/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Find \(f(n)\) when \(n=2^{k},\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(f(n)\) when \(n=2^{k},\) where \(f\) satisfies the recurrence relation \(f(n)=8 f(n / 2)+n^{2}\) with \(f(1)=1\)

Short Answer

Expert verified
f(2^k) = 2^{2k}((k + 1) - 1)

Step by step solution

01

- Analyze the recurrence relation

The given recurrence relation is: \[ f(n) = 8f\left(\frac{n}{2}\right) + n^2 \]and the initial condition is: \[ f(1) = 1 \]We need to find \(f(n)\) when \( n = 2^k \).
02

- Substitute \( n = 2^k \) into the relation

Substitute \( n = 2^k \). This gives us:\[ f(2^k) = 8f\left(\frac{2^k}{2}\right) + (2^k)^2 \]which simplifies to:\[ f(2^k) = 8f(2^{k-1}) + 2^{2k} \]
03

- Solve the base case

The base case is when \( k = 0 \):\[ f(1) = 1 \]
04

- Solve the recurrence relation for increasing \(k\)

For \( k = 1 \):\[ f(2) = 8f(1) + 2^2 \ = 8(1) + 4 \ = 12 \]For \( k = 2 \):\[ f(4) = 8f(2) + 4^2 \ = 8(12) + 16 \ = 112 \]Notice the pattern and general form as we continue this process.
05

- Identify the pattern and generalize

From Step 4, the results are:\[ f(2^0) = 1 \ f(2^1) = 12 \ f(2^2) = 112 \]We can hypothesize and validate through induction or pattern recognition that:\[ f(2^k) = (k + 1)2^{2k} - (k2^{2k+1} - k2^{k+1}) + \ ... \ = (2^{2k})(k + 1) - 1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

discrete mathematics
Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. This means that discrete mathematics deals with countable, often finite sets, such as integers, graphs, and statements in logic. It serves as a foundational element for computer science and enables the handling of algorithms and computational problems. In the context of this exercise, discrete mathematics helps us understand and solve recurrence relations, which are pivotal in predicting sequences and patterns, especially within computer algorithms.
recursive functions
Recursive functions are functions that refer back to themselves to define their output. Solving problems using recursion involves defining a function in terms of smaller instances of itself. For example, the recurrence relation in the exercise is defined as: f(n) = 8f(n/2) + n^2 Here, the function f(n) can be broken down into smaller parts until it fits an initial base case, making it easier to handle complex computations. Recursion is a fundamental concept in computer science and mathematics because it provides a mechanism to break down and solve problems systematically by dividing them into simpler sub-problems.
mathematical induction
Mathematical induction is a powerful technique used to prove statements about all natural numbers. It consists of two steps: the base case and the inductive step. In this exercise, after establishing the base case for f(1)=1, each subsequent step involves assuming the hypothesis for an instance k and then proving it for k+1. Here's a more detailed look: - **Base Case:** When k=0, f(1)=1 is given. - **Inductive Step:** Assume f(2^k) holds true, and then prove it for f(2^(k+1)). This method validates the pattern and general formula we've identified for f(2^k). Induction is highly effective in proving the correctness of recursive algorithms and relations, ensuring they hold for every instance in a sequence.
base case analysis
Base case analysis is the first step in solving recursive problems and proving statements via induction. It establishes a simple, non-recursive case which serves as a foundation for further recursion. In the provided exercise, the base case is given by: f(1) = 1 Establishing the base case is crucial because every recursive call reduces the problem to a smaller instance until it reaches this fundamental case. Without a proper base case, a recursive function could end up in an infinite loop. In mathematical induction as well, the base case confirms that the initial step of our proof is accurate, setting the groundwork for the inductive step to follow.

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Most popular questions from this chapter

This exercise deals with the problem of finding the largest sum of consecutive terms of a sequence of n real numbers. When all terms are positive, the sum of all terms provides the answer, but the situation is more complicated when some terms are negative. For example, the maximum sum of consecutive terms of the sequence \(-2,3,-1,6,-7,4\) is \(3+(-1)+6=8 .\) (This exercise is based on \([\mathrm{Be} 86] .\) Recall that in Exercise 56 in Section 8.1 we developed a dynamic programming algorithm for solving this problem. Here, we first look at the brute-force algorithm for solving this problem; then we develop a divide- and-conquer algorithm for solving it. a) Use pseudocode to describe an algorithm that solves this problem by finding the sums of consecutive terms starting with the first term, the sums of consecutive terms starting with the second term, and so on, keeping track of the maximum sum found so far as the algorithm proceeds. b) Determine the computational complexity of the algorithm in part (a) in terms of the number of sums computed and the number of comparisons made. c) Devise a divide-and-conquer algorithm to solve this problem. [Hint: Assume that there are an even number of terms in the sequence and split the sequence into two halves. Explain how to handle the case when the maximum sum of consecutive terms includes terms in both halves.] d) Use the algorithm from part (c) to find the maximum sum of consecutive terms of each of the sequences: \(-2,4,-1,3,5,-6,1,2 ; 4,1,-3,7,-1,-5, \quad 3,-2 ;\) and \(-1,6,3,-4,-5,8,-1,7\) e) Find a recurrence relation for the number of sums and comparisons used by the divide-and-conquer algorithm from part (c). f ) Use the master theorem to estimate the computational complexity of the divide-and-conquer algorithm. How does it compare in terms of computational complexity with the algorithm from part (a)?

a) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s. b) What are the initial conditions? c) How many bit strings of length seven contain two consecutive 0s?

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Show that the coefficient \(p_{d}(n)\) of \(x^{n}\) in the formal power series expansion of \((1+x)\left(1+x^{2}\right)\left(1+x^{3}\right) \cdots\) equals the number of partitions of \(n\) into distinct parts, that is, the number of ways to write \(n\) as the sum of positive inte- gers, where the order does not matter but no repetitions are allowed.

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Show that if \(n\) is a positive integer, then the number of partitions of \(n\) into distinct parts equals the number of partitions of \(n\) into odd parts with repetitions allowed; that is, \(p_{o}(n)=p_{d}(n) .[\text { Hint: Show that the generating }\) functions for \(p_{o}(n)\) and \(p_{d}(n)\) are equal. \(]\)

If \(G(x)\) is the generating function for the sequence \(\left\\{a_{k}\right\\}\) what is the generating function for each of these sequences? a) \(2 a_{0}, 2 a_{1}, 2 a_{2}, 2 a_{3}, \cdots\) b) \(0, a_{0}, a_{1}, a_{2}, a_{3}, \ldots\) (assuming that terms follow the pattern of all but the first term) c) \(0,0,0,0, a_{2}, a_{3}, \ldots\) (assuming that terms follow the pattern of all but the first four terms) d) \(a_{2}, a_{3}, a_{4}, \ldots\) e) \(a_{2}, 2 a_{2}, 3 a_{3}, 4 a_{4}, \ldots[\text { Hint: Calculus required here. }]\) f) \(a_{0}^{2}, \quad 2 a_{0} a_{1}, \quad a_{1}^{2}+2 a_{0} a_{2}, \quad 2 a_{0} a_{3}+2 a_{1} a_{2}, \quad 2 a_{0} a_{4}+\) \(\quad 2 a_{1} a_{3}+a_{2}^{2}, \ldots\)

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