n) = \Theta(n^{\log_2 3}),$ where $n^{\log_2 3} \approx n^{1.585}$.

This is the asymptotic solution to the recurrence relation.

Would you like a detailed breakdown of each calculation or have further questions?

Related Problems

1. Solve the recurrence $T(n) = 2T(n/2) + n$.
2. Find the solution for $T(n) = 4T(n/2) + n^2$.
3. Solve $T(n) = T(n/2) + n \log n$ using the Master Theorem.
4. Determine the solution to $T(n) = 3T(n/3) + n$.
5. Use the recursion tree method to solve $T(n) = 3T(n/4) + n$.

Tip

For divide-and-conquer recurrences, start by identifying $a$, $b$, and $f(n)$ terms to quickly assess whether the Master Theorem applies.