Elementary functions

 Below we list some frequently used elementary functions.

1. Algebraic functions
2. Exponential functions
3. Logarithm
4. Trigonometric functions
5. Inverse trigonometric functions
6. Hyperbolic functions

Algebraic functions

A polynomial of $x$

$$f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$$

where $a_n,a_{n-1},\cdots ,a_0\in \mathbb{R}$ is a continuous function on $\mathbb{R}$. Such functions are called polynomial functions.

If $g(x)$ and $h(x)$ are polynomial functions such that $h(x)\ne 0$, the function $f(x)$ defined by

$$f(x)=\frac{g(x)}{h(x)}$$

is continuous on $\{x\mid x\in \mathbb{R},h(x)\ne 0\}$. Such functions are called rational functions. If $h(x)=1$ then $f(x)=g(x)$ so any polynomial functions are also rational functions (i.e., polynomial functions are a special case of rational functions).

We can ``algebraically'' define functions other than polynomial or rational functions. For example, $f(x)=\sqrt{x}$ is not a rational function, but satisfies an algebraic equality

$$[f(x){]}^2-x=0.$$

Definition (Algebraic function)

The continuous function $f(x)$ is said to be an algebraic function if there exist polynomial functions $g_0(x),g_1(x),\cdots ,g_n(x)$ such that the following identity is satisfied:

$$g_n(x)[f(x){]}^n+g_{n-1}(x)[f(x){]}^{n-1}+\cdots +g_1(x)f(x)+g_0(x)=0.$$

Example. Rational functions are a special case of algebraic functions. □

Example. Let us prove that the function $f(x)=\sqrt{x}$ defined on $x\ge 0$ is continuous.

First, consider the case when $a>0$. For all $\epsilon >0$, let us define $\delta =\epsilon \sqrt{a}$. Then, if $x\ge 0$ and $0<|x-a|<\delta$,

$$\begin{array}{rcl}|\sqrt{x}-\sqrt{a}|& =& \left|\frac{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}{\sqrt{x}+\sqrt{a}}\right|\\ & =& \frac{|x-a|}{\sqrt{x}+\sqrt{a}}\\ & \le & \frac{|x-a|}{\sqrt{a}}<\frac{\delta }{\sqrt{a}}=\epsilon .\end{array}$$

Next, consider the case when $a=0$. For any $\epsilon >0$, take $\delta ={\epsilon }^2>0$. If $0<x<\delta$, then $|\sqrt{x}-0|=\sqrt{x}<\sqrt{\delta }=\epsilon$. Thus, $\underset{x\to +0}{lim}\sqrt{x}=0$. Therefore, $f(x)=\sqrt{x}$ is continuous at all $x\ge 0$. □

Exponential functions

Definition (Exponential function)

Let $a>0$ be a real number. The function defined by

$$f(x)=a^x$$

is called an exponential function with base $a$. In particular, when we simply say the exponential function, the base is $e$ (Napier's constant).

Exponential functions are continuous everywhere on $\mathbb{R}$. If $a>1$, then $a^x$ is a strictly increasing function. If $0<a<1$, then $a^x$ is a strictly decreasing function. If $a=1$, then $a^x=1$ for all $x\in \mathbb{R}$.

But what do we mean by $a^x$, exactly? Review how we introduced $e^x$ through $\exp (x)$.

See also: $\log$ and $e$

Logarithm

First, we provide the following theorem without proof.

Theorem

1. Let $f(x)$ be a continuous and strictly monotone increasing function defined on an interval $I$. Then $f(x)$ has the inverse $f^{-1}(x)$ which is also continuous and strictly monotone increasing.
2. Let $f(x)$ be a continuous and strictly monotone decreasing function defined on an interval $I$. Then $f(x)$ has the inverse $f^{-1}(x)$ which is also continuous and strictly monotone decreasing.

Remark. If $f(x)$ is a strictly monotone function (either increasing or decreasing) on an interval $I$, then the function is a bijection from $I$ to $f(I)$. For every bijection, there exists an inverse map that is also bijective. □

Using this theorem, we can see that for each exponential function $f(x)=a^x$, there is its inverse function $f^{-1}(x)$ which we define as the logarithmic function with base $a$ denoted $f^{-1}(x)={\log }_a(x)$. When the base is $e$ (Napier's constant), ${\log }_e(x)$, this is the logarithm we have defined earlier and we often omit the base to write simply $\log (x)$ or $\ln (x)$.

Theorem 

We have the following limits.

1. $$\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1.$$
2. $$\underset{x\to 0}{lim}\frac{e^x-1}{x}=1.$$

Proof. 

1. $$\underset{x\to 0}{lim}\frac{\log (1+x)}{x}=\underset{x\to 0}{lim}\log (1+x{)}^{\frac{1}{x}}=\log e=1.$$
2. Let $t=e^x-1$. Then $x=\log (1+t)$ As $x\to 0$, $t\to 0$ so $$\underset{x\to 0}{lim}\frac{e^x-1}{x}=\underset{t\to 0}{lim}\frac{t}{\log (1+t)}=1$$ using the result of Part 1.

■

Trigonometric functions

We already know $\sin$ and $\cos$. The tangent function is defined as

$$\tan x=\frac{\sin x}{\cos x},x\in \mathbb{R}\setminus \{(n+\frac{1}{2})\pi \mid n\in \mathbb{Z}\}.$$

Quiz. Why do we exclude the points $x=(n+\frac{1}{2})\pi ,n\in \mathbb{Z}$, from the domain of the tangent function? □

$\sin$ and $\cos$ have the fundamental period of $2\pi$ whereas $\tan$ has the fundamental period of $\pi$. That is, for $n\in \mathbb{Z}$,

$$\begin{array}{rcl}\sin (x+2\pi n)& =& \sin x,\\ \cos (x+2\pi n)& =& \cos x,\\ \tan (x+\pi n)& =& \tan x.\end{array}$$

We give here the following result without proof (for the moment):

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1.$$

Example. $$\begin{array}{rcl}\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}& =& \underset{x\to 0}{lim}\frac{(1-\cos x)(1+\cos x)}{x^2(1+\cos x)}\\ & =& \underset{x\to 0}{lim}\frac{1-{\cos }^{2}x}{x^2(1+\cos x)}\\ & =& \underset{x\to 0}{lim}{\left(\frac{\sin x}{x}\right)}^2\frac{1}{1+\cos x}=\frac{1}{2}.\end{array}$$

Inverse trigonometric functions

Since $\sin$, $\cos$, and $\tan$ are periodic functions (and hence not monotone), they don't have inverse functions. Nevertheless, by restricting their domains, we may define the inverse functions.

$\sin x$ is strictly monotone increasing on the closed interval $[-\frac{\pi }{2},\frac{\pi }{2}]$. Therefore it has an inverse function on this domain, which we define as $\arcsin x$. In other words, we consider $\sin x$ as a function

$$\sin :[-\frac{\pi }{2},\frac{\pi }{2}]\to [-1,1]$$

so its inverse function is

$$\arcsin :[-1,1]\to [-\frac{\pi }{2},\frac{\pi }{2}].$$

Similarly, we restrict the domain of $\cos x$ to $[0,\pi ]$ to define its inverse, which we call $\arccos x$:

$$\cos :[0,\pi ]\to [-1,1]$$

and

$$\arccos :[-1,1]\to [0,\pi ].$$

We restrict the domain of $\tan x$ to the open interval $(-\frac{\pi }{2},\frac{\pi }{2})$ to define its inverse, which we call $\arctan x$:

$$\tan :(-\frac{\pi }{2},\frac{\pi }{2})\to (-\mathrm{\infty },\mathrm{\infty }),$$

and

$$\arctan :(-\mathrm{\infty },\mathrm{\infty })\to (-\frac{\pi }{2},\frac{\pi }{2}).$$

Note that these definitions of inverse trigonometric functions are not unique. We could restrict the domains of the trigonometric functions differently. For example, we could restrict the domain of $\tan$ as

$$\tan :(\frac{\pi }{2},\frac{3\pi }{2})\to (-\mathrm{\infty },\mathrm{\infty })$$

to define

$$\arctan :(-\mathrm{\infty },\mathrm{\infty })\to (-\frac{\pi }{2},\frac{3\pi }{2}).$$

Example. Let us find the value of $\arcsin (\sin \frac{5\pi }{6})$. First note that $\sin \frac{5\pi }{6}=\sin \frac{\pi }{6}$ (why?) and $\frac{\pi }{6}\in [-\frac{\pi }{2},\frac{\pi }{2}]$. Therefore $\arcsin (\sin \frac{5\pi }{6})=\frac{\pi }{6}$. □

Hyperbolic functions

The hyperbolic cosine, hyperbolic sine, and hyperbolic tangent are defined, respectively, by

$$\begin{array}{rcl}\cosh x& =& \frac{e^x+e^{-x}}{2},\\ \sinh x& =& \frac{e^x-e^{-x}}{2},\\ \tanh x& =& \frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}.\end{array}$$

The domain of these functions is $\mathbb{R}$.

Why are their names similar to trigonometric functions? Note that

$${\cosh }^2\theta -{\sinh }^2\theta =1.$$

Recall that the equation $x^2-y^2=1$ represents the unit hyperbola on ${\mathbb{R}}^2$. 

See also: Unit hyperbola (Wikipedia) 

Compare this with

$${\cos }^2\theta +{\sin }^2\theta =1,$$

and $x^2+y^2=1$ represents the unit circle on ${\mathbb{R}}^2$.

Therefore, on ${\mathbb{R}}^2$, while $(\cos \theta ,\sin \theta )$ corresponds to a point on the unit circle centered at the origin, $(\cosh \theta ,\sinh \theta )$ corresponds to a point on the unit hyperbola.

Also, compare the following relations with the definitions of the hyperbolic functions: 

$$\begin{array}{rcl}\cos x& =& \frac{e^{ix}+e^{-ix}}{2},\\ \sin x& =& \frac{e^{ix}-e^{-ix}}{2i},\\ \tan x& =& \frac{\sin x}{\cos x}=\frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}.\end{array}$$

Example. 

$$\begin{array}{rcl}\frac{1}{{\sinh }^{2}x}-\frac{1}{{\tanh }^{2}x}& =& \frac{4}{(e^x-e^{-x}{)}^2}-\frac{(e^x+e^{-x}{)}^2}{(e^x-e^{-x}{)}^2}\\ & =& \cdots =-1.\end{array}$$

(You should fill in the details.) □

Example. 

$$\sinh (x+y)=\sinh x\cosh y+\cosh x\sinh y,$$

and

$$\sinh (x-y)=\sinh x\cosh y-\cosh x\sinh y.$$

(You should prove these equalities.) □