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Chapter 9: Problem 50

a. For all real numbers \(u, \log _{2} u0\). b. Interpret the statement of part (a) using \(O\)-notation.

Short Answer

Expert verified
We showed that \(\log_2 x < nx^{\frac{1}{n}}\) for any positive integer \(n\) and real numbers \(x>0\) by applying the given inequality \(\log_2 u0\).

Step by step solution

01

Apply the given inequality to the problem.

The inequality given is \(\log_2 u < u\) for all real numbers \(u\). Now, we will use this fact to show that \(\log_2 x < nx^{\frac{1}{n}}\) for any positive integer \(n\) and real numbers \(x>0\).
02

Rewrite the desired inequality in terms of a new variable v.

Let \(v = x^{\frac{1}{n}}\), and consider the function \(f(v) = nv\). Our goal is to show that \(\log_2 x < f(v)\). By the change of variables above, this is equivalent to \(\log_2 (v^n) < nv\).
03

Use properties of logarithms to simplify the inequality.

By the properties of logarithms, we can rewrite the expression \(\log_2 (v^n)\) as \(n\log_2 v\). Thus, the inequality now becomes \(n\log_2 v < nv\).
04

Apply the given fact to the new variable v.

The inequality \(\log_2 u < u\) is true for all real numbers \(u\) and so it must also be true for \(u = v\), which implies that \(\log_2 v < v\).
05

Multiply the inequality by n.

Now, we can multiply both sides of the inequality \(\log_2 v < v\) by the positive integer \(n\). Since \(n>0\), we can multiply without changing the direction of the inequality. Thus, we get \(n\log_2 v < nv\).
06

Relate the inequality back to x and n.

Recall that we defined \(v = x^{\frac{1}{n}}\), so \(n\log_2 v < nv\) is equivalent to \(\log_2 x < nx^{\frac{1}{n}}\). Therefore, we have shown that the inequality holds for any positive integer \(n\) and real numbers \(x>0\).
07

Interpret the inequality using Big-O notation.

Big-O notation is used in computer science to describe the growth of functions. In this context, the inequality \(\log_2 x < nx^{\frac{1}{n}}\) can be interpreted as follows: The function \(\log_2 x\) has a growth rate strictly less than \(nx^{\frac{1}{n}}\) for all positive integers \(n\) and real numbers \(x>0\). This can be expressed using Big-O notation as follows: \[ \log_2 x = O(nx^{\frac{1}{n}}) \] which means that \(\log_2 x\) is dominated by the function \(nx^{\frac{1}{n}}\), for all positive integers \(n\) and real numbers \(x>0\).

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Most popular questions from this chapter

In each of 10-14 assume \(f\) and \(g\) are real-valued functions defined on the same set of nonnegative real numbers. Prove that if \(f(x)\) is \(O(h(x))\) and \(g(x)\) is \(O(k(x))\), then \(f(x) g(x)\) is \(O(h(x) k(x))\)

\(k(n)=\left\lfloor n^{1 / 2}\right\rfloor\) for each integer \(n \geq 0\)

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.

a. Show that for any real number \(x\), $$ \text { if } x>1 \text { then }\left|x^{4}\right| \leq\left|23 x^{4}+8 x^{2}+4 x\right| . $$ b. Show that for any real number \(x\), $$ \text { if } x>1 \text { then }\left|23 x^{4}+8 x^{2}+4 x\right| \leq 35\left|x^{4}\right| \text {. } $$ c. Use the \(\Omega\) - and \(O\)-notations to express the results of parts (a) and (b). d. What can you deduce about the order of \(23 x^{4}+8 x^{2}+\) \(4 x\) ?

Exercises \(36-39\) refer to the following algorithm to compute the value of a real polynomial. Algorithm 9.3.3 Term-by-Term Polynomial Evaluation [This algorithm computes the value of the real polynomial \(a[n] x^{n}+a[n-1] x^{n-1}+\cdots+a[2] x^{2}+a[1] x+a[0]\) by computing each term separately, starting with \(a[0]\), and adding it on to an accumulating sum.] Input: \(n\) [a nonnegative integer], \(a[0], a[1], a[2], \ldots, a[n]\) [an array of real numbers], \(x[\) a real number \(]\) Algorithm Body: polyval := \(a[0]\) for \(i:=1\) to \(n\) term : = \(a[i]\) for \(j:=1\) to \(i\) term \(:=\) term \(\cdot x\) next \(j\) polyval := polyval + term next \(i\) [At this point polyval \(=a[n] x^{n}+a[n-1] x^{n-1}\) \(\left.+\cdots+a[2] x^{2}+a[1] x+a[0] .\right]\) Output: polyval [a real number] Trace Algorithm \(9.3 .3\) for the input \(n=3, a[0]=2, a[1]=\) \(1, a[2]=-1, a[3]=3\), and \(x=2\).
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