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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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    What is the official definition of differential you are referring to? – user Sep 15 '18 at 08:57
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    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 –  Sep 15 '18 at 08:58
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    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. – user Sep 15 '18 at 09:00
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    Refer to https://en.wikipedia.org/wiki/Differential_(mathematics) – user Sep 15 '18 at 09:00
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    and to https://en.wikipedia.org/wiki/Differential_of_a_function – user Sep 15 '18 at 09:03
  • and to https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio/21209#21209 – user Sep 15 '18 at 09:05
  • edited the title. just wanted to know that my understanding of if you divide a an interval of length L into a bunch of dxs then the total number of xs could be seen as n = L/dx –  Sep 15 '18 at 09:05
  • Here is a short description of the difference between derrivative and differentiable. – rtybase Sep 15 '18 at 09:07
  • The initial statement: "Given an interval of size $L$, we could express $L$ as follows :$L=\alpha.dx$" is completely unclear and it seems to be not true. What do you mean exactly? – user Sep 15 '18 at 09:08
  • I mean that if you have, for example, a rope of length $L$ and you divide it into bits of size $dx$ then you could express its length as follows $L = \alpha . dx$ i.e $L = dx + dx + dx + ...$ –  Sep 15 '18 at 09:10
  • The number of $x$s cannot be invoked, that is meaningless. Furthermore, introducing two "variables" $L$ and $\alpha$ gives no insight in what a differential is. –  Sep 15 '18 at 09:22
  • I meant $dx$ by differential, the infinitesimal change –  Sep 15 '18 at 09:30

2 Answers2

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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

user
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  • 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 : $\sum \frac{f(x)}{n} = \frac{1}{L} \sum f(x).dx = \frac{1}{L} \int f(x).dx$ –  Sep 15 '18 at 16:21
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    This the concept of Riemann sum, take a look here https://math.stackexchange.com/q/750953/505767 – 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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    In the last maybe you should use $\approx$? – user Sep 15 '18 at 09:56
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    @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). –  Sep 15 '18 at 09:58
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    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 $x$. – user Sep 15 '18 at 10:03
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    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. – user Sep 15 '18 at 10:04
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    @gimusi: it is non-official. But I have subsituted approximation with equality. –  Sep 15 '18 at 10:29
  • Indeed as you noted I’m not claiming that it is official, I’ve only claimed that your notation is a little bit confusing at a first sight because I’ m not used to it. After your explanation now it’s clear what it stands for. – user Sep 15 '18 at 10:29
  • Yes equality is then the appropriate term in that context. – user Sep 15 '18 at 10:36