To solve the recurrence relation $T(n) = 9T\left(\frac{n}{6}\right) + n \log^2 n$ using the Master Theorem, we will first analyze the related simpler recurrence $T(n) = 9T\left(\frac{n}{6}\right) + n$.

Step 1: Compare with the Standard Form

The standard form of a divide-and-conquer recurrence is: $T(n) = aT\left(\frac{n}{b}\right) + f(n)$ For the given recurrence $T(n) = 9T\left(\frac{n}{6}\right) + n$, we have:

• $a = 9$
• $b = 6$
• $f(n) = n$

Step 2: Calculate the Critical Exponent $\log_b a$

The next step is to calculate the value of $\log_b a$, where $b = 6$ and $a = 9$. $\log_b a = \log_6 9$ This can be simplified using the change of base formula: $\log_6 9 = \frac{\log 9}{\log 6} = \frac{\log(3^2)}{\log(2 \cdot 3)} = \frac{2\log 3}{\log 2 + \log 3}$

Step 3: Analyze $f(n) = n$

Next, we compare $f(n) = n$ to $n^{\log_b a}$.

We know that $\log_6 9 > 1$ because $\log_6 6 = 1$ and $\log_6 9 = \log_6 (6 \times \frac{3}{2})$ will yield a value greater than 1. Therefore, $n^{\log_b a}$ grows faster than $f(n) = n$.

Step 4: Apply the Master Theorem

In the context of the Master Theorem, there are three cases to consider:

1. Case 1: If $f(n) = O\left(n^{\log_b a - \epsilon}\right)$ for some $\epsilon > 0$, then $T(n) = \Theta\left(n^{\log_b a}\right)$.
2. Case 2: If $f(n) = \Theta\left(n^{\log_b a}\right)$, then $T(n) = \Theta\left(n^{\log_b a} \log n \right)$.
3. Case 3: If $f(n) = \Omega\left(n^{\log_b a + \epsilon}\right)$ for some $\epsilon > 0$ and if $a f\left(\frac{n}{b}\right) \leq c f(n)$ for some constant $c < 1$, then $T(n) = \Theta\left(f(n)\right)$.

For our case:

• Since $f(n) = n$ and $n^{\log_b a}$ grows faster than $n$, this fits Case 1.

Thus, $T(n) = \Theta\left(n^{\log_b a}\right)$.

Step 5: Consider the Original Recurrence

Now, let's consider the original recurrence $T(n) = 9T\left(\frac{n}{6}\right) + n \log^2 n$. Here, $f(n) = n \log^2 n$.

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

• Case 1: If $f(n) = O\left(n^{\log_b a - \epsilon}\right)$, $T(n) = \Theta(n^{\log_b a})$.
• Case 2: If $f(n) = \Theta\left(n^{\log_b a}\right)$, $T(n) = \Theta\left(n^{\log_b a} \log n \right)$.
• Case 3: If $f(n) = \Omega\left(n^{\log_b a + \epsilon}\right)$, $T(n) = \Theta(f(n))$.

In this case, $f(n) = n \log^2 n$ grows faster than $n^{\log_b a}$, so we are in Case 3.

Final Solution

Therefore, for the given recurrence $T(n) = 9T\left(\frac{n}{6}\right) + n \log^2 n$, the solution is: $T(n) = \Theta\left(n \log^2 n\right)$

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

1. What is the significance of the base in logarithmic functions?
2. How does the Master Theorem apply to non-divide-and-conquer algorithms?
3. Can the Master Theorem be used for recurrences with additive constants?
4. How do we handle recurrences that don’t fit into the Master Theorem cases?
5. What are other methods to solve recurrences besides the Master Theorem?

Tip: Always check the conditions for the Master Theorem carefully; some recurrences may require different methods if they don’t strictly fit the form.