I'm trying to gain an intuition for the use of nonstandard analysis over the limit approach.

Traditionally the motivation for derivatives is that the derivative of a point $(x,f(x))$ for some function $f$ is the slope of the line connecting that point to some nearby point, $(x+\alpha, f(x+\alpha)), \alpha > 0$, and making $\alpha$ smaller and smaller to the limit of $x$, hence getting a tangent line.

So maybe this means (I know its wrong but follow my train of thought) that the derivative of a point $x$ of some function $D: \mathbb{Z}\rightarrow \mathbb{Z}, \text{ where } \mathbb{Z} \text{ is the integers}$, as the slope of the line connecting $x$ to $x+min(\mathbb{Z^+})$, where of course $min(\mathbb{Z^+})$ = 1. So the derivative of such a function $D$ is $$D' = \frac{D(x+1) - D(x)}{1}.$$ For example, if $D(x) = x^2$ on the naturals then $D' = (x+1)^2 - x^2 = 2x+1$. Obviously if this same function was defined on the reals then $D' = 2x$.

So at this point it seems like more generally, the derivative for a function $G: \mathbb{F}\rightarrow \mathbb{F}, \text{ where } \mathbb{F} \text{ is some field/ring/group}$, is

$$G' = \frac{G(x+dx) - D(x)}{dx}, dx = min(\mathbb{F^+})$$ (again I know this isn't true, but follow my intuition). In words, my thinking so far is that the derivative at $x$ is to take the slope of the line from $x$ to $x+dx$ where $dx$ is the smallest nonzero element in the field or ring.

I think this until I consider a function $L: \mathbb{R} \rightarrow \mathbb{R}$, defined on the reals. I quickly realize that there is no minimal non-zero element to do my derivative calculation as described above.

BUT, I decide this isn't a problem if I assume there is some larger field to which the reals is a subset, $\mathbb{R} \subset \mathbb{F_a}$, and I assume this larger field $F_a$ contains an element $dx \not\in \mathbb{R}$ that is smaller than any non-zero element in $\mathbb{R}$, allowing me to proceed with my derivative calculation as above, as long as I remember that $dx$ is not a real number anymore.

So for the function $f(x) = x^2$ defined over the reals, I now calculate $$f' = \frac{(x+dx)^2 - f(x)}{dx} = \frac{x^2 - 2x(dx) - (dx)^2 - x^2}{dx} = \frac{2x(dx) + (dx)^2}{dx} = \frac{dx(2x+dx)}{dx} = 2x+dx$$

So now I have a result that is some number with a real part ($2x$) and a non-real part ($dx$), but given that the non-real part is smaller than any non-zero real number, if I just ignore that term and consider only the real-part, I will still have the correct slope (in the limit), so I just take $2x$ as my final derivative.

Is that a reasonable way to think about it?

  • $\begingroup$ Aside from the minimum nonzero part, the early users of calculus thought of $d\!x$ as an infinitesimal without fully developing a theory of infinitesimal numbers as an extension of the reals. $\endgroup$ – Somos Nov 25 '18 at 13:45
  • $\begingroup$ From the point of view of the internal set theory, there is no extension, just a distinction of standard and non-standard elements of $\Bbb R$ with some non-standard axioms. $\endgroup$ – LutzL Dec 7 '18 at 20:51

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