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 \(x\)

\[f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\]

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 \(h(x) \neq 0\), the function \(f(x)\) defined by

\[f(x) = \frac{g(x)}{h(x)}\]

is continuous on \(\{x \mid x \in \mathbb{R}, h(x) \neq 0\}\). 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 \(x \geq 0\) is continuous.

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

\[\begin{eqnarray*} |\sqrt{x} - \sqrt{a}| &=& \left| \frac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{\sqrt{x} + \sqrt{a}}\right|\\ &=& \frac{|x - a|}{\sqrt{x} + \sqrt{a}}\\ &\leq & \frac{|x - a|}{\sqrt{a}} < \frac{\delta}{\sqrt{a}} = \varepsilon. \end{eqnarray*}\]

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

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. \[\lim_{x\to 0}\frac{\log(1 + x)}{x} = 1.\]
2. \[\lim_{x\to 0}\frac{e^x - 1}{x} = 1.\]

Proof. 

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

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Trigonometric functions

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

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

Quiz. Why do we exclude the points \(x = \left(n + \frac{1}{2}\right)\pi, n \in \mathbb{Z}\), 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 \(n\in\mathbb{Z}\),

\[ \begin{eqnarray} \sin(x + 2\pi n) &=& \sin x,\\ \cos(x + 2\pi n) &=& \cos x,\\ \tan(x + \pi n) &=& \tan x. \end{eqnarray} \]

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

\[\lim_{x\to 0}\frac{\sin x}{x} = 1.\]

Example. \[ \begin{eqnarray} \lim_{x\to 0}\frac{1 - \cos x}{x^2} &=& \lim_{x\to 0}\frac{(1 - \cos x)(1 + \cos x)}{x^2(1 + \cos x)}\\ &=& \lim_{x\to 0}\frac{1 - \cos^2 x}{x^2(1 + \cos x)}\\ &=& \lim_{x\to 0}\left(\frac{\sin x}{x}\right)^2\frac{1}{1 + \cos x} = \frac{1}{2}. \end{eqnarray} \]

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 \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). 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

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

so its inverse function is

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

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 \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) to define its inverse, which we call \(\arctan x\):

\[\tan: \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to (-\infty, \infty),\]

and

\[\arctan:   (-\infty, \infty) \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).\]

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

\[\tan: \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \to (-\infty, \infty)\]

to define

\[\arctan:   (-\infty, \infty) \to \left(-\frac{\pi}{2}, \frac{3\pi}{2}\right).\]

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

Hyperbolic functions

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

\[ \begin{eqnarray} \cosh x &=& \frac{e^x + e^{-x}}{2},\\ \sinh x &=& \frac{e^x - e^{-x}}{2},\\ \tanh x &=& \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}. \end{eqnarray} \]

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: 

\[\begin{eqnarray} \cos x &=& \frac{e^{ix} + e^{-ix}}{2},\\ \sin x &=& \frac{e^{ix} - e^{-ix}}{2i},\\ \tan x &=& \frac{\sin x}{\cos x} = \frac{e^{ix} - e^{-ix}}{i(e^{ix} + e^{-ix})}. \end{eqnarray}\]

Example. 

\[\begin{eqnarray*} \frac{1}{\sinh^2x} - \frac{1}{\tanh^2x} &=& \frac{4}{(e^x - e^{-x})^2} -\frac{(e^x + e^{-x})^2}{(e^x - e^{-x})^2}\\ &=& \cdots = -1. \end{eqnarray*}\]

(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.) □