Differential and Integral Calculus
6. Elementary functions
This chapter gives some background to the concept of a function. We also consider some elementary functions from a (possibly) new viewpoint. Many of these should already be familiar from high school mathematics, so in some cases we just list the main properties.
Functions
Definition: Function
A function \(f\colon A\to B\) is a rule that determines for each element \(a\in A\) exactly one element \(b\in B\). We write \(b=f(a)\).
Definition: Domain and codomain
In the above definition of a function \(A=D_f\) is the domain (of definition) of the function \(f\) and \(B\) is called the codomain of \(f\).
Definition: Image of a function
The image of \(f\) is the subset
\(f[A]= \{ f(a) \mid a\in A\}\) of \(B\). An alternative name for image is range.
For example, \(f\colon \mathbb{R}\to\mathbb{R}\), \(f(x)=x^2\), has codomain \(\mathbb{R}\), but its image is \(f[\mathbb{R} ] =[0,\infty[\).
The function in the previous example can also be defined as \(f\colon \mathbb{R}\to [0,\infty[\), \(f(x)=x^2\), and then the codomain is the same as the image. In principle, this modification can always be done, but it is not reasonable in practice.
Example: Try to do the same for \(f\colon \mathbb{R}\to\mathbb{R}\), \(f(x)=x^6+x^2+x\), \(x\in\mathbb{R}\).
If the domain \(A\subset \mathbb{R}\) then \(f\) is a function of one (real) variable: the main object of study in this course.
If \(A\subset \mathbb{R}^n\), \(n\ge 2\), then \(f\) is a function of several variables (a multivariable function)
Inverse functions
Definition: Injection, surjection and bijection
A function \(f\colon A \to B\) is- injective (one-to-one) if it has different values at different points; i.e. \[x_1\neq x_2 \Rightarrow f(x_1)\neq f(x_2),\] or equivalently \[f(x_1)= f(x_2) \Rightarrow x_1=x_2.\]
- surjective (onto) if its image is the same as codomain, i.e. \(f[A]=B\)
- bijective (one-to-one and onto) if it is both injective and surjective.
Observe: A function becomes surjective if all redundant points of the codomain are left out. A function becomes injective if the domain is reduced so that no value of the function is obtained more than once.
Another way of defining these concepts is based on the number of solutions to an equation:
Definition
For a fixed \(y\in B\), the equation \(y=f(x)\) has
- at most one solution \(x\in A\) if \(f\) is injective
- at least one solution \(x\in A\) if \(f\) is surjective
- exactly one solution \(x\in A\) if \(f\) on bijective.
Definition: Inverse function
If \(f\colon A \to B\) is bijective, then it has an inverse \(f^{-1}\colon B \to A\), which is uniquely determined by the condition \[y=f(x) \Leftrightarrow x = f^{-1}(y).\]
The inverse satisfies \(f^{-1}(f(a))=a\) for all \(a\in A\) and \(f(f^{-1}(b))=b\) for all \(b\in B\).
The graph of the inverse is the mirror image of the graph of \(f\) with respect to the line \(y=x\): A point \((a,b)\) lies on the graph of \(f\) \(\Leftrightarrow\) \(b=f(a)\) \(\Leftrightarrow\) \(a=f^{-1}(b)\) \(\Leftrightarrow\) the point \((b,a)\) lies on the graph of \(f^{-1}\). The geometric interpretation of \((a,b)\mapsto (b,a)\) is precisely the reflection with respect to \(y=x\).
If \(A \subset \mathbb{R}\) and \(f\colon A\to \mathbb{R}\) is strictly monotone, then the function \(f\colon A \to f[A]\) has an inverse.
If here \(A\) is an interval and \(f\) is continuous, then also \(f^{-1}\) is is continuous in the set \(f[A]\).
Theorem: Derivative of the inverse
Let \(f\colon \, ]a,b[\, \to\, ]c,d[\) be differentiable and bijective, so that it has an inverse \(f^{-1}\colon \, ]c,d[\, \to\, ]a,b[\). As the graphs \(y=f(x)\) and \(y=f^{-1}(x)\) are mirror images of each other, it seems geometrically obvious that also \(f^{-1}\) is differentiable, and we actually have \[ \left(f^{-1}\right)'(x)=\frac{1}{f'(f^{-1}(x))}, \] if \(f'(f^{-1}(x))\neq 0\).
Differentiate both sides of the equation \begin{align} f(f^{-1}(x)) &= x \\ \Rightarrow f'(f^{-1}(x))\left(f^{-1}\right)'(x) &= Dx = 1, \end{align} and solve for \(\left(f^{-1}\right)'(x)\).
\(\square\)
Note. \(f'(f^{-1}(x))\) is the derivative of \(f\) at the point \(f^{-1}(x)\).
Transcendental functions
Trigonometric functions
Unit of measurement of an angle = rad: the arclength of the arc on the unit circle, that corresponds to the angle.
\(\pi\) rad = \(180\) degrees, i.e. \(1\) rad = \(180/\pi \approx 57,\! 3\) degrees
The functions \(\sin x, \cos x\) are defined in terms of the unit circle so that \((\cos x,\sin x)\), \(x\in [0,2\pi]\), is the point on the unit circle corresponding to the angle \(x\in\mathbb{R}\), measured counterclockwise from the point \((1,0)\). \[\tan x = \frac{\sin x}{\cos x}\ (x\neq \pi /2 +n\pi),\] \[\cot x = \frac{\cos x}{\sin x}\ (x\neq n\pi)\]
Periodicity: \[\sin (x+2\pi) = \sin x,\ \cos (x+2\pi)=\cos x,\] \[\tan (x+\pi) = \tan x\]
\(\sin 0 = 0\), \(\sin (\pi/2)=1\)
\(\cos 0=1\), \(\cos (\pi/2)= 0\)
Parity: \(\sin\) and \(\tan\) are odd functions, \(\cos\) is an even function: \[\sin (-x) = -\sin x,\] \[\cos(-x) = \cos x,\] \[\tan (-x) = -\tan x.\]
\(\sin^2 x + \cos^2 x = 1\) for all \(x\in\mathbb{R}\)
Proof: Pythagorean Theorem.
Addition formulas:
\(\sin (x+y) = \sin x \cos y +\cos x \sin y\)
\(\cos (x+y) = \cos x \cos y -\sin x \sin y\)
Basic properties (from the unit circle!)
Proof: Geometrically, or more easily with vectors and matrices.
Derivatives: \[ D(\sin x) = \cos x,\ \ D(\cos x) = -\sin x \]
Example
It follows that the functions \(y(t)=\sin (\omega t)\) and \(y(t)=\cos (\omega t)\) satisfy the differential equation \[ y''(t)+\omega^2y(t)=0, \] that models harmonic oscillation. Here \(t\) is the time variable and the constant \(\omega>0\) is the angular frequency of the oscillation. We will see later that all the solutions of this differential equation are of the form \[ y(t)=A\cos (\omega t) +B\sin (\omega t), \] with \(A,B\) constants. They will be uniquely determined if we know the initial location \(y(0)\) and the initial velocity \(y'(0)\). All solutions are periodic and their period is \(T=2\pi/\omega\).
where \(t\) is the elapsed time in seconds
Arcus functions
The trigonometric functions have inverses if their domain and codomains are chosen in a suitable way.
The Sine function \[ \sin \colon [-\pi/2,\pi/2]\to [-1,1] \] is strictly increasing and bijective.
The Cosine function \[ \cos \colon [0,\pi] \to [-1,1] \] is strictly decreasing and bijective.
The tangent function \[ \tan \colon ]-\pi/2,\pi/2[\, \to \mathbb{R} \] is strictly increasing and bijective.
Arcus functions
Inverses: \[\arctan \colon \mathbb{R}\to \ ]-\pi/2,\pi/2[,\] \[\arcsin \colon [-1,1]\to [-\pi/2,\pi/2],\] \[\arccos \colon [-1,1]\to [0,\pi]\]
This means: \[x = \tan \alpha \Leftrightarrow \alpha = \arctan x \ \ \text{for } \alpha \in \ ]-\pi/2,\pi/2[ \] \[x = \sin \alpha \Leftrightarrow \alpha = \arcsin x \ \ \text{for } \alpha \in \, [-\pi/2,\pi/2] \] \[x = \cos \alpha \Leftrightarrow \alpha = \arccos x \ \ \text{for } \alpha \in \, [0,\pi] \]
The graphs of \(\tan\) and \(\arctan\).
Derivatives of the arcus functions
\[D \arctan x = \frac{1}{1+x^2},\ x\in \mathbb{R} \tag{1}\] \[D\arcsin x = \frac{1}{\sqrt{1-x^2}},\ -1 < x < 1 \tag{2}\] \[D\arccos x = \frac{-1}{\sqrt{1-x^2}},\ -1 < x < 1 \tag{3}\]
Note. The first result is very useful in integration.
Here we will only prove the first result (1). By differentiating both sides of the equation \(\tan(\arctan x)=x\) for \(x\in \mathbb{R}\): \[\bigl( 1+\tan^2(\arctan x)\bigr) \cdot D(\arctan x) = D x = 1\] \[\Rightarrow D(\arctan x)= \frac{1}{1+\tan^2(\arctan x)}\] \[=\frac{1}{1+x^2}.\]
The last row follows also directly from the formula for the derivative of an inverse.
Example
Show that \[ \arcsin x +\arccos x =\frac{\pi}{2} \] for \(-1\le x\le 1\).
Example
Derive the addition formula for tan, and show that \[ \arctan x+\arctan y = \arctan \frac{x+y}{1+xy}. \]
Solutions: Voluntary exercises. The first can be deduced by looking at a rectangular triangle with the length of the hypotenuse equal to 1 and one leg of length \(x\).
Introduction: Radioactive decay
Let \(y(t)\) model the number of radioactive nuclei at time \(t\). During a short time interval \(\Delta t\) the number of decaying nuclei is (approximately) directly proportional to the length of the interval, and also to the number of nuclei at time \(t\): \[ \Delta y = y(t+\Delta t)-y(t) \approx -k\cdot y(t)\cdot \Delta t. \] The constant \(k\) depends on the substance and is called the decay constant. From this we obtain \[ \frac{\Delta y}{\Delta t} \approx -ky(t), \] and in the limit as \(\Delta t\to 0\) we end up with the differential equation \(y'(t)=-ky(t)\).
Exponential function
Definition: Euler's number
Euler's number (or Napier's constant) is defined as \[e = \lim_{n\to \infty} \left( 1+\frac{1}{n}\right) ^n = 1+1+\frac{1}{2!}+\frac{1}{3!} +\frac{1}{4!} +\dots \] \[\approx 2,\! 718281828459\dots\]
Definition: Exponential function
The Exponential function exp: \[ \exp (x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}= \lim_{n\to \infty} \left( 1+\frac{x}{n}\right) ^n = e^x. \] This definition (using the series expansion) is based on the conditions \(\exp'(x)=\exp(x)\) and \(\exp(0)=1\), which imply that \(\exp^{(k)}(0)=\exp(0)= 1\) for all \(k\in\mathbb{N}\), so the Maclaurin series is the one above.
The connections between different expressions are surprisingly tedious to prove, and we omit the details here. The main steps include the following:
Define \(\exp\colon\mathbb{R}\to\mathbb{R}\), \[ \exp (x) =\sum_{k=0}^{\infty}\frac{x^k}{k!}. \] This series converges for all \(x\in\mathbb{R}\) (ratio test).
Show: exp is differentiable and satisfies \(\exp'(x)=\exp(x)\) for all \(x\in \mathbb{R}\). (This is the most difficult part but intutively rather obvious, because in practice we just differentiate the series term by term like a polynomial.)
It has the following properties \(\exp (0)=1\), \[ \exp (-x)=1/\exp (x) \text{ and } \exp (x+y)=\exp (x)\, \exp(y) \] for all \(x,y\in \mathbb{R}\).
These imply that \(\exp (p/q)=(\exp (1))^{p/q}\) for all rational numbers \(p/q\in \mathbf{Q}\).
By continuity \[ \exp (x) =(\exp (1))^x \] for all \(x\in \mathbb{R}\).
Since \[ \exp (1) = \sum_{k=0}^{\infty}\frac{1}{k!} =\lim_{n\to \infty} \left( 1+\frac{1}{n}\right) ^n=e, \] we obtain the form \(e^x\).
\(\square\)?
Corollary
It follows from above that \(\exp\colon\mathbb{R}\to\, ]0,\infty[\) is strictly increasing, bicective, and \[ \lim_{x\to\infty}\exp(x) = \infty,\ \lim_{x\to-\infty}\exp(x) = 0,\ \lim_{x\to\infty}\frac{x^n}{\exp (x)} = 0 \text{ for all } n\in \mathbf{N}. \]
From here on we write \(e^x=\exp(x)\). Properties:
- \(e^0 = 1\)
- \(e^x >0\)
- \(D(e^x) = e^x\)
- \(e^{-x} = 1/e^x\)
- \((e^x)^y = e^{xy}\)
- \(e^xe^y =e^{x+y}\)
Differential equation \(y'=ky\)
Theorem
Let \(k\in\mathbb{R}\) be a constant. All solutions \(y=y(x)\) of the ordinary differenial equation (ODE) \[ y'(x)=ky(x),\ x\in \mathbb{R}, \] are of the form \(y(x)=Ce^{kx}\), where \( C\) is a constant. If we know the value of \(y\) at some point \(x_0\), then the constant \(C\) will be uniquely determined.
Suppose that \(y'(x)=ky(x)\). Then \[D(y(x)e^{-kx})= y'(x)e^{-kx}+y(x)\cdot (-ke^{-kx})\] \[= ky(x)e^{-kx}-ky(x)e^{-kx}=0\] for all \(x\in\mathbf{R}\), so that \(y(x)e^{-kx}=C=\) constant. Multiplying both sides with \(e^{kx}\) we obtain \(y(x)=Ce^{kx}\).
\(\square\)
Euler's formula
Definition: Complex numbers
Imaginary unit \(i\): a strange creature satisfying \(i^2=-1\). The complex numbers are of the form \(z=x+iy\), where \(x,y\in \mathbb{R}\). We will return to these later.
Theorem: Euler's formula
If we substitute \(ix\) as a variable in the expontential fuction, and collect real terms separately, we obtain Euler's formula \[e^{ix}=\cos x+i\sin x.\]
\(\square\)
As a special case we have Euler's identity \(e^{i\pi}+1=0\). It connects the most important numbers \(0\), \(1\), \(i\), \(e\) ja \(\pi\) and the three basic operations sum, multiplication, and power.
Using \(e^{\pm ix}=\cos x\pm i\sin x\) we can also derive the expressions \[ \cos x=\frac{1}{2}\bigl( e^{ix}+e^{-ix}\bigr),\ \sin x=\frac{1}{2i}\bigl( e^{ix}-e^{-ix}\bigr), \ x\in\mathbb{R}. \]
Logarithms
Definition: Natural logarithm
Natural logarithm is the inverse of the exponential function: \[ \ln\colon \ ]0,\infty[ \ \to \mathbb{R} \]
Note. The general logarithm with base \(a\) is based on the condition \[ a^x = y \Leftrightarrow x=\log_a y \] for \(a>0\) and \(y>0\).
Beside the natural logarithm, in applications also appear the Briggs logarithm with base 10: \(\lg x = \log_{10} x\), and the binary logarithm with base 2: \({\rm lb}\, x =\log_{2} x\).
Usually (e.g. in mathematical software) \(\log x\) is the same as \(\ln x\).
Properties of the logarithm:
- \(e^{\ln x} = x\) for \(x>0\)
- \(\ln (e^x) =x\) for \(x\in\mathbb{R}\)
- \(\ln 1=0\), \(\ln e = 1\)
- \(\ln (a^b) = b\ln a\) if \(a>0\), \(b\in\mathbb{R}\)
- \(\ln (ab) = \ln a+\ln b\), if \(a,b>0\)
- \(D\ln |x|=1/x\) for \(x\neq 0\)
These follow from the corresponding properties of exp.
Example
Substituting \(x=\ln a\) and \(y=\ln b\) to the formula
\(e^xe^y =e^{x+y}\) we obtain \(ab =e^{\ln a+\ln b},\)
so that \(\ln (ab) = \ln a +\ln b\).Hyperbolic functions
Definition: Hyperbolic functions
Hyperbolic sine sinus hyperbolicus \(\sinh\), hyperbolic cosine cosinus hyperbolicus \(\cosh\) and hyperbolic tangent \(\tanh\) are defined as \[\sinh \colon \mathbb{R}\to\mathbb{R}, \ \sinh x=\frac{1}{2}(e^x-e^{-x})\] \[\cosh \colon \mathbb{R}\to [1,\infty[,\ \cosh x=\frac{1}{2}(e^x+e^{-x})\] \[\tanh \colon \mathbb{R}\to \ ]-1,1[, \ \tanh x =\frac{\sinh x}{\cosh x}\]
Properties: \(\cosh^2x-\sinh^2x=1\); all trigonometric have their hyperbolic counterparts, which follow from the properties \(\sinh (ix)=i\sin x\), \(\cosh (ix)=\cos x\). In these formulas, the sign of \(\sin^2\) will change, but the other signs remain the same.
Derivatives: \(D\sinh x=\cosh x\), \(D\cosh x=\sinh x\).
Hyperbolic inverse functions: the so-called area functions; area and the shortening ar refer to a certain geometrical area related to the hyperbola \(x^2-y^2=1\): \[\sinh^{-1}x=\text{arsinh}\, x=\ln\bigl( x+\sqrt{1+x^2}\, \bigr) ,\ x\in\mathbb{R} \] \[\cosh^{-1}x=\text{arcosh}\, x=\ln\bigl( x+\sqrt{x^2-1}\, \bigr) ,\ x\ge 1\]
Derivatives of the inverse functions: \[D \sinh^{-1}x= \frac{1}{\sqrt{1+x^2}} ,\ x\in\mathbb{R} \] \[D \cosh^{-1}x= \frac{1}{\sqrt{x^2-1}} ,\ x > 1.\]