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In an attempt to explain the concept of the infinitesimal change, I have defined it as such :
Given an interval of size $L$, we could express $L$ as follows :$L=\alpha.dx$ thus, theoretically, $\alpha = \frac{L}{dx}$ could be seen as the total number of $x$s since $dx$ is infinitesimally small.
Do you think this to be a conceptually wrong explanation?

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    $\begingroup$ What is the official definition of differential you are referring to? $\endgroup$ – user Sep 15 '18 at 8:57
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    $\begingroup$ not really referring to anything, i'm just looking to know whether i got it right or wrong if i try explaining it this way $\endgroup$ – user531476 Sep 15 '18 at 8:58
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    $\begingroup$ Since you are talking about "a personal definition of the differential" you should clarify what is the other not personal definition of the differential you are reffering to, what is the context, otherwise your claim is really unclear. $\endgroup$ – user Sep 15 '18 at 9:00
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    $\begingroup$ Refer to en.wikipedia.org/wiki/Differential_(mathematics) $\endgroup$ – user Sep 15 '18 at 9:00
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    $\begingroup$ and to en.wikipedia.org/wiki/Differential_of_a_function $\endgroup$ – user Sep 15 '18 at 9:03
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Let consider the interval of length $L$ divided in $n$ equal parts of length $\Delta x$ therefore we have

$$\Delta x= \frac L n \iff L=n\cdot \Delta x$$

we can then define the infinitesimal interval as

$$dx:=\lim_{n\to \infty }\Delta x= \lim_{n\to \infty }\frac L n= \frac L {\lim_{n\to \infty }n}=0$$

but it is meaningless write $$dx=\frac L \infty$$

since $\infty$ is not a number

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  • $\begingroup$ let it as a variable that is by definition the total number of $x$s and you could work out with it the average value of a function for example : $\sum \frac{f(x)}{n} = \frac{1}{L} \sum f(x).dx = \frac{1}{L} \int f(x).dx$ $\endgroup$ – user531476 Sep 15 '18 at 16:21
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    $\begingroup$ This the concept of Riemann sum, take a look here math.stackexchange.com/q/750953/505767 $\endgroup$ – user Sep 15 '18 at 16:35
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In concrete terms, an infinitesimal change is so small that the linear approximation holds.

The local behavior of a smooth function can be described as

$$f(x+h)=f(x)+ah+r(x,h)$$

where $a$ is a constant and $r$ a remainder term that vanishes when $h$ tends to zero, $\lim_{h\to0}\dfrac{r(x,h)}h=0$.

Then $h$ infinitesimal means that we can just drop the remainder term and "admit" the equality

$$f(x+h)=f(x)+ah.$$

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    $\begingroup$ In the last maybe you should use $\approx$? $\endgroup$ – user Sep 15 '18 at 9:56
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    $\begingroup$ @gimusi: precisely not. When $h$ is infinitesimal, the equality holds exactly, by convention (this is another way to say that we replace the function by its differential). $\endgroup$ – Yves Daoust Sep 15 '18 at 9:58
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    $\begingroup$ Ah ok, I'm not used to this notation, I write $f(x+h)=f(x)+ah+o(h)$ and (for $f$ not linear) $f(x+h)\approx f(x)+ah$ and $l(x+h)=f(x)+ah$ as a linear approximation of $f(x)$ at $x$. $\endgroup$ – user Sep 15 '18 at 10:03
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    $\begingroup$ I mean it seems a little confusing use the same symbol f to indicate the function and its linear approximation. But it is a detail of course. $\endgroup$ – user Sep 15 '18 at 10:04
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    $\begingroup$ @gimusi: it is non-official. But I have subsituted approximation with equality. $\endgroup$ – Yves Daoust Sep 15 '18 at 10:29

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