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Chapter 16: Problem 14

Show that \(100 n^{3}+50 n^{2}+75\) is in \(\mathrm{O}\left(20 n^{3}\right)\) by finding a positive \(K\) that satisfies Equation (16.1) in this section

Short Answer

Expert verified
Question: Show that \(100n^3 + 50n^2 + 75\) is in \(O(20n^3)\). Answer: We can find a positive constant \(K = 6\) and another positive constant \(N_0 = 1\) such that \(100n^3 + 50n^2 + 75 \le K \cdot 20n^3\) for all \(n > N_0\), proving that \(100n^3 + 50n^2 + 75\) is in \(O(20n^3)\).

Step by step solution

01

Choose an appropriate K

We will try to find a suitable \(K\) by analyzing the inequality \(100n^3 + 50n^2 + 75 \le K \cdot 20n^3\). Divide both sides of the inequality by \(20n^3\) to simplify: \(\frac{100n^3 + 50n^2 + 75}{20n^3} \le K\). Now, we simplifiy the left side of the inequality: \(\frac{5n^3 + (50/20)n^2 + (75/20)n^3}{n^3} = 5 + \frac{5}{2} \frac{n^2}{n^3} + \frac{15}{4} \frac{1}{n^3} \le K\). As \(n\) gets larger, the \(\frac{n^2}{n^3}\) term goes to 0, and the \(\frac{1}{n^3}\) term goes to 0 as well. Therefore, we can choose \(K\) as 6, as it is greater than the sum of \(5\) and the remaining terms for large enough \(n\).
02

Prove the selected K satisfies the inequality

Now we must show that our chosen \(K=6\) indeed satisfies the inequality \(100n^3 + 50n^2 + 75 \le K \cdot 20n^3\) for some constant \(N_0\). For \(n \ge 1\), we have that \(\frac{5}{2} \frac{n^2}{n^3} \le \frac{5}{2}\) and \(\frac{15}{4} \frac{1}{n^3} \le \frac{15}{4}\). So, for all \(n \ge 1\), the inequality holds: \(5 + \frac{5}{2} \frac{n^2}{n^3} + \frac{15}{4} \frac{1}{n^3} \le 5 + \frac{5}{2}+\frac{15}{4} < 6\). Hence, \(100n^3 + 50n^2 + 75 \le 6 \cdot 20n^3\) for all \(n \ge 1\). Therefore, we have shown that \(100n^3 + 50n^2 + 75\) is in \(O(20n^3)\) with \(K=6\) and the constant \(N_0 = 1\).

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Most popular questions from this chapter

One of the sorting algorithms that we discussed works like this: It begins by putting the initial portion of the array consisting of the first two elements in sorted order. Then the third element is moved to its correct position, relative to the first two elements. At that point the first three elements are in sorted order. This process continues with the fourth and subsequent elements until the entire array is sorted. Which algorithm is this?

On average, with an array of 20,000 elements, how many comparisons will the sequential search perform? (Assume the items being searched have equal probability of being found at any of the positions in the array.)

With an array of 20,000 elements, what is the maximum number of comparisons the binary search will perform?

If a sequential search is performed on an array, and it is known that some items are searched for more frequently than others, how can the contents of the array be reordered to improve the average performance of the search?

One of the sorting algorithms that we discussed works like this: The smallest value in the array is located and moved to element 0. Then the next smallest value is located and moved to element 1\. This process continues until all of the elements have been placed in their proper order. Which algorithm is this?
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