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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Question
![**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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3632b9ca-0e22-48c8-943d-c9e08fc0f04c%2Fb3a7dd83-724e-48e3-986f-1c2af943c8e6%2Fgz0gknu_processed.png&w=3840&q=75)
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
Step by step
Solved in 2 steps with 1 images
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