Use the Master Theorem to give a big-O estimate for f(n) = 2f (1) + 4.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Problem Statement:**

Use the Master Theorem to give a big-O estimate for the recurrence relation:

\[ f(n) = 2f\left(\frac{n}{3}\right) + 4. \] 

**Explanation:**

This is a typical divide-and-conquer recurrence relation of the form:

\[ T(n) = aT\left(\frac{n}{b}\right) + f(n), \]

where:
- \( a = 2 \) is the number of recursive calls,
- \( b = 3 \) is the factor by which the subproblem size is reduced,
- \( f(n) = 4 \) is the cost of combining the results of the subproblems. 

**Master Theorem Application:**

For the Master Theorem, compare \( f(n) = 4 \) with \( n^{\log_b a} = n^{\log_3 2} \). Calculate \( \log_3 2 \approx 0.631 \).

- Since \( f(n) = 4 = \Theta(n^0) \), and \( n^{\log_b a} = n^{0.631} \), we have:

  \( f(n) \) is \( \Theta(n^c) \) where \( c = 0 \).

- The Master Theorem states:

  \[
  T(n) = 
  \begin{cases} 
  \Theta(n^{\log_b a}) & \text{if } f(n) = O(n^{\log_b a - \epsilon}) \text{, for some } \epsilon > 0 \\
  \Theta(n^{\log_b a} \log n) & \text{if } f(n) = \Theta(n^{\log_b a}) \\
  \Theta(f(n)) & \text{if } f(n) = \Omega(n^{\log_b a + \epsilon}) \text{, for some } \epsilon > 0 
  \end{cases}
  \]

- Here, \( f(n) = \Theta(n^0) \) and \( n^{\log_b a} = \Theta(n^{0.631}) \).

- Therefore, \( f(n) = O(n^{\log_b a - \epsilon}) \) implies \( \epsilon = 0.631 \).

**Conclusion:**

Using
Transcribed Image Text:**Problem Statement:** Use the Master Theorem to give a big-O estimate for the recurrence relation: \[ f(n) = 2f\left(\frac{n}{3}\right) + 4. \] **Explanation:** This is a typical divide-and-conquer recurrence relation of the form: \[ T(n) = aT\left(\frac{n}{b}\right) + f(n), \] where: - \( a = 2 \) is the number of recursive calls, - \( b = 3 \) is the factor by which the subproblem size is reduced, - \( f(n) = 4 \) is the cost of combining the results of the subproblems. **Master Theorem Application:** For the Master Theorem, compare \( f(n) = 4 \) with \( n^{\log_b a} = n^{\log_3 2} \). Calculate \( \log_3 2 \approx 0.631 \). - Since \( f(n) = 4 = \Theta(n^0) \), and \( n^{\log_b a} = n^{0.631} \), we have: \( f(n) \) is \( \Theta(n^c) \) where \( c = 0 \). - The Master Theorem states: \[ T(n) = \begin{cases} \Theta(n^{\log_b a}) & \text{if } f(n) = O(n^{\log_b a - \epsilon}) \text{, for some } \epsilon > 0 \\ \Theta(n^{\log_b a} \log n) & \text{if } f(n) = \Theta(n^{\log_b a}) \\ \Theta(f(n)) & \text{if } f(n) = \Omega(n^{\log_b a + \epsilon}) \text{, for some } \epsilon > 0 \end{cases} \] - Here, \( f(n) = \Theta(n^0) \) and \( n^{\log_b a} = \Theta(n^{0.631}) \). - Therefore, \( f(n) = O(n^{\log_b a - \epsilon}) \) implies \( \epsilon = 0.631 \). **Conclusion:** Using
Expert Solution
Introduction

As per the question we are given a recurrence relationship as the following :

f(n) = 2f(n/3) + 4

Now we have to estimate the asymptomatic behaviour of f(n) as n → ∞ in big-O notation using the Master Theorem

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