7.2. Difference Quotient Functions

Now we are going to develop a general idea from the results of our introductory example. As the secant gradients played a major role there our first step will be to asign to any function  f, with respect to a preselected point a, a survey of all these secant gradients.

Consider:

• ${\displaystyle m_a:A\backslash \{a\}\to ℝ}$  and  ${\displaystyle m_a(x)=\frac{f(x)-f(a)}{x-a}}$ .
  As constructed, the preselected number a  is no member of the domain of any difference quotient function.
   

• Occasionally we use ${\displaystyle m_{f,a}}$ instead of  ${\displaystyle m_a}$ to focus on the relation to  f.

• The values of  ${\displaystyle m_a}$ are the gradient numbers of the secants.

  Sometimes the value ${\displaystyle m_a(x)}$ is used for measuring the alteration behaviour of a function and is called the rate of change (or average rate of change) of  f between a and x in this context.

• A special notation is used in physics: When studying a moving particle for example, the distance covered within t units of time is usually denoted by ${\displaystyle s(t)}$. The rates of change ${\displaystyle \frac{s(t_2)-s(t_1)}{t_2-t_1}}$ respective to the function ${\displaystyle t\mapsto s(t)}$, i.e. the quotients

  ${\displaystyle \frac{\text{distance covered}}{\text{time spent}}}$,

  are usually denoted by ${\displaystyle \mathrm{\Delta }v=\frac{\mathrm{\Delta }s}{\mathrm{\Delta }t}}$

  i

  v for velocity from the Latin word velocitas.

  and are addressed as the particle's average speed between the timepoints ${\displaystyle t_1}$ and ${\displaystyle t_2}$ or the waypoints ${\displaystyle s(t_1)}$ and ${\displaystyle s(t_2)}$ respectively.
   

  
   

The following example shows the difference quotient functions of several standard maps. The transformations carried out in addition are used in the next chapter.

Example:  We calculate the difference quotient functions in respect of a for a linear function  $mX+b$  with an arbitrary a:   ${\displaystyle m_a=\frac{mX+b-(ma+b)}{X-a}=\frac{m(X-a)}{X-a}=m\frac{X-a}{X-a}}$ . [7.2.2]   for a monomial  $X^n$  with arbitrary a and $n\in \mathbb{N}$ :   ${\displaystyle \begin{array}{ll}{m}_a\hfill & =\frac{X^n-a^n}{X-a}\hfill \\ \hfill & \underset{(\ast )}{=}\frac{(X-a)(X^{n-1}+a{X}^{n-2}+\cdots +a^{n-2}X+a^{n-1})}{X-a}\hfill \\ \hfill & =\frac{(X-a)\sum _{i=0}^{n-1}{a}^i{X}^{n-i-1}}{X-a}\text{ .}\hfill \end{array}}$ [7.2.3] We show the equality (*) by induction on n.   for the reciprocal function  ${\displaystyle \frac{1}{X}}$ with $a\ne 0$ :   ${\displaystyle m_a=\frac{\frac{1}{X}-\frac{1}{a}}{X-a}=\frac{a-X}{aX(X-a)}=-\frac{X-a}{aX(X-a)}}$ . [7.2.4]   for the root function  $\sqrt{X}$ with $a\ge 0$ :   ${\displaystyle m_a=\frac{\sqrt{X}-\sqrt{a}}{X-a}=\frac{X-a}{(\sqrt{X}+\sqrt{a})(X-a)}}$ . [7.2.5]   for the absolute value function  $|X|$  for $a=0$ :   ${\displaystyle m_0=\frac{|X|-|0|}{X-0}=\frac{|X|}{X}}$ . [7.2.6]

It is always rewarding to check how far newly built notions cooperate with basic arithmetics. In many cases this leads to powerful technics. The following proposition shows that the mapping  ${\displaystyle f\mapsto m_{f,a}}$  respects all of them.

Proposition:  Let  $A,B\subset ℝ$ and $a\in A\cap B$. For  $f:A\to ℝ$ and $g:B\to ℝ$ we have:   1.${\displaystyle m_{f+g,a}=m_{f,a}+m_{g,a}}$ [7.2.7] 2.${\displaystyle m_{f-g,a}=m_{f,a}-m_{g,a}}$ [7.2.8] 3.${\displaystyle m_{f\cdot g,a}=m_{f,a}\cdot g+f(a)\cdot m_{g,a}}$ [7.2.9] 4.${\displaystyle m_{\frac{f}{g},a}=\frac{m_{f,a}\cdot g-f\cdot m_{g,a}}{g(a)\cdot g}}$ [7.2.10] Proof:   1. ►  ${\displaystyle m_{f+g,a}=\frac{(f+g)-(f+g)(a)}{X-a}=\frac{f-f(a)}{X-a}+\frac{g-g(a)}{X-a}=m_{f,a}+m_{g,a}}$ . 2. ►  The calculation is a copy of 1. 3. ►  We need the basic adding zero trick, this time: $0=-f(a)g+f(a)g$.   ${\displaystyle \begin{array}{ll}{m}_{f\cdot g,a}\hfill & =\frac{fg-fg(a)}{X-a}\hfill \\ \hfill & =\frac{fg-f(a)g+f(a)g-f(a)g(a)}{X-a}\hfill \\ \hfill & =\frac{(f-f(a))g}{X-a}+\frac{f(a)(g-g(a))}{X-a}\hfill \\ \hfill & =m_{f,a}\cdot g+f(a)\cdot m_{g,a}\text{ .}\hfill \end{array}}$ 4. ►  Again we add zero now in the shape of  $0=fg-fg$.   ${\displaystyle \begin{array}{ll}{m}_{\frac{f}{g},a}\hfill & =\frac{\frac{f}{g}-\frac{f}{g}(a)}{X-a}\hfill \\ \hfill & =\frac{g(a)f-f(a)g}{(X-a)g(a)g}\hfill \\ \hfill & =\frac{fg-f(a)g-fg+g(a)f}{(X-a)g(a)g}\hfill \\ \hfill & =\frac{\frac{(f-f(a))\cdot g}{X-a}-\frac{f\cdot (g-g(a))}{X-a}}{g(a)g}\hfill \\ \hfill & =\frac{m_{f,a}\cdot g-f\cdot m_{g,a}}{g(a)\cdot g}\text{ .}\hfill \end{array}}$

7.1.

7.3.