Question

What simple rigorous ways are there to incorporate infinitesimals into the number system and are they then useful for basic Calculus?

Answer 1

Here's about the simplest way, but the explanation gets a little long...

Explanation:

The Real numbers (`RR`) are an example of a field.

They are a set containing distinct elements called `0` and `1`, equipped with addition `+` and multiplication `*` that behave in the ways that you are used to. In technical language, they form an abelian group under addition, the non-zero elements form an abelian group under multiplication and multiplication is distributive over addition.

The simplest way to add infinitesimals to any field `F` is to add one, then add anything else you need to make the resulting set closed under addition, multiplication, additive inverse, and multiplicative inverse of non-zero elements. If `F` is an ordered field (as the Real numbers are), then you can extend `F` in a way consistent with the ordering.

Given a field `F`, we will construct a field `bar(F)` containing infinitesimal elements. We will also specify that `F` is an ordered field, having a `<` binary relation.

First note that if `bar(F)` is a field containing infinitesimals, then it will contain the reciprocals of those elements and thus infinite elements too. Rather than adding an infinitesimal, I will start by adding a designed infinite element, let's call it `H` (short for Huge). `H` satisfies:

`x < H` for all `x in F`

In particular `overbrace(1+1+...+1)^"N times" < H` for any positive integer `N`. So `bar(F)` is a non-Archimedean field.

If `P(x)` is any polynomial with coefficients in `F`, then `P(H)` must also be in `bar(F)`. If `Q(x)` is another non-zero polynomial with coefficients in `F`, then `(P(H))/(Q(H))` must also be in `bar(F)`.

Now notice something a little subtle: If `P(H) != 0` then the multiplicative inverse of `(P(H))/(Q(H))` must be `(Q(H))/(P(H))`, so we must have:

`(P(H))/(Q(H))*(Q(H))/(P(H)) = (P(H)Q(H))/(Q(H)P(H)) = 1`.

This requires that we be able to 'cancel out' common factors between the numerator and denominator.

As a result, we consider:

`(P(H))/(Q(H)) = (R(H))/(S(H)) <=> P(H)S(H) = Q(H)R(H)`

That's our field `bar(F)` or in particular `bar(RR)`

More formally, it's the set of rational functions with coefficients in `F` modulo the equivalence relation:

`(P(t))/(Q(t)) -= (R(t))/(S(t)) <=> P(t)S(t) = Q(t)R(t)`

Answer 2

Here's a little discussion of application...

Explanation:

Let `bar(RR)` be the field of rational expressions `(P(t))/(Q(t))` with Real coefficients modulo the equivalence relation:

`(P(t))/(Q(t)) -= (R(t))/(S(t)) <=> P(t)S(t) = Q(t)R(t)`

Let `epsilon = 1/t` be our designated infinitesimal.

Is this field useful for simple Calculus?

Consider `f(x) = x^3`

If `a in RR` then:

`f(a) = a^3`

`f(a+epsilon) = (a+epsilon)^3 = a^3+3a^2epsilon+3a epsilon^2+epsilon^3`

So the average slope of `f(x)` over the interval `[a, a+epsilon]` is:

`(f(a+epsilon)-f(a))/((a+epsilon)-a) = ((a^3+3a^2epsilon+3aepsilon^2+epsilon^3)-a^3)/epsilon = 3a^2+3aepsilon+epsilon^2`

We can introduce another concept to our field `bar(RR)`, which is that of "standard part". The standard part of a finite number in `bar(RR)` is the part formed by throwing away any infinitesimal terms.

For example, the standard part of `3a^2+3aepsilon+epsilon^2` is `3a^2`, which is the derivative of `f(x) = x^3` at `x=a`.

Has this helped any?

It kind of replaces the limit process with infinitesimal arithmetic and taking "standard part".

Note that this discussion has been a little informal. It is fairly easy but lengthy to make it more rigorous, but I do not really like this "standard part" mechanism.

In addition, note that the number system defined in this way does not include `sqrt(epsilon)`. For such objects, you need more sophisticated fields incorporating infinitesimals, such as the Levi-Civita field.

Note that even our simple extension `bar(RR)` supports some interesting ideas:

If:

`f(x) = { (0, x < 0), (1/2, x = 0), (1, x > 0) :}`

Then we can 'approximate' the derivative of `f(x)` with:

`f'(x) = { (1/epsilon, x in [-epsilon/2, epsilon/2]), (0, x !in [-epsilon/2, epsilon/2]) :}`