William Boggs
2021-12-19
Answered

Show that $x}^{3$ is $O\left({x}^{4}\right)$ but that $x}^{4$ is not $O\left({x}^{3}\right)$ .

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Jack Maxson

Answered 2021-12-20
Author has **25** answers

Defintions

Big-O Notation:$f\left(x\right)$ is $O\left(g\left(x\right)\right)$ is there exists constants C and k such that

$\left|f\left(x\right)\right|\le C\left|g\left(x\right)\right|$

Whener x>k

Note: Choises of C and k are not unique.

Solution:

First part

$f\left(x\right)={x}^{3}$

$g\left(x\right)={x}^{4}$

For convevience sake, we will choose k=1 and thus x>1.

$\left|f\left(x\right)\right|=\left|{x}^{3}\right|$

$={x}^{3}$

$1\cdot {x}^{3}$

$<x\cdot {x}^{3}$

$={x}^{4}$

$=\left|{x}^{4}\right|$

Thus we need to choose C to be at least 1. Let us then take C=1

By the definition of Big-O notation,$f\left(x\right)={x}^{3}$ is $O\left({x}^{4}\right)$ with k=1 and C=1.

Second part

$f\left(x\right)={x}^{4}$

$g\left(x\right)={x}^{3}$

Let us assume that$f\left(x\right)={x}^{4}$ is $O\left({x}^{3}\right)$ . Then there exist constants C and k such that:

$\left|{x}^{4}\right|\le C\left|{x}^{3}\right|$

when x>k

Let us assume$C\ge 0$ , which is a safe assumption.

$\left|{x}^{4}\right|\le C\left|{x}^{3}\right|=\left|C{x}^{3}\right|$

We have then obtained a contradiction, because$\left|{x}^{4}\right|>\left|C{x}^{3}\right|$ when x>C

Thus$f\left(x\right)={x}^{4}$ is not $O\left({x}^{3}\right)$

Big-O Notation:

Whener x>k

Note: Choises of C and k are not unique.

Solution:

First part

For convevience sake, we will choose k=1 and thus x>1.

Thus we need to choose C to be at least 1. Let us then take C=1

By the definition of Big-O notation,

Second part

Let us assume that

when x>k

Let us assume

We have then obtained a contradiction, because

Thus

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