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While reading calculus books, I see sections on differentials which refer to "infinitessimals" in a very loose way, alluding to the fact that this view on calculus is not the standard, but makes a lot of intuitive sense. However, the next "level" of rigor I've seen is in differential geometry books, where differentials are well-defined in a standard way, but require a depth of knowledge just to understand that definition.

My question is: is there a middle ground here, where we can rigorously and formally define the concept of a differential $\text{d}x$ for an independent variable $x$, that is accessible to students fresh out of undergraduate or first-year-graduate analysis?

I emphasize independent because I've heard many times that a differential is defined as $\text{d}y = f^\prime(x)\text{d}x$, but without defining $\text{d}x$ this definition is pointless.

Edit: I think what I'm looking for is basically a simpler explanation of $\{\text{d}x^i\}$ as a dual basis for $\{e_i\}$, without talking about tangent and cotangent spaces and, if possible, without talking about bases at all. When we say $\text{d}x^i(e_j) = \delta_{ij}$, what is the definition of $\text{d}x^i$ that is used?

Alex Jones
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    The way you learn calculus in an introductory class, $dx$ is merely a remnant of $\Delta x$ and the limiting process; a reminder of which variable you're differentiating / integrating with respect to, although some times they try to tell you that it's an infinitesimal of some kind. By the way, this question popped up as related. – Arthur Apr 23 '17 at 07:42
  • By the way, in my opinion, most calculus books introduce differential in a bad and messy way. They always lead students mess up the symbol $dx$ and $\frac{dy}{dx}$, thinking the $dx$ are the same in the denominator of $\frac{dy}{dx}$(actually this is NOT a fraction, but a bad-style symbol). In fact, $\frac{dy}{dx}$ is a historical symbol and a kind of abuse of notation. But there seems no other invented another one and become popular. And changing the habit of million people accustomed to this notation is difficult. – Eric Apr 23 '17 at 07:42
  • Yes, I understand and agree with both of you. Standard calculus texts do an awful job at explaining the concept of a differential, and I remember my high school calculus teachers telling me that the real definition is over our heads and so we just have to deal with it unless we continue studying math in college. In some sense, the purpose of my question is to determine if it really is that abstract that we can't even begin to explain it to high school and even beginning graduate students can't fully understand if they haven't taken a graduate differential geometry course. – Alex Jones Apr 23 '17 at 07:48
  • Gateux derivative is easy to understand for me. Frechet was the one used in the best DG book I read. Makes me think of the d as a macro for microscopic changes. – Emil Apr 23 '17 at 07:49
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    The Gateaux derivative is basically just a dressed up version of $\text{d}y=f^\prime(x)\text{d}x$, in that it defines the differential of a function but not an independent variable, as I'm looking for. – Alex Jones Apr 23 '17 at 07:53
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    Personally, I just think of it as convenient notation. Manipulations involving $dx$ or $dy$ can be justified rigorously using nothing other than 'standard' analysis techniques. – MathematicsStudent1122 Apr 23 '17 at 08:41
  • @AlexanderJ93 The only definition I have seen is the one on wikipedia where it is defined as a limit. Very hands on definition. There is no $f'$ in the definition there. – Emil Apr 23 '17 at 09:15
  • I think the biggest obstacle here is that introductory calculus materials don't do justice to the notion of variable, nor to the distinction between a real-valued expression and a function. –  Apr 24 '17 at 05:35
  • Incidentally, just defining $\mathrm{d}y = f'(x) \mathrm{d}x$ (when $y = f(x)$) isn't nearly as ill-defined as you make it sound. It needs more to be a complete definition but the basic method is fundamentally sound. For example, compare to the "one construction of" Kähler differentials that is commonly used in algebraic geometry. –  Apr 25 '17 at 02:44

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I emphasize independent because I've heard many times that a differential is defined as $\text{d}y = f^\prime(x)\text{d}x$, but without defining $\text{d}x$ this definition is pointless.

One way would be to define the (first) differential $\mbox{d}\,f$ of a function $f$ as a map: $$\mbox{d}\,f:(x,\Delta x) \mapsto f'(x) \Delta x$$ Note that this mapping takes a point $x$ and an increment $\Delta x$. The number which is associated with a point $x$ and increment $\Delta x$, is how much the $y$-value changes on the tangent line (the linearisation of the function), which is an approximation of the change in the real function value.

The explicit dependency of $\mbox{d}\,f$ on $x$ and the increment $\Delta x$ is often omitted and for $y=f(x)$, this also leads to the following notation: $$\mbox{d}y = f'(x) \Delta x$$

Applying this to the function $f(x)=x$, which is differentiable everywhere with $f'(x) = 1$, leads to the following relation where we take $y=x$: $$\mbox{d}y = f'(x) \Delta x \to \mbox{d}x = 1 \Delta x \implies \mbox{d}x = \Delta x$$ This can be seen as a motivation to replace $\Delta x$ by $\mbox{d}x$, giving the more common: $$\mbox{d}y = f'(x) \, \mbox{d}x$$ As a bonus, this nicely "agrees" with the Leibniz notation for derivatives: $$\frac{\mbox{d}y}{\mbox{d}x} = f'(x)$$

StackTD
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  • The not-so-satisfying thing about this definition is that it's just a notation. $\text{d}x = \Delta x = $ what? $\Delta x$ is an increment, so I guess it's just a positive real number. So, by this definition, $\text{d}x$ is any positive real number. That doesn't give the feeling of structure that I think either the differential geometry or nonstandard analysis definitions give to $\text{d}x$. – Alex Jones Apr 23 '17 at 18:00
  • @AlexanderJ93: $\Delta x$ isn't an increment; $\Delta x$ is a variable in its own right. This is, incidentally, almost literally the usual definition of differential from differential geometry, since $T\mathbb{R} \cong \mathbb{R}^2$; the tangent space to $x$ is made of the points $(x, \epsilon)$. Following along the lines of StackTD's answer describes also reproduces exactly the definition of the differential from Keisler's NSA book. –  Apr 24 '17 at 05:17
  • I have basically no experience in differential geometry or nonstandard analysis, but I was under the impression that $\text{d}x$ was some kind of linear functional, and an element of the dual basis for the canonical basis of $\mathbb{R}^n$. – Alex Jones Apr 25 '17 at 03:20
  • @AlexanderJ93: $f'(x) \Delta x$ is linear in $\Delta x$. Put differently, for each value of $x$, the mapping $\Delta x \to f'(x) \Delta x$ is a linear functional. The bivariate function $(x, \Delta x)$ is just a repackaging of the covector field that assigns to each $x$ a linear functional. –  Apr 25 '17 at 03:27
  • I understand that $\text{d}y$ is a linear in $\Delta x$, but if $y=x$ so that $\text{d}x=\Delta x$, then $\text{d}x(\Delta x)$ is just an identity function. Just so you know where I'm coming from, do Carmo's "Differential Forms and Applications" states that $\text{d}x^i(e_j) = \frac{\partial x^i}{\partial x^j} = \delta_{ij}$. However, it doesn't give a definition of $\text{d}x^i$ for any other elements of $\mathbb{R}^n$. – Alex Jones Apr 25 '17 at 03:42
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    @Hurkyl: I think I see now... $\text{d}x^i(v) = \sum_{k=1}^n \frac{\partial x^i}{\partial x^k}\cdot v_k$? So in the one-dimensional case it is an identity function, and the basis is just 1, so $\text{d}x(1)=1$ means that it's a dual basis for $\mathbb{R}$. I think it makes sense now, thanks! – Alex Jones Apr 25 '17 at 03:53
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My preferred way is through nonstandard analysis, which requires some medium-duty logic to set up but then allows you to treat $\mathrm{d}x$ simply as an infinitesimal. There are textbooks of introductory analysis which do it this way, such as Pétry's Analyse Infinitésimale: une présentation non standard. It's within the grasp of an undergraduate, though to do it properly (justifying that one can rigorously make infinitesimals exist) is not trivial.

Patrick Stevens
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    While a valid way to go about it, it somewhat sidesteps the issue by creating a parallel version of the subject where infinitessimals exist. I'm looking for something within the standard framework -- essentially a simpler explanation of the differential form definition. – Alex Jones Apr 23 '17 at 08:43
  • Actually, nonstandard analysis doesn't do anything new here. For a function $f$, the idea of defining $\mathrm{d}f$ to be a bivariate function $\mathrm{d}f(x, e) = f'(x) e$ is completely standard. Suppressing the $e$ so that you get $\mathrm{d}f(x) = f'(x) \mathrm{d}x$ is a simple notational convention, as is adapting it to related variables instead of functions (i.e. things like $\mathrm{d}y$ and the aforementioned $\mathrm{d}x$). In addition to suppressing $e$ you could alternatively plug in a standard choice for it: Keisler's book uses the variable $\Delta x$. –  Apr 24 '17 at 05:04
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The notation $dx$ suggests that we had a coordinate map $x:M\to \mathbb R$, $p \mapsto x(p)$.

Then $dx(p)$ is the map derivative, a push-forward between tangent spaces, or, when $M$ (and $\mathbb R$) is a normed vector space, $dx_p$ is called the Fréchet derivative. $M$ itself could be another copy of $\mathbb R$, but holding on to this generality allows me to emphasize the point $p\in M$.

$dx_p$ is a linear map mapping (tangent) vectors in $M$ to vectors (increments $\Delta x$ sometimes, or simply a real number - that's why it's also called a [basis] 1-form) in $\mathbb R$.

You then consider a map $f:\mathbb R\to\mathbb R$ and a composite map $y:M \to \mathbb R$, $y=f\circ x$, and use the chain rule to find $dy(p)=df(x(p))\circ dx(p).$

A confusing moment happens when we decide to drop $p$, as if reusing $x$ to also denote the target element in $\mathbb R$, $x=x(p)$(!?), and write $dy=df(x)\circ dx.$

Finally we introduce notation $f'(x)=df(x)$ and this map being simply a scalar multiplication for a given $x(p)$ we replace the composition by the multiplication, $dy=f'(x)dx.$

rych
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