I initially had these questions in my mind (as I have seen these being used by physicists and chemists though I wanted to know the perspective of a mathematician. For the use of such see this post of mine.)

  • While finding the derivative (say) $\frac {dF(x)}{dx}$ we actually mean that $\frac {dF(x)}{dx} = \lim_{h\to 0} \frac {F(x+h)-F(x)}{h} $. But what does the term (say) $dP$ mean independent of the denominator (I don't know what to say it)?

  • Is it in any way right to say that $dP$ is constant?

While trying to find about these I came to this through an answer to this Math SE post and there in the note it says

Now (I hope you're enjoying this as much as I am) another person wrote, in response to that note, saying: "I noticed a note at the bottom of the page on differentials, saying that Real Mathematicians don't use differentials, that they aren't "rigorous." In fact, a Real Mathematician, Abraham Robinson, in the 1960's proved a rigorous formulation of differentials—a formulation in which you can with full confidence do algebra with infinitely small and infinitely large quantities. It is a branch of mathematics known as "Nonstandard Analysis"; it is actually used by lots of mathematicians because proofs are simpler and theorems less wordy in the "non-standard" formulation. Some people have gone through and written whole introductory calc texts that abandon limits altogether in favor of the much simpler dx, though they get very little attention. There is no shame in using differentials."

So I just wanted to know how these questions are seen through the eyes of non-standard analysis.

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    $\begingroup$ The rigorous formulation is that $dx$ is a one-form, i.e. a mapping from $TM$ to $C^\infty(M)$ for a smooth manifold $M$ ($M= \mathbb{R}^d$ is just an example of this with standard topology at maximal atlas) $\endgroup$ – Mathphys meister Nov 23 '19 at 12:58
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    $\begingroup$ No, do not say $dP$ is constant. If you wait until you learn about differential forms, then you may understand what Dani said. It is incorrect, but useful for beginners, to think of $dy/dx$ as a quotient. Even in nonstandard analysis, the quotient $dy/dx$ is not "equal" to the derivative, but merely infinitely close. But that is an alternate way of saying the thing with limits in standard calculus. $\endgroup$ – GEdgar Nov 23 '19 at 13:36
  • $\begingroup$ dy/dx can be seen as the quotient of two quantities with dy depending on dx by f(dx)=dy. Then after one takes the limit as $dx \to{0}$, the limit of dy/dx becomes the derivative of y=f(x) wrt x $\endgroup$ – Divide1918 Nov 23 '19 at 15:42
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    $\begingroup$ When you're reading physics textbooks, you can just think of $dx$ as a very tiny but finite number. And $df$ is the tiny (but finite) change in the value of a function $f$ as its input changes from $x$ to $x + dx$. In this viewpoint, $f'(x) \approx df/dx$. The equations we derive this way are only approximate, but if we are careful then we can hope that "in the limit" as $dx$ approaches $0$ we will get exact equality. Physicists can be a bit cavalier about this type of argument, but actually the intuition is quite clear. No need for differential forms or nonstandard analysis in basic physics. $\endgroup$ – littleO Nov 24 '19 at 3:17

The commentors have focused on one approach to defining differentials, as forms. A layman's description of this approach would be that while in $y = f(x), x$ and $y$ are coordinates along the curve, the differentials $dx, dy$ are coordinates along the tangent line to $y = f(x)$.

This is the most common rigorous approach to differentials to be taught to students. But it is not the only approach. In fact, it is one of the most restrictive definitions, unable to easily handle many, many things that differentials are used for. For example, the differential of arclength in 3 dimensions is $$ds = \sqrt{dx^2 + dy^2 + dz^2}$$ You will see this widely used. But it makes no sense if $ds, dx, dy, dz$ are differential forms, which do not support the square root.

Another approach I've seen, developed by Solomon Leader, defined differentials in terms of the Kurzweil-Henstock integral. Unfortunately, it has been so long since I've seen it that I no longer remember the definition. It handled differentials much more naturally than differential forms does, allowing such constructions as the arclength differential without any problems. I have encountered other definitions of differentials as well.

However, Non-Standard Analysis is something entirely different for any of those. You can find some good sources for them online, for example, here. My own knowledge is rather limited. Mikhail Katz, who used to post in these forums was (and probably still is) a big supporter of NSA, but unfortunately, he has been inactive for over a year.

For NSA, differentials are simply infinitesimal numbers. These are simply extra numbers we add to the numberline. The numberline with the new numbers is called the Hyperreals. All hyperreals have a "standard part", an "infinite part", and an "infinitesimal part". The standard part is just an ordinary real number. The infinite part is a new number larger (in absolute value) than any real number, and the infinitesimal part is smaller in absolute value than any real number other than $0$ (which is the only number which is both real and infinitesimal). If the infinite part of the hyperreal is $0$, the number is finite, and if the standard part is $0$, the number is infinitesimal. If both infinite and infinitesimal parts are $0$, the number is real.

This is all well and good, and a quite straight-forward construction. Where NSA gets a little wonky is in evaluating functions for these new numbers. If I have a function $f$ defined on the real numbers, how do I determine what values it should have for hyperreals? NSA declares that every real function $f$ has a unique "natural" extension $f^*$ to the hyperreals. What is it? In general, NSA cannot tell us. It simply declares that one exists by fiat. Now there are some principles that allow you to calculate the functions we use regularly. The actions of addition, multiplication, etc., are built into the hyperreal construction, so we can handle polynomials and exponentials and ratios. But consider the Dirichlet function: $$f(x) = \begin{cases}1, & x\text{ is rational}\\0, & x\text{ is irrational}\end{cases}$$ What is its natural extension? Do it take all non-real numbers to $0$, because they are not rational. Does it take them to $1$, because they are not irrational? Does it base its value on the standard part, ignoring the other two parts? All of these extend $f$ to the hyperreals, as do many others. Which one is "natural"?

It is because of this ambiguousness that I am not comfortable with NSA. However, and please understand this clearly, this is a conceptual issue on my part, not a logical issue with NSA. Far (far) better mathematicians than me have reviewed the underpinnings of NSA, and found it acceptable. And I even have some inkling of why, though I haven't looked deeply myself. And quite frankly, as long as it is logically consistent, I think any mathematical theory deserves study.

Doubtless some advocates of Non-Standard Analysis will be happy to point out the failings of my account of NSA here, and in particular, my comments and example. And I will be happy to listen, though I may not always agree.


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