Let's solve the recurrence relation given and verify the solution.

The recurrence relation is: $T(n) = 3T\left(\frac{n}{2}\right) + n \log n$ with the base case $T(1) = 1$.

Step 1: Let $n = 2^k$

Since $n = 2^k$, we have $k = \log_2 n$. The recurrence can then be rewritten as: $T(2^k) = 3T\left(2^{k-1}\right) + 2^k \cdot k$ where $k$ replaces $\log n$ (since $\log_2 n = k$).

Step 2: Expand the Recurrence

Let's expand the recurrence relation by substituting recursively.

1. First expansion: $T(2^k) = 3T(2^{k-1}) + k \cdot 2^k$
2. Second expansion: $T(2^{k-1}) = 3T(2^{k-2}) + (k-1) \cdot 2^{k-1}$ Substituting this into the first expansion gives: $T(2^k) = 3\left(3T(2^{k-2}) + (k-1) \cdot 2^{k-1}\right) + k \cdot 2^k$ Simplifying: $= 3^2 T(2^{k-2}) + 3(k-1) \cdot 2^k + k \cdot 2^k$ $= 3^2 T(2^{k-2}) + \left(3(k-1) + k\right) \cdot 2^k$
3. Third expansion: $T(2^{k-2}) = 3T(2^{k-3}) + (k-2) \cdot 2^{k-2}$ Substituting: $T(2^k) = 3^3 T(2^{k-3}) + \left(3^2(k-2) + 3(k-1) + k\right) \cdot 2^k$

Step 3: General Pattern and Sum of Terms

After $j$ steps, we observe: $T(2^k) = 3^j T(2^{k-j}) + 2^k \sum_{i=0}^{j-1} 3^i (k - i)$ When $j = k$, we reach the base case $T(2^{k-k}) = T(1) = 1$: $T(2^k) = 3^k \cdot T(1) + 2^k \sum_{i=0}^{k-1} 3^i (k - i)$ Since $T(1) = 1$, this simplifies to: $T(2^k) = 3^k + 2^k \sum_{i=0}^{k-1} 3^i (k - i)$

Step 4: Approximate the Summation

The summation $\sum_{i=0}^{k-1} 3^i (k - i)$ is complex, but it grows asymptotically as $O(3^k)$. Therefore: $T(n) = O(n^{\log_2 3}) = O(n^{1.585})$

Final Answer

So, the asymptotic solution to the recurrence relation is: $T(n) = \Theta(n^{\log_2 3}) \approx \Theta(n^{1.585})$

This corrects your initial approach. If you would like further details on the summation step, let me know.

Follow-Up Questions

1. How can the Master Theorem be applied to verify this solution?
2. What is the approximate value of $\log_2 3$ and why is it significant?
3. How would the recurrence change if the term $n \log n$ were replaced with $n$?
4. What does the solution imply about the time complexity in practical scenarios?
5. Can this recurrence relation be solved by substitution method?

Tip:

When solving recurrence relations, the Master Theorem is a useful shortcut if the recurrence fits its format.