/*! 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 48 Show that if \(n\) is a variable... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 48

Show that if \(n\) is a variable that takes positive integer values, then \(2^{n}\) is \(O(n !)\).

Short Answer

Expert verified
To show that \(2^n\) is \(O(n!)\), we need to find a constant \(C\) and a positive integer value \(n_0\) such that \(2^n \le C \cdot n!\), for all \(n \ge n_0\). By observing that for \(n \ge 4\), \(2^n \le n!\), we can choose \(C = 1\) and \(n_0 = 4\). Thus, we can conclude that \(2^n\) is \(O(n!)\).

Step by step solution

01

Rewrite the condition for Big-O notation

To show that \(2^n\) is \(O(n!)\), we need to find a constant \(C\) and a positive integer value \(n_0\) such that: \(2^n \le C \cdot n!\), for all \(n \ge n_0\)
02

Finding an appropriate constant

The main idea here is to find a constant \(C\) so that the inequality holds. Let's examine the first few values of \(n\) and their corresponding values for \(2^n\) and \(n!\): 1. \(n = 1 : 2^1 = 2 , 1! = 1\) 2. \(n = 2 : 2^2 = 4 , 2! = 2\) 3. \(n = 3 : 2^3 = 8 , 3! = 6\) 4. \(n = 4 : 2^4 = 16, 4! = 24\) 5. \(n = 5 : 2^5 = 32, 5! = 120\) We can observe that for \(n \ge 4\), \(2^n \le n!\). Therefore, let's choose \(C = 1\) and \(n_0 = 4\).
03

Confirming the inequality

Now, let's confirm that for all \(n \ge n_0\), the inequality \(2^n \le C \cdot n!\) holds: 1. For \(n = 4\), \(2^4 = 16\) and \(C \cdot n! = 1 \cdot 4! = 24\). We have \(2^4 \le 1 \cdot 4!\), which is true. 2. For \(n \ge 5\), we can see that as \(n\) grows, the difference between \(n!\) and \(2^n\) also grows larger, since the factorial function grows much faster than the exponential function. Thus, the inequality will continue to hold for all \(n \ge 5\).
04

Concluding the proof

We have found a constant \(C = 1\) and a positive integer value \(n_0 = 4\) such that \(2^n \le C \cdot n!\) for all \(n \ge n_0\). Therefore, we can conclude that \(2^n\) is \(O(n!)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Asymptotic Analysis
Understanding the performance of algorithms is crucial in computer science, and that's where asymptotic analysis comes into play. This type of analysis allows us to describe the running time of algorithms in terms of input size, ignoring constants and low-order terms.

For instance, when we say an algorithm is O(n), we mean that in the worst case, the algorithm will not take more time than a constant times n, where n represents the size of the input.

By analyzing algorithms asymptotically, we can focus on their efficiency trends rather than getting mired in intricate calculations for each possible scenario. For the exercise mentioned, 2^n = O(n!) indicates that the exponential function 2^n does not grow faster than the factorial function of n when n becomes very large, a conclusion that is at the heart of asymptotic analysis.
Factorial Function
The factorial function, denoted by n!, plays a key role in mathematics, particularly in combinations and permutations. It's defined as the product of all positive integers up to n. For example, 5! equals 5 x 4 x 3 x 2 x 1, which is 120.

What's intriguing about the factorial function is its incredibly rapid growth rate. Even for small values of n, the factorial can be quite large. This is why factorial functions can quickly outgrow exponential functions like 2^n, which we saw in the exercise we're examining.

It's not that the exponential function grows slowly鈥攐n the contrary, but when compared to factorial growth, it doesn't hold up. This is crucial to understand why in the exercise, 2^n is found to be effectively 'smaller' than n! for sufficient large values of n.
Exponential Growth
When we talk about exponential growth, we're referring to a process where quantities increase rapidly over time鈥攏ormally, at a rate proportional to their current value. This is represented mathematically as a^n, where a is a constant and n is a variable, often time or another progressive increment.

Exponential functions grow quickly; doubling with each increment in the case of 2^n. However, the pivotal takeaway from the textbook exercise is that despite its rapid increase, exponential growth can still be outpaced by other types of growth, such as that of the factorial function.

Understanding this dynamism is not just academically fascinating but also practically significant in fields like computer science, biology, and economics, where the concept of exponential growth is frequently applied to model real-world phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Define binary relations \(R\) and \(S\) from \(\mathbf{R}\) to \(\mathbf{R}\) as follows: $$ \begin{aligned} &R=\left\\{(x, y) \in \mathbf{R} \times \mathbf{R} \mid x^{2}+y^{2}=4\right\\} \quad \text { and } \\ &S=\\{(x, y) \in \mathbf{R} \times \mathbf{R} \mid x=y\\} . \end{aligned} $$ Graph \(R, S, R \cup S\), and \(R \cap S\) in the Cartesian plane.

Suppose a computer takes 1 nanosecond \(\left(=10^{-9}\right.\) second \()\) to execute each operation. Approximately how long will it take for the computer to execute the following numbers of operations? Convert your answers into seconds, minutes, hours, days, weeks, or years, as appropriate. For example, instead of \(2^{50}\) nanoseconds, write 13 days. a. \(\log _{2} 200\) b. 200 c. \(200 \log _{2} 200\) d. \(200^{2}\) e. \(200^{8}\) f. \(2^{200}\)

Draw the graphs of \(y=2\lfloor x\rfloor\) and \(y=\lfloor 2 x\rfloor\) for all real numbers \(x\). What can you conclude from these graphs? Graph each of the functions defined in \(5-8\) below.

Show that the function \(g: \mathbf{R} \rightarrow \mathbf{R}\) defined by the formula \(g(x)=-(x / 3)+1\) is decreasing on the set of all real numbers.

In 4-9, express each statement using \(\Omega-, O-\), or \(\Theta\)-notation. \(\left|x^{7 / 2}\right| \leq\left|\frac{\left(x^{2}-7\right)^{2}\left(10 x^{1 / 2}+3\right)}{x+1}\right|\) for all real numbers \(x>4\). (Use \(\Omega\)-notation.)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.