Solve T(n) = 9T(n/6) + n log^2
 n, by master method. (10 marks)

Let's solve the recurrence relation \( T(n) = 9T\left(\frac{n}{6}\right) + n \log^2 n \) using the **Master Theorem** for divide-and-conquer recurrences.

The recurrence is of the form:
\[
T(n) = aT\left(\frac{n}{b}\right) + f(n)
\]
Where:
- \( a = 9 \)
- \( b = 6 \)
- \( f(n) = n \log^2 n \)

To apply the Master Theorem, we compare \( f(n) \) with \( n^{\log_b a} \).

### Step 1: Calculate \( \log_b a \)
We need to compute \( \log_b a = \log_6 9 \). Using the change of base formula:
\[
\log_6 9 = \frac{\log 9}{\log 6}
\]
We approximate:
\[
\log 9 \approx 0.9542, \quad \log 6 \approx 0.7782
\]
Thus:
\[
\log_6 9 \approx \frac{0.9542}{0.7782} \approx 1.226
\]
So, \( n^{\log_6 9} \approx n^{1.226} \).

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

We now compare the asymptotic behavior of \( f(n) = n \log^2 n \) with \( n^{\log_6 9} \):
- \( f(n) = n \log^2 n \) grows slightly slower than \( n^{1.226} \), because \( n \log^2 n \) is \( n \) times a polylogarithmic term, while \( n^{1.226} \) has a strictly larger exponent of \( n \).

### Step 3: Apply Master Theorem

The Master Theorem cases are:
1. **Case 1**: If \( f(n) = O(n^{\log_b a - \epsilon}) \) for some \( \epsilon > 0 \), then \( T(n) = \Theta(n^{\log_b a}) \).
2. **Case 2**: If \( f(n) = \Theta(n^{\log_b a} \log^k n) \) for some \( k \geq 0 \), then \( T(n) = \Theta(n^{\log_b a} \log^{k+1} n) \).
3. **Case 3**: If \( f(n) = \Omega(n^{\log_b a + \epsilon}) \) for some \( \epsilon > 0 \) and if \( a f\left(\frac{n}{b}\right) \leq c f(n) \) for some constant \( c < 1 \) and sufficiently large \( n \), then \( T(n) = \Theta(f(n)) \).

In this case, \( f(n) = n \log^2 n \) is asymptotically smaller than \( n^{\log_6 9} \) but not polynomially smaller. Therefore, we are in **Case 2** of the Master Theorem. Hence, the solution to the recurrence is:

\[
T(n) = \Theta(n^{\log_6 9} \log^3 n)
\]

### Final Answer:
The solution to the recurrence \( T(n) = 9T(n/6) + n \log^2 n \) by the Master Method is:
\[
T(n) = \Theta(n^{1.226} \log^3 n)
\]

Would you like any additional details or clarifications?

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Here are 5 related questions:
1. How does the Master Theorem handle logarithmic factors in \( f(n) \)?
2. What conditions are needed for each case of the Master Theorem to apply?
3. How do we handle non-standard functions like \( \log^2 n \) in recurrence relations?
4. Can the Master Theorem be applied to all recurrences?
5. How does the change of base formula help in solving recurrences with non-integer bases?

**Tip**: When solving recurrence relations, carefully compare the growth rates of the terms to choose the correct case of the Master Theorem.

Learn how to solve the recurrence relation T(n) = 9T(n/6) + n log^2 n using the Master Theorem for divide-and-conquer recurrences. Understand the application of the Master Theorem and its relevance in asymptotic analysis. Suitable for advanced undergraduate students and those studying algorithms and asymptotic complexity.

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