Chapter 3: Q49E (page 217)

Show that if $f_{1} (x)$ is $Θ (g_{1} (x))$ , is $Θ (g_{2} (x))$ , anddata-custom-editor="chemistry" $f_{2} (x) \neq 0$ and $g_{2} (x) \neq 0$ for all real numbers x > 0 then, ${(f_{1} / f)}_{2} (x)$ is $Θ (g_{1} / g_{2} (x))$ .

Short Answer

Expert verified

It is given that, $f_{1} (x$ ) is $Θ (g_{1} (x))$ , $f_{2} (x)$ is $Θ (g_{2} (x))$ , then we have to prove that $(f_{1} / f_{2}) (x)$ is $Θ (g_{1} / g_{2}) (x)$ .

Step by step solution

Step 1:

For $f_{1} (x)$ is $Θ (g_{1} (x))$ , applying the definition of Big-theta Notation

\begin{aligned} |f_{1} (x)| \leq a_{1} |g_{1} (x)| if x > b_{1} \\ |f_{1} (x)| \geq a_{2} |g_{1} (x)| if x > b_{2} \end{aligned}

Where, $a_{1}, a_{2}, b_{1} a n d b_{2}$ and are constants.

For $f_{2} (x)$ is data-custom-editor="chemistry" $Θ (g_{2} (x))$ , applying the definition of Big-theta Notation

role="math" localid="1668528171320"

\begin{aligned} |f_{2} (x)| \leq a_{3} |g_{2} (x)| if x > b_{3} \\ |f_{2} (x)| \geq a_{4} |g_{2} (x)| if x > b_{4} \end{aligned}

Where, $a_{3}, a_{4}, b_{3} a n d b_{4}$ and are constants.

Step 2:

We can take, b = max $(b_{1}, b_{2}, b_{3}, b_{4})$

Dividing by $|f_{1} (x)| \geq a_{2} |g_{1} (x)| by |f_{2} (x)| \geq a_{4} |g_{2} (x)|$

$\Rightarrow \frac{|f_{1} (x)|}{|f_{2} (x)|} \leq \frac{a_{1} |g_{1} (x)|}{a_{3} |g_{2} (x)|}$

We get, $|(\frac{f_{1}}{f_{2}}) (x)| \leq \frac{a_{1}}{a_{3}} |(\frac{g_{1}}{g_{2}}) (x)|$

Hence, $(\frac{f_{1}}{f_{2}}) (x) is O ((\frac{g_{1}}{g_{2}}) (x)) with b = max (b_{1}, b_{2}, b_{3}, b_{4}) and a = a_{1} / a_{3}$

$\Rightarrow \frac{|f_{1} (x)|}{|f_{2} (x)|} \geq \frac{a_{1} |g_{1} (x)|}{a_{3} |g_{2} (x)|}$

We get, $|(\frac{f_{1}}{f_{2}}) (x)| \geq \frac{a_{1}}{a_{3}} ∣ \frac{g_{1}}{g_{2}}) (x) ∣$

Hence, $(\frac{f_{1}}{f_{2}}) (x) is Ω ((\frac{g_{1}}{g_{2}}) (x))$ with b = maxrole="math" localid="1668528483523" $(b_{1}, b_{2}, b_{3}, b_{4}) and a = a_{2} / a_{4}$

Step 3:

By using $(\frac{f_{1}}{f_{2}}) (x)$ is $O ((\frac{g_{1}}{g_{2}}) (x)) \cdot (f_{1} f_{2}) (x) is (\frac{f_{1}}{f_{2}}) (x) is Ω ((\frac{g_{1}}{g_{2}}) (x))$

Hence, by using the definition of Big-Theta Notation, we can conclude that $Θ (f_{1} / f_{2} (x))$ is $Θ (g_{1} / g_{2} (x))$ .

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Show that if $f_{1} (x)$ is $Θ (g_{1} (x))$ , is $Θ (g_{2} (x))$ , anddata-custom-editor="chemistry" $f_{2} (x) \neq 0$ and $g_{2} (x) \neq 0$ for all real numbers x > 0 then, ${(f_{1} / f)}_{2} (x)$ is $Θ (g_{1} / g_{2} (x))$ .

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Step 2:

Step 3:

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Show that iff1(x)is Θ(g1(x)), is Θ(g2(x)), anddata-custom-editor="chemistry" f2(x)≠0andg2(x)≠0for all real numbers x > 0 then, (f1/f)2(x)is Θ(g1/g2(x)).

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Step 1:

Step 2:

Step 3:

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Most popular questions from this chapter

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Show that if $f_{1} (x)$ is $Θ (g_{1} (x))$ , is $Θ (g_{2} (x))$ , anddata-custom-editor="chemistry" $f_{2} (x) \neq 0$ and $g_{2} (x) \neq 0$ for all real numbers x > 0 then, ${(f_{1} / f)}_{2} (x)$ is $Θ (g_{1} / g_{2} (x))$ .