soft question - What is the point of making dx an infinitesimal hyperreal?

It seems fairly common to describe $\mathrm{d}x$ in nonstandard analysis as an infinitesimal. But after thinking it through (and then skimming Keisler's text), I can't see the point and actually think it's misleading!

First, let me clearly point out that $\mathrm{d}y$ is not being used here as to express a "difference in $y$"; this post is following the convention that $\Delta y$ is used for such things, and $\mathrm{d}y$ is reserved for the differential.

That is, suppose $y = x^2$. A "change in $y$" is the quantity $\Delta y$ given by, after fixing some change $\Delta x$ in $x$:
$$\Delta y = (x + \Delta x)^2 - x^2 = 2 x (\Delta x) + (\Delta x)^2$$

This is not what $\mathrm{d}y$ is. We simply have $\mathrm{d}y = 2x \,\mathrm{d} x$. In general, if $y = f(x)$, then $\mathrm{d} y$ is simply defined to be $f'(x) \,\mathrm{d} x$. No differences, infinitesimal approximations, or anything of that flavor is going on here; $\mathrm{d} y$ is nothing more than a vessel for carrying around a copy of $f'(x)$.

(and $\mathrm{d} x$ was simply defined to be an independent, infinitesimal variable)

A typical application of a differential is that in a definite integral $\int_a^b f(x) \,\mathrm{d}x$, we might decide to write down a Riemann sum with $H$ evenly spaced partitions for some infinite $H$, and substitute in the notation $\mathrm{d}x$ with the width of an interval $\frac{b-a}{H}$ to get
$$\int_a^b f(x) \,\mathrm{d}x \approx \sum_{i=1}^H f\left(a + \frac{b-a}{H}i \right) \frac{b-a}{H}$$
However, if I encode $\mathrm{d}x$ as an infinitesimal, then write $\int_a^b f(x) \epsilon$, there's no way to figure out what that means. You might write $H = (b-a)/\epsilon$ and write down the Riemann sum above, but that gives the wrong answer if I encoded $\mathrm{d}x$ as $2 \epsilon$. The best you can do is to undo the encoding; e.g.
$$\int_a^b f(x) \epsilon \approx \sum_{i=1}^H f\left(a + \frac{b-a}{H}i \right) \frac{\epsilon}{\mathrm{d}x} \frac{b-a}{H}$$

Thus, the encoding of the differential form as an infinitesimal does not seem to do anything useful for this application of differentials. But maybe we can do other interesting arithmetic with them. However, I don't think there's any application of quantities like $(\mathrm{d}y)^2$ or $1 + \mathrm{d}y$ or $\sin(\mathrm{d}y)$ — it's the quantities like $(\Delta y)^2$ or $1 + \Delta y$ or $\sin(\Delta y)$ that we play with.

Instead, the only useful operations seem to be the ordinary differential form operations — things like adding two differential forms or multiplying a differential form by a function.

In sum, the only application of this definition seems to be to allow one to say that $\frac{\mathrm{d}y}{\mathrm{d}x}$ is the ratio of two hyperreal-valued variables — but even in ordinary analysis we can understand that as a ratio of differential forms!

Furthermore, insistence that $\mathrm{d}x$ be infinitesimal appears to be completely irrelevant; you could do the same thing in standard analysis simply by removing the constraint that $\mathrm{d}x$ be infinitesimal. In fact, to some extent, people do do the same thing; e.g. defining the differential of a function $f$ to be the function $\mathrm{d}f(x,e) = f'(x) e$.

So, I pose my question — what is the point of making $\mathrm{d}x$ an infinitesimal hyperreal?

Answer

I assert that that there is no intrinsic reason to make $\mathrm{d}x$ an infinitesimal. However, conventions can force us to doing so; for example, it is a consequence of:

• The functional form of the differential: $\mathrm{d}f(x,y) = f'(x) y$


• The habit of designating a particular variable (e.g. $x$) as special



• The habit of implicitly partially evaluating differentials at the special difference $\Delta x$

And by making the second argument implicit and fixed, if we use $\mathrm{d}f(x)$ to implicitly mean $\mathrm{d}f(x, \Delta x)$, the notation lends itself to the variable form $\mathrm{d}y$ to be used in place of $\mathrm{d}f(x)$ whenever $y = f(x)$.

Under these conventions, if $i$ is the identity function $i(x) = x$, then
$$\mathrm{d}x = \mathrm{d}i(x) = \mathrm{d}i(x, \Delta x) = \Delta x$$

thus identifying the differential $\mathrm{d}x$ with the variable $\Delta x$ which, conventionally, is infinitesimal-valued.