6.3 Other ways of defining functions

Polynomial expressions give us a very wide class of functions and these functions have the convenient property that they can be expressed as a single formula. But not all functions are polynomial functions and not all functions can be expressed as a single formula. For example, the graph below has a sudden change of direction at $x=0$ and therefore its corresponding function cannot be expressed as a polynomial²² 2 This is related to the fact that polynomials are “smooth”. This is a term you will encounter in your future Analysis courses..

So how do we express the function that corresponds to graphs like this?

6.3.1 Defining a function by cases

A mathematical function defined by cases specifies different formulas or rules for different parts of its domain. These are also known as piecewise functions. Instead of using a single expression, the function is broken into pieces depending on conditions such as inequalities, intervals, or specific values of the input.

Example 6.23.

Consider the function $f:\mathbb{R}\to\mathbb{R}$ where

f(x)=\begin{cases}x^{2}&\text{if }x\geq 0,\\ -x^{2}&\text{if }x<0.\end{cases}

This means:

•

when $x\geq 0$ , the function outputs $x^{2}$ ;
•

when $x<0$ , the function outputs $-x^{2}$ .

We can sketch this in a graph below:

In general, the structure of a piecewise function is as follows:

f(x)=\begin{cases}\text{expression A}&\text{if condition A},\\ \text{expression B}&\text{if condition B},\\ \quad\vdots&\quad\vdots\end{cases}

The conditions must cover the whole domain of the function and they should not overlap (this is to ensure that each input has a unique output).

Definition 6.24.

The modulus or absolute value function is defined by $|\cdot|:\mathbb{R}\to[0,\infty)$ where

|x|=\begin{cases}x,&\text{if }x\geq 0,\\ -x,&\text{if }x<0.\end{cases}

Note that the notation for the modulus function is the same as the notation we used for the cardinality of a set. The way to tell them apart is to identify the type of object that lies between the two vertical lines. If this object represents a real number, then this notation represents the modulus function.

Exercise 6.25.

Consider the function $f:\mathbb{R}\to\mathbb{R}$ where $f(x)=x-|x|$ .

1.

Use the definition of the modulus function to re-write $f$ as a piecewise function.
2.

Sketch the graph of the function $f$ .

Solution (please try for yourself before looking)

1.

When $x\geqslant 0$ , $|x|=x$ . Thus, for this range of $x$ , $f(x)=x-x=0$ . However, when $x<0$ , $|x|=-x$ which means $f(x)=x-(-x)=2x$ for this range of $x$ . Hence, we can write $f$ as

$f(x)=\begin{cases}0,\text{ if }x\geqslant 0,\\ 2x,\text{ if }x<0.\end{cases}$
2.

A sketch of the graph $f$ is as follows:

Exercise 6.26.

The Dirichlet function $\bm{1}_{\mathbb{Q}}:\mathbb{R}\to\{0,1\}$ is defined by the formula

\bm{1}_{\mathbb{Q}}(x)=\begin{cases}1&\text{if $x\in\mathbb{Q}$}\\ 0&\text{otherwise.}\end{cases}

Let $r$ be a positive rational number. Prove that $\bm{1}_{\mathbb{Q}}$ is $r$ -periodic.

Solution (please try for yourself before looking)

If $r$ is a positive rational number, $x+r\in\mathbb{Q}$ for every $x\in\mathbb{Q}$ and $y+r\notin\mathbb{Q}$ for every $y\notin\mathbb{R}$ (as we saw in Block 1). Hence, $\bm{1}_{\mathbb{Q}}(x+r)=\bm{1}_{\mathbb{Q}}(x)=1$ for every $x\in\mathbb{Q}$ and $\bm{1}_{\mathbb{Q}}(y+r)=\bm{1}_{\mathbb{Q}}(y)=0$ for every $y\notin\mathbb{Q}$ .

6.3.2 Defining a function in words

Example 6.27.

The integer part function $\left[\;\cdot\;\right]:\mathbb{R}\to\mathbb{Z}$ takes a real number $x$ as its input and returns the integer part of $x$ . (For a definition of the integer part of a real number, see Section 2.1.)

Example 6.28.

The function $\cos:\mathbb{R}\to\mathbb{R}$ can be defined by the following process.

Consider the unit circle $x^{2}+y^{2}=1$ .

Let $\theta$ be any real number and let $L$ denote the straight line that radiates from the origin to form the angle $\theta$ radians anti-clockwise with the positive $x$ -axis. The $x$ -coordinate of the intersection point between $L$ and the unit circle is defined to be the value of $\cos(\theta)$ .

The function $\sin:\mathbb{R}\to\mathbb{R}$ can be defined in a similar way but where the value of the function is given by the $y$ -coordinate of the intersection point on the unit circle.

Exercise 6.29.

Assume that each $n\in\mathbb{N}$ , is written as a product of primes

n=2^{p_{2}(n)}3^{p_{3}(n)}5^{p_{5}(n)}\cdots,

where each $p_{k}(n)$ is a non-negative integer, which depends on $n$ . Let $f:\mathbb{N}\to\mathbb{N}_{0}$ be defined as $f(n)=p_{2}(n)$ .

How do you know that $f$ is a function?

Solution (please try for yourself before looking)

This is a function since the Fundamental Theorem of Arithmetic tells us every integer $n\geqslant 2$ can be written as a product of primes and any prime that doesn’t appear in the prime decomposition of $n$ just has a power of 0, which our codomain $\mathbb{N}_{0}$ allows. Also, $1$ can be written as a product of prime powers where every power is 0. So for every $x\in\mathbb{N}$ , there exists some $y\in\mathbb{N}_{0}$ where $f(x)=y$ and this $y$ is unique as $p_{2}(n)$ is defined to be largest $y\in\mathbb{N}_{0}$ such that $2^{y}|n$ .

6.3.3 Not all formulas define functions

When defining functions, we cannot simply write down any old algebraic expression and use this as the rule.

Exercise 6.30.

Suppose we try to define a function $g:\mathbb{Q}\to\mathbb{Z}$ by the formula

g\left(\frac{a}{b}\right)=a+b.

Why is this not a function?

[Inspired by Why aren’t all functions well-defined?]

Solution (please try for yourself before looking)

According to the formula, $g\bigl{(}\frac{1}{2}\bigr{)}=3$ and $g\bigl{(}\frac{2}{4}\bigr{)}=6$ . But $\frac{1}{2}=\frac{2}{4}$ and so if $g$ were a function then we would require $g\bigl{(}\frac{1}{2}\bigr{)}=g\bigl{(}\frac{2}{4}\bigr{)}$ .

Example 6.31.

Let $h:\mathbb{Q}\to\mathbb{Z}$ where

h\left(\frac{a}{b}\right)=\frac{a^{2}+b^{2}}{ab}.

This is not a function because the codomain is $\mathbb{Z}$ but $h\left(\frac{1}{2}\right)=\frac{1^{2}+2^{2}}{1\cdot 2}=\frac{5}{2}\notin% \mathbb{Z}$ .