Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

Show that if $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0},$ , where $a_{0}, a_{1}, . . . . ., a_{n - 1},$ and $a_{n}$ are real numbers and an $a_{n} \neq 0$ , then data-custom-editor="chemistry" $f (x)$ is $Θ (x^{n})$ . Big-O, big-Theta, and big-Omega notation can be extended to functions in more than one variable. For example, the statement $f (x, y)$ is $O (g (x, y))$ means that there exist constants C, $k_{1}$ , and $k_{2}$ such that $| f (x, y) | \leq C | g (x, y) |$ whenever $x > k_{1}$ and $x > k_{2}$ .

It is given that $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ , where $a_{0}, a_{1}, . . . . ., a_{n - 1},$ and $a_{n}$ are real numbers then we have to prove that $f (x)$ is $Θ (x^{n})$ .

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TABLE OF CONTENTS

Step 1:

$\begin{aligned} f (x) = a_{n} x^{n} + a_{n - 1} x^{x^{n - 1} + \dots + a_{1} x + a_{0}} \\ | f (x) | = |a_{n} x^{n} + a_{n - 1} 1^{x^{n - 1} + \dots + a_{1} x + a_{0}}| \end{aligned}$

Assume b=max(1, $a_{0}, a_{1}, . . . . ., a_{n - 1}$ )

Thus x> max(1, $a_{0}, a_{1}, . . . . ., a_{n - 1}$ )

\begin{aligned} \Rightarrow | f (x) | \leq |a_{n} x^{n} + x x^{n - 1} + \dots + x x + x| \\ \Rightarrow | f (x) | \leq |a_{n} x^{n} + x^{n} + \dots + x^{2} + x| \\ \Rightarrow | f (x) | \leq |a_{n} x^{n} + x^{n} + \dots + x^{n} + x^{n}| \\ \Rightarrow | f (x) | \leq |a_{n} x^{n} + n x^{n}| \\ \Rightarrow | f (x) | \leq |a_{n} + n| x^{n} ∥ \end{aligned}

Hence, it can be said that f(x) is $O (x^{n})$ with constants b=max(1, $a_{0}, a_{1}, . . . . ., a_{n - 1}$ ) and C= $| a_{n} + n |$

Step 2:

$\begin{aligned} f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} \\ | f (x) | = |a_{n} x^{x^{n} + a_{n - 1} 1^{x^{n - 1}} + \dots + a_{1} x + a_{0}}| \end{aligned}$

Assume b=min (1, $a_{0}, a_{1}, . . . . ., a_{n - 1}$ )

Thus x> min (1, $a_{0}, a_{1}, . . . . ., a_{n - 1}$ )

\begin{aligned} \Rightarrow | f (x) | \geq |b x^{n} + b x^{n - 1} + \dots + b x + b| \\ \Rightarrow | f (x) | \geq |b (x^{n} + x^{n - 1} + \dots + x^{1} + x)| \\ \Rightarrow | f (x) | \geq | b | |x^{n} + x^{n - 1} + \dots + x^{1} + x| \end{aligned}

Hence, it can be said that f(x) is $Ω (x^{n})$ with constants b=min (1, $a_{0}, a_{1}, . . . . ., a_{n - 1}$ ) and C= $|b|$

Step 3:

As we know that $f (x)$ is $O (x^{n})$ and $f (x)$ is $Ω (x^{n})$ .

Hence, by applying the definition of Big-Theta Notation, f (x) is $Θ (x^{n})$ .

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