I was learning about derivatives and I saw that the slope of the secant line between two points is the average rate of change of the function between the two points. But, the average, as we normally define it, is the sum of values divided by the number of values. I would like to know how these two ways of defining an average for the rate of change interconnect and are same
1
-
1$\begingroup$ $$\overline{\text{f}\space'}=\frac{1}{\text{b}-\text{a}}\int\limits_\text{a}^\text{b}\text{f}\space'\left(x\right)\space\text{d}x=\frac{\text{f}\left(\text{b}\right)-\text{f}\left(\text{a}\right)}{\text{b}-\text{a}}\tag1$$ $\endgroup$Jan Eerland– Jan Eerland2025-11-27 13:31:06 +00:00Commented Nov 27 at 13:31
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Suppose we are talking about a differentiable function $f$ with $y=f(x)$. If the closed interval in $x$ is divided up into $n$ equal parts $[x_0,x_1]$, $[x_1,x_2]$ ... , $[x_{n-1},x_n]$ then the gradients of the parts are approximately $\frac{y_1-y_0}{x_1-x_o}$, $\frac{y_2-y_1}{x_2-x_1}$, ... ,$\frac{y_n-y_{n-1}}{x_n-x_{x-1}}$. The denominators are all equal, so we can the numerators, getting $y_n-y_0$. We need to divide by the common denominator and also divide by $n$ the number of intervals, so we need to divide by $x_n-x_0$. But this quotient is the gradient of the secant.
Of course this uses approximations for the $n$ individual steps, but as $n$ is made larger the approximations become better. Because the sum of the approximations is always the same, the limit of the approximate sums has the same value as well. So the gradient of the secant is the average of the infinitely many infinitely small lines comprising the curve.
-
$\begingroup$ Thanks! One clarification: You can use a limit here right? Because we have infinite points and to accommodate for that, in the summation there has to be a limit, am I right? $\endgroup$Tasd 541– Tasd 5412025-11-27 13:52:13 +00:00Commented Nov 27 at 13:52
-
$\begingroup$ Yes. In this case we have a sequence of sums, but all the terms in the sequence are equal, so the limit is the same as well. $\endgroup$Peter– Peter2025-11-27 14:02:36 +00:00Commented Nov 27 at 14:02
-
$\begingroup$ Could you show how I would manipulate that mathematically? $\endgroup$Tasd 541– Tasd 5412025-11-27 14:06:48 +00:00Commented Nov 27 at 14:06
-
$\begingroup$ It is just a constant sequence. The limit of the sequence {5, 5, 5, ... } is 5. $\endgroup$Peter– Peter2025-11-27 14:09:53 +00:00Commented Nov 27 at 14:09