Let's solve the recurrence relation $T(n) = 3T(n/2) + n \log n$ with the base case $T(1) = 1$ using the Master Theorem.

Step 1: Identify the Components

The recurrence has the form:

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

where:

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

Step 2: Calculate $n^{\log_b a}$

We first compute $\log_b a$:

$\log_b a = \log_2 3$

This means:

$n^{\log_b a} = n^{\log_2 3}$

Step 3: Compare $f(n)$ with $n^{\log_b a}$

Next, we need to compare $f(n) = n \log n$ with $n^{\log_2 3}$:

1. Finding $\log_2 3$: Since $\log_2 3$ is approximately $1.585$, we have:

$n^{\log_2 3} = n^{1.585}$

2. Comparing $n \log n$ and $n^{1.585}$:

We see that:

$f(n) = n \log n = o(n^{\log_2 3}) \quad \text{as } n \to \infty$

This means $f(n)$ grows slower than $n^{\log_b a}$.

Step 4: Apply the Master Theorem

Since $f(n)$ is polynomially smaller than $n^{\log_b a}$:

$f(n) = o\left(n^{\log_b a - \epsilon}\right) \quad \text{for some } \epsilon > 0$

We can apply Case 1 of the Master Theorem:

$T(n) = \Theta(n^{\log_2 3})$

Step 5: Final Result

Thus, we conclude:

$T(n) = \Theta(n^{\log_2 3}) = \Theta(n^{1.585})$

Summary

Your approach started correctly with the recurrence relation but had some errors in the derivation of constants and final expressions. The correct solution is:

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

Would you like more details on any specific part of the solution? Here are some related questions:

1. What are the implications of the result for algorithms using this recurrence?
2. How would you solve a similar recurrence with different coefficients?
3. Can you explain the different cases of the Master Theorem?
4. How does the $f(n)$ term affect the overall complexity?
5. What would happen if $f(n)$ were instead $n^2$?