/*! 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 2 Find a real number \(c\) and an ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a real number \(c\) and an \(N_{0} \in \mathbb{N}\) such that \(n^{3}+5 n^{2}+2 n

Short Answer

Expert verified
Choose \(c = 2\) and \(N_0 = 6\).

Step by step solution

01

Simplify the Inequality

Simplify the inequality \(n^3 + 5n^2 + 2n < cn^3\) by dividing all terms by \(n^3\): \[1 + \frac{5}{n} + \frac{2}{n^2} < c\] for \(n \geq N_0\).
02

Analyze the Limit as \(n\) Approaches Infinity

Consider the behavior of \(\frac{5}{n}\) and \(\frac{2}{n^2}\) as \(n\) becomes very large. Both \(\frac{5}{n}\) and \(\frac{2}{n^2}\) approach 0. Thus, the inequality simplifies in the limit to: \[1 < c\].
03

Choose \(c\)

Since \(1 < c\) for the inequality to hold, choose \(c\) slightly greater than 1, such as \(c = 2\). This choice satisfies \(1 < c\).
04

Determine \(N_0\)

To ensure \(1 + \frac{5}{n} + \frac{2}{n^2} < 2\), evaluate \[1 + \frac{5}{n} + \frac{2}{n^2} < 2\], which simplifies to \[\frac{5}{n} + \frac{2}{n^2} < 1\]. When \(n = 6\), \(\frac{5}{6} + \frac{2}{36} \approx 0.833 + 0.056 = 0.889\), which is indeed less than 1. Hence, choose \(N_0 = 6\).

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Key Concepts

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

Inequality Simplification
Simplifying inequalities is like breaking a complex problem into smaller, more manageable pieces, allowing us to better handle the situation. When faced with the inequality \(n^3 + 5n^2 + 2n < cn^3\), the goal is to simplify the terms in a way that makes it easier to solve. To do this, we divide every term in the inequality by the highest power of \(n\) present, which is \(n^3\). This process is beneficial because it reduces each term to something simpler and often reveals the core idea of the inequality in a cleaner mathematical expression. As a result, our inequality becomes \[1 + \frac{5}{n} + \frac{2}{n^2} < c\]. This step eliminates the variable's highest power and suggests that for very large \(n\), values of \(\frac{5}{n}\) and \(\frac{2}{n^2}\) diminish in significance, making it easier to assess the inequality.
Limit Behavior
The behavior of a function as it approaches a limit helps us understand long-term trends, such as what happens as values become very large or very small. Here, our interest is in what happens as \(n\), a natural number, grows infinitely large. By examining \(\frac{5}{n}\) and \(\frac{2}{n^2}\) as \(n\) gets larger, we find that these terms become very small. Specifically, - \(\frac{5}{n}\) approaches zero because the numerator stays constant while the denominator grows.- \(\frac{2}{n^2}\) also approaches zero, but at a faster rate since \(n^2\) grows even more rapidly than \(n\) alone.With both terms approaching zero as \(n\) increases, the inequality simplifies to \(1 < c\). This limit behavior is crucial as it shows that the inequality simplifies in a straightforward manner when concentrating on the largest term, \(1\), and ignoring the diminishing fractions.
Natural Numbers
Natural numbers, denoted \(\mathbb{N}\), are a fundamental concept in mathematics. They include all positive integers starting from 1 (\{1, 2, 3, 4, \ldots\}). They are often used in mathematical problems dealing with whole, countable units. In the exercise, the problem defines \(n\) as a natural number, which means it must be a positive integer. This influences the concepts of limits and inequalities since we only consider discrete values increasing without bound.When selecting \(N_0\) in step 4, we ensure the inequality holds for all \(n \geq N_0\). In this case, selecting \(N_0 = 6\) ensures that any \(n\) greater than or equal to 6 will satisfy the inequality due to the decreasing influence of terms \(\frac{5}{n}\) and \(\frac{2}{n^2}\). We test this with \(n = 6\) as an anchor point, confirming the conditions are met with these fixed natural numbers.

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Most popular questions from this chapter

Show that: (a) \(\sin (n) \in O(1)\) - that is, the sine function is asymptotically dominated by the constant function \(1 .\) (b) \(n \sin (n) \in \mathbf{O}(n)\)

Prove or find a counterexample. If \(\mathbf{O}(G)=\mathbf{O}(F)\) and \(\mathbf{O}(H) \subset \mathbf{O}(F)\), then for all but finitely many \(n \in \mathbb{N}, H(n)

Prove that \(\sqrt{n} \in O(n)\) and that \(n \notin O(\sqrt{n})\).

It is tempting to compute the complexity of an algorithm by counting statements, as we did with the BubbleSort example, but only keeping track of the number of steps along the way up to \(O\left(^{*}\right)\). This turns out not to work with loops. For example, it is possible that each time through the loop, the number of statements executed is in \(\mathbf{O}(1)\). but that the number of statements executed by a loop of length \(n\) is not in \(\mathbf{O}(n)\). Find an example. (Hint: Each time through the loop, the number of statements executed may be in \(\mathrm{O}(1)\), but the constants \(c\) and \(N_{0}\) may change).

For each of the problems (a)-(d) below: (i) Write an algorithm in pseudocode to solve the problem (be sure your algorithm works correctly if \(m=0\) or \(n=0\); it should not make any assignments to clements of the array), and (ii) Calculate how many assignment statements and how many comparisons the algorithm causes to be executed as a function of \(m\) and \(n\). In this case, count assignments to and comparisons of index variables, as well as assignments to and comparisons of positions in the array. Simplify your answers. (a) Initialize all the elements of an \(m \times n\) array to \(0 .\) (b) Initialize all the elements of an \(m \times n\) array that lie on or above the diagonal to \(0 .\) (Here, by "diagonal" we mean positions \([r, c]\) where \(r=c .)\) (c) Initialize all the elements of an \(m \times n\) array that lic above the diagonal to \(0 .\) (d) Initialize all the elements on the diagonal of an \(m \times n\) array on the diagonal to \(0 .\)

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