do in subsitution, use Claim: cnlog(base6)9 - c'n(log^2)n

To solve the recurrence relation \( T(n) = 9T\left(\frac{n}{3}\right) + n\log^2{n} \) using the **substitution method**, we proceed as follows:

### Recurrence Relation
Given:
\[
T(n) = 9T\left(\frac{n}{3}\right) + n\log^2{n}
\]

### Hypothesis for Induction
We assume a solution of the form:
\[
T(n) \leq c n \log_6{9} - c' n (\log^2{n})
\]
for some constants \( c \) and \( c' \), and we need to verify this assumption.

### Base Case
First, check if this assumption holds for a small value of \( n \) (e.g., \( n = 1 \)). Since we're primarily interested in large \( n \), the base case is typically straightforward and involves setting \( n \) to the smallest value that makes sense in the context of the recurrence (usually \( n = 1 \) or \( n = 3 \)).

### Inductive Step
Assume that the hypothesis holds for some \( n \geq 1 \), i.e., 
\[
T\left(\frac{n}{3}\right) \leq c \cdot \frac{n}{3} \log_6{9} - c' \cdot \frac{n}{3} \log^2{\frac{n}{3}}
\]

Now substitute this into the recurrence relation:
\[
T(n) = 9T\left(\frac{n}{3}\right) + n\log^2{n}
\]
\[
\leq 9 \left( c \cdot \frac{n}{3} \log_6{9} - c' \cdot \frac{n}{3} \log^2{\frac{n}{3}} \right) + n\log^2{n}
\]
\[
= 3cn \log_6{9} - 3c'n \log^2{\frac{n}{3}} + n \log^2{n}
\]

### Simplification
Let's simplify \( \log^2{\frac{n}{3}} \) using logarithmic properties:
\[
\log{\frac{n}{3}} = \log{n} - \log{3}
\]
\[
\log^2{\frac{n}{3}} = (\log{n} - \log{3})^2 = \log^2{n} - 2\log{n}\log{3} + \log^2{3}
\]

Substitute back:
\[
T(n) \leq 3cn \log_6{9} - 3c'n (\log^2{n} - 2\log{n}\log{3} + \log^2{3}) + n \log^2{n}
\]
\[
= 3cn \log_6{9} - 3c'n \log^2{n} + 6c'n\log{n}\log{3} - 3c'n\log^2{3} + n \log^2{n}
\]

Combine like terms:
\[
T(n) \leq n\log^2{n}(1 - 3c') + 3cn \log_6{9} + 6c'n\log{n}\log{3} - 3c'n\log^2{3}
\]

### Choosing \( c \) and \( c' \)
- To satisfy the recurrence, choose \( c' = \frac{1}{3} \), so that \( 1 - 3c' = 0 \) eliminates the negative term.
- The remaining terms should satisfy the original hypothesis:
\[
T(n) \leq n\log^2{n} \times 0 + 3cn \log_6{9} + 6c'n\log{n}\log{3} - 3c'n\log^2{3}
\]

Thus:
\[
T(n) \leq cn\log_6{9} + O(n\log{n}) 
\]

### Conclusion
We have shown that for sufficiently large \( c \) and appropriate choice of \( c' \), the assumed form holds, which leads to the final bound:
\[
T(n) = O(n \log_6{9}) = O(n^2)
\]
Since \( \log_6{9} = \frac{\log{9}}{\log{6}} \), this still gives a polynomial \( T(n) = O(n^2) \) as expected.

---

Would you like more details or have any questions about this method?

Here are five related questions that might deepen your understanding:

1. Why is the logarithm base important when analyzing the recurrence relation?
2. How do you handle recurrence relations where \( f(n) \) is a non-polynomial function?
3. What is the significance of the constants \( c \) and \( c' \) in the substitution method?
4. How does the substitution method differ from the Master Theorem in solving recurrences?
5. Can you apply the substitution method to a recurrence relation that results in an exponential function?

**Tip:** When using the substitution method, it's important to carefully choose the constants in your hypothesis to ensure that the inequality holds true for the entire range of \( n \).

Learn how to solve the recurrence relation T(n) = 9T(n/3) + n*log^2(n) using the substitution method. This problem involves analyzing and solving a complex recurrence relation using mathematical induction and logarithmic properties.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation