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In many fields of physics and engineering, when we want to describe an infinitesimal, for example, the electric field, we could say $dE(x_i,y_i) = e^{jkr}...dx_0dy_0$

Since derivatives are not fractions, does it still make sense, in a mathematically rigorous way to talk about infinitesimals and would this be the correct notation of describing it? What would be the rigorous way of dealing with infinitesimals?

--Edit--

The history of this question went from "duplicate to an irrelevant question" but once reopened was closed for being "missing context" despite proper english, a clear yes or no question with already great answers (which would be impossible if the question was indecipherable). It seems like those with points have been abusing their powers an incorrectly closing questions that they simply don't like.

JobHunter69
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    This question has been asked on this site many times. – A. Thomas Yerger May 01 '18 at 03:52
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    I call things like "$dx$" differentials, not infinitesimals. – Angina Seng May 01 '18 at 03:53
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    @AlfredYerger Usually when a question has been asked many times, one marks it as a duplicate. Where is the question that is the duplicate of mine? – JobHunter69 May 01 '18 at 03:58
  • @LordSharktheUnknown Ok – JobHunter69 May 01 '18 at 04:04
  • Try reading any text on nonstandard analysis. A very short answer: infinitesimals can be defined rigorously but it isn’t east to do. When they are defined, the derivative can be treated as a quotient of infinitesimals. Again, any text on nonstandard analysis is a good start. – Ittay Weiss May 01 '18 at 05:29
  • @IttayWeiss what's that? – JobHunter69 May 01 '18 at 16:27
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    Rigour is overated. There are times when being rigorous is important, but most of the times it's just not even in pure maths. Treating differentials as if they are fractions is a very useful, simple and intuitive method. Most of the time it’s just a tool to arrive at an ODE or an integral describing some physical process which you can check is correct by other means so being rigorous in the derivation is not important. I would focus on learning how to do formal manipulations with differentials and learn the limitations instead of insiting on being rigorous. But that is just my 2cents. – Winther May 02 '18 at 16:50

3 Answers3

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There are basically two rigorous ways to deal with differentials. One is to treat them as differential forms. This is kind of an algebraic way of doing things, it sets rules for how you can manipulate differentials without trying to describe them as, say, "limits of small differences".

The other way is nonstandard analysis, of which there are at least two incompatible types. One of those is the one from which that name originated, which originally used the idea of a nonstandard model (from model theory) to construct a self-consistent theory containing infinite and infinitesimal "hyperreal" numbers. This originated with Robinson. A different formalism with the same semantics (which is probably easier to understand for non-logicians) was made later by Nelson.

An entirely different semantics arises in smooth infinitesimal analysis. SIA is somewhat alien to "mainstream" mathematicians, because it works in a field which has nonzero nilpotent elements (e.g. $dx \neq 0$ but $(dx)^2=0$). Such a thing is a contradiction in terms in classical logic, so this subject requires a weaker logic called intuitionistic logic in order to function (and even then $dx \neq 0$ is really "it cannot be proven that $dx=0$", a weaker statement).

Honestly, most mathematicians, scientists, and engineers don't need either one. It is better to learn methods for manipulating differentials in formal (i.e. "regarding only form", which is sort of like "non-rigorous") calculations. Optionally you can also learn proofs in standard analysis (which use finite but arbitrarily small numbers). These never explicitly use differentials.

Ian
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  • Can't one think of differentials as changes in the best linear approximation to a function? That's how I think of them. – Allawonder May 02 '18 at 16:25
  • @Allawonder That depends which meaning you intend. Thinking of $\Delta y$ as the change in the tangent line of a function over an interval of length $\Delta x$ does not actually require a new type of object, you simply have the relationship $\Delta y = f'(x) \Delta x$ among ordinary real numbers. On the other hand, the notation $dy=f'(x) dx$ only tells you how to define $dy$ in terms of $dx$, but it does not say what $dx$ itself is. – Ian May 02 '18 at 16:26
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    The statement $\Delta y=f'(x)\Delta x$ is not true. The right hand side is a differential while the left hand side is a difference. By definition, a differential only ever approximates a difference, never be exactly equal to it -- except the function is itself linear.

    I see you don't really understand me.

     – Allawonder May 02 '18 at 16:36
  • @Allawonder $\Delta y = f'(x) \Delta x$ is true along the tangent line, which is what I said. From this perspective you then go back and think about $\Delta x \to 0$ in order to understand the relationship between $\Delta y$ for the tangent line and $\Delta f$. But this never really involved differentials anywhere, because you never send $\Delta x \to 0$. – Ian May 02 '18 at 16:37
  • Well, of course, trivially. But no one uses the variable $y$ as used by you in the LHS to represent the tangent linear function, but the function $f$ whose differential is being expressed. – Allawonder May 02 '18 at 16:52
  • @Allawonder The point is that without introducing a new type of object, the only nice relationship you can get is the one along the tangent line. $dy=f'(x) dx$ tells you what $dy$ is in terms of $dx$ without telling you what $dx$ is. You might say "$dx$ is the differential of the identity function" but the differential of the identity function is defined in terms of itself. – Ian May 02 '18 at 16:54
  • No, no. There's no new object anywhere. A differential is sometimes defined as a function $df=df(x, dx)$. If that's what you mean by new object, I agree. But even at that we know what $dx$ is -- it is a finite change in the independent variable. This follows from the fact that for a linear function $g$, $$dg=\Delta g.$$ – Allawonder May 02 '18 at 16:59
  • @Allawonder Yes, so you're saying the first version of my first response to you and just writing it with different notation. The change in the dependent variable along the tangent line (which I called $\Delta y$) is $f'(x)$ times the change in the independent variable (which I called $\Delta x$). That does not yet say anything about the original function $f(x)$. – Ian May 02 '18 at 17:00
  • No. What I'm saying is that for any differentiable function $f=f(x)=y$, its differential is given as $$df(x)=dy=f'(x)dx,$$ where there's no strange new object anywhere since both $df$ and $dx$ are finite quantities. – Allawonder May 02 '18 at 17:03
  • @Allawonder This is only different from what I am saying in that you are introducing the symbol $df$ to refer to what I called $\Delta y$. – Ian May 02 '18 at 17:05
  • No, because $\Delta f$ is a difference, not a differential. As you can easily verify, $$df\approx \Delta x,$$ contrary to what you wrote. – Allawonder May 02 '18 at 17:07
  • @Allawonder I didn't write $\Delta f$. I wrote $\Delta y$ and I referred to it as "the change in the tangent line of a function over an interval of length $\Delta x$". – Ian May 02 '18 at 17:08
  • Well, that's so, but in that case your claim reduces to the differentials of linear functions alone. I was talking about any differentiable function. – Allawonder May 02 '18 at 17:11
  • @Allawonder No, in my notation $\Delta y$ is the change in the tangent line of $f$ at $x$ over $[x,x+\Delta x]$, which makes sense regardless of what $f$ is as long as it is differentiable at $x$. I hesitate to use $d$'s for this concept because I am inclined to reserve that notation for things that are at least "intuitively infinitesimal" if not literally infinitesimal. My motivation is that it tends to cause semantic arguments much like this one. However, I concede that $\Delta y$ is probably not good notation, at least if I don't draw the corresponding picture. – Ian May 02 '18 at 17:17
  • I don't see how an infinitesimal can be intuitive. But I think I'm beginning to see where the confusion came from -- your refusal to use the standard notation. I don't see any confusion it causes since the change in the tangent linear function ($y$ in your notation) can be thought of as related to the change in $f$ about $x$. Indeed it is, as is usually shown in beginning calculus. Yes, the notation might be initially confusing, but once one discovers that it leads to the identity $$df=\frac{df}{dx}dx,$$ I don't see why one should hesitate to use the $d$ notation any longer since it eventually – Allawonder May 02 '18 at 17:32
  • ...eventually all fits, although once again it has the tendency to confuse the beginning student (why are they using $d$'s in another way). But once one sees that the $d$ in the derivative operator $d/dx$ can never be alone, I don't see how any confusion should ensue. – Allawonder May 02 '18 at 17:36
  • @Allawonder What I mean by "intuitively infinitesimal" is that it is going to zero without actually being zero at any time. Thus to our intuition it is infinitesimal but it is not literally infinitesimal because we are not literally dealing with infinitesimals anywhere. Experienced analysts call this "epsilon" in informal parlance, and scientists and engineers usually mean this sort of thing when they write differentials. This conflict is why I avoid using $dx$ for finite numbers. – Ian May 02 '18 at 18:02
  • No, no, no. You're jumbling up many things here. Yes, an infinitesimal is a quantity that vanishes in the limit, but differentials are not infinitesimals! Also, no one writes $dx$, say, to mean a quantity whose limiting value is $0$. The differential of a function is a fixed finite quantity at each point of its domain that gives us qualitative information (and sometimes even sufficiently approximate quantitative information) of how it behaves near that point. – Allawonder May 02 '18 at 18:09
  • @Allawonder Apparently we have very different exposure. Physicists and engineers routinely write things like $dx,dV,dA$ in diagrams as labels of infinitesimal length elements etc. They write equations between these quantities which are not true for finite quantities because they neglect nonlinear corrections. Then they identify these infinitesimal elements with the differential $dx$ or similar appearing in an integral or derivative. Mathematicians in my experience often speak in similar language when they talk informally. Also note that the OP makes the same assertion (first sentence). – Ian May 02 '18 at 18:40
  • @Allawonder One such diagram can be seen at https://www.math24.net/barometric-formula/ where one works out the force balance on an infinitesimal packet of gas. – Ian May 02 '18 at 18:41
  • Out of curiosity (I haven't read any SIA except on S.E.), how does SIA handle the second derivative, if it says that dx^2 = 0? – johnnyb May 02 '18 at 21:41
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    @johnnyb There are different "tiers" of infinitesimals. A nilsquare infinitesimal say $\epsilon_1$ measures the first derivative: $f(x+\epsilon_1)=f(x)+f'(x)\epsilon_1$ (exact equality). A nilcube but not nilsquare infinitesimal say $\epsilon_2$ measures the second derivative: $f(x+\epsilon_2)=f(x)+f'(x)\epsilon_2+f''(x)\epsilon_2^2/2$ (again exact equality). (Note that in the answer I was careful to say "e.g." and not "i.e.") – Ian May 02 '18 at 21:56
  • @Ian, it is incorrect to assert as you do that in SIA one has "$dx \neq 0$ but $(dx)^2=0$". The reason this is incorrect is because in SIA one cannot assert that $dx\neq0$. Since the background logic is intuitionistic, this does not mean that $dx$ is zero; one can't prove that either. – Mikhail Katz May 03 '18 at 09:09
  • @MikhailKatz I am aware of that, but this is the intuition, and this is not meant to be a careful primer on infinitesimals. Saying $\neg \neg \neg (dx=0)$ would not be illuminating to someone unfamiliar with intuitionistic logic. – Ian May 03 '18 at 11:27
  • (incidentally I didn't downvote). Still, it may be better to say "dx cannot be shown to be zero" so as to avoid a claim that is outright false. – Mikhail Katz May 03 '18 at 11:40
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One can deal with infinitesimals properly, although it is far from trivial.

But, more importantly, equalities of the kind $$\tag1 dy=f(x)\,dx$$ are just notation. What I mean is that you never do any algebra nor any other operation with $(1)$: you don't add it, you don't multiply it, you don't take its square root.

So, whether $dy$ and $dx$ in $(1)$ have meaning as objects is irrelevant. One uses an expression as $(1)$ to indicate a substitution in an integral; and, in an integral, $dx$ and $dy$ are notation and not mathematical objects per se.

Martin Argerami
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  • $dx$ and $dy$ are certainly mathematical objects in an integral,if by "integral" you mean an antidifferential. Unfortunately,there are multiple different operations that $\int$ means, and antidifferential is only one of them. For instance, $\int dy = \int (dx + dz)$, all of these are mathematical objects, and this antidifferentiates into $y = x + z + C$. Likewise, the formula for integrating an arc treats these as objects that can be square rooted. I.e., $L = \int \sqrt{dx^2 + dy^2}$, where $L$ is arc length. It has to be transformed in order to be solved, but that is true of a lot of things. – johnnyb May 02 '18 at 15:00
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    It's still notation. You don't do operations with them, and one can happily write integrals without them, as mathematicians have done for a very long time. Let me ask you: what kind of object is $dx$? What information does $dx$ convey in $\int f(x),dx$ when one can write $\int f$? – Martin Argerami May 02 '18 at 15:12
  • Everything is notation. If you do it without the $dx$ you are probably doing antiderivatives rather than antidifferentiation. The reason that everything seems like "just notation" is that there are multiple, similar concepts that the same symbols have been used for. As an example, $\int f$ usually is used to mean the antiderivative - a combination of multiply by $dx$ then take the antidifferential. $\int f(x),dx$ means the antidifferential itself. The kind of object that $dx$ is? It is an infinitesimal. – johnnyb May 02 '18 at 15:41
  • As to what information it conveys, take my simple example of: $\int dy = \int(dx + dz)$. If you keep the differentials, it becomes $y = x + z + C$, which is true, and can be checked by differentiation. If you leave out the differentials, it becomes: $\int 1 = \int(1 + 1)$, which would integrate to $1x + C = 2x + C$. The differentials are acting as true mathematical objects, and doing real work in the equation. – johnnyb May 02 '18 at 15:42
  • It looks like you don't know much about integration. "Antiderivatives rather than antidifferentiation"? What does that even mean? You say that $dx$ is an infinitesimal; and what is an infinitesimal? What things can you do with $dx$ (which you are still not defining)? What is $\sqrt{dx}$? – Martin Argerami May 02 '18 at 15:49
  • What is $\sqrt{2}$? You can give an approximation but not a number. The best we can write its value is $\sqrt{2}$. If you want the decimal approximation of $dx$ it is $0.00000\ldots$. Not exact, but a close decimal approximation. Differentiation: $d(x^2) = 2x,dx$ Derivative: $D_x(x^2) = 2x$. Derivatives are actually much easier if you separate differentiation and finding the derivative. It allows a single methodology that combines explicit and implicit differentiation as well as multivariable differentiation. (1) differentiate, (2) solve for $\frac{dy}{dx}$. – johnnyb May 02 '18 at 15:58
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    You are still not saying what $dx$ is. A number like $\sqrt 2$ can be defined properly, most commonly by using Dedeking cuts or Cauchy sequences; and its properties can be formally established. In particular, $(\sqrt 2)^2=2$, and if you approximate it with rational numbers (which you also need to define, by the way), their squares are closed to $2$. So you tell me that $dx$ is close to zero; how close? Hard to say if you don't define it. – Martin Argerami May 02 '18 at 16:01
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    Just to point out, $dx$ is not a specific number. It is in the hyperreals, but it is just like $\sqrt{x}$ - a function of $x$ within a certain space. The properties of the hyperreals have been established through the transfer principle. I'm not a specialist, but most books on nonstandard analysis cover the basics. $dx$ is shorthand for $d(x)$. So, if $y = x^2$, then $dy = 2x,dx$. So, let's say that $x = 3$ and $d(x)$ is, say, $\epsilon$, then $dy = 2\cdot 3\cdot \epsilon = 6\epsilon$. In short, the differential converts a formula to a linear graph in the infinitesimal hyperreals. – johnnyb May 02 '18 at 16:09
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I join those who would rather call them differentials instead of infinitesimals. While there is a rigorous way of using infinitesimals as they were historically used (say, by Leibniz), I prefer to think in terms of differentials.

The differential of function $f=f(x)=y$ at point $x$, wlog, is the amount of change in (not the function $f$ itself) but in its best linear approximation when the independent variable is given a change $\Delta x$. This is written $df(x)=dy$ and is given by $$df(x)=dy=f'(x)\Delta x.$$ If $f$ is the identity function, then we obtain $df=dx=\Delta x$, so that we usually replace $\Delta x$ with $dx$ in the definition to have $$dy=f'(x)dx.$$ It turns out that this justifies manipulations with differentials as historically practised (and even today by physicists, engineers and differential geometers), so that one has $$dy=\frac{dy}{dx}dx.$$ This is one way to scrupulously practise this use of so-called "infinitesimals".

Allawonder
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