reconsider comparison n^logba with n log^2 n 

Let's revisit the comparison between \( n^{\log_b a} \) and \( n \log^2 n \) more carefully to apply the Master Theorem correctly.

### Given Recurrence:
\[
T(n) = 9T\left(\frac{n}{6}\right) + n \log^2 n
\]

We have:
- \( a = 9 \)
- \( b = 6 \)
- \( f(n) = n \log^2 n \)

### Step 1: Calculate \( n^{\log_b a} \)
First, we compute \( \log_b a \). As previously computed:
\[
\log_b a = \log_6 9 = \frac{\log 9}{\log 6} \approx 1.226
\]
Thus, \( n^{\log_b a} = n^{1.226} \).

### Step 2: Compare \( f(n) = n \log^2 n \) with \( n^{\log_b a} = n^{1.226} \)

Now, we compare the growth of \( f(n) = n \log^2 n \) with \( n^{1.226} \).

- \( f(n) = n \log^2 n \) grows **slower** than \( n^{1.226} \), because logarithmic factors grow much more slowly than polynomial terms. Specifically, \( n^{1.226} \) grows faster than \( n \log^2 n \) for large \( n \).

### Step 3: Apply the Master Theorem

#### Case 1: \( f(n) = O(n^{\log_b a - \epsilon}) \)
This case applies when \( f(n) \) is polynomially smaller than \( n^{\log_b a} \). Here, \( f(n) = n \log^2 n \) grows more slowly than \( n^{1.226} \), but it is not polynomially smaller. Therefore, **Case 1 does not apply**.

#### Case 2: \( f(n) = \Theta(n^{\log_b a} \log^k n) \)
This case applies when \( f(n) \) matches the growth rate of \( n^{\log_b a} \) up to a logarithmic factor. However, in this case, \( f(n) \) is **not** of the form \( n^{1.226} \log^k n \); it's strictly slower. Therefore, **Case 2 does not apply** either.

#### Case 3: \( f(n) = \Omega(n^{\log_b a + \epsilon}) \)
This case applies when \( f(n) \) grows faster than \( n^{\log_b a} \). Clearly, \( f(n) = n \log^2 n \) grows **slower** than \( n^{1.226} \), so **Case 3 does not apply**.

### Correct Conclusion

Given that \( f(n) = n \log^2 n \) grows slower than \( n^{\log_b a} \), but not by a polynomial factor, we actually fall into **Case 1 of the Master Theorem**, contrary to the previous conclusion. Therefore:

\[
T(n) = \Theta(n^{\log_b a}) = \Theta(n^{1.226})
\]

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

Would you like further clarifications on any step of this process?

---

Here are 5 related questions:
1. Why does \( f(n) = n \log^2 n \) grow slower than \( n^{1.226} \)?
2. Can logarithmic terms affect the application of the Master Theorem?
3. How do we determine which case of the Master Theorem applies?
4. What is the significance of the \( \log_b a \) term in the Master Theorem?
5. How does the growth of logarithmic functions compare with polynomial functions?

**Tip**: In divide-and-conquer recurrences, logarithmic factors often make the recurrence behave slower than pure polynomial terms, affecting the application of the Master Theorem.

Explore the comparison between n^log_b a and n log^2 n using the Master Theorem. Understand how logarithmic factors affect divide-and-conquer recurrences. Suitable for advanced mathematics enthusiasts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation