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

May 25, 2016

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

#### Explanation:

The Real numbers ($\mathbb{R}$) are an example of a field.

They are a set containing distinct elements called $0$ and $1$, equipped with addition $+$ and multiplication $\cdot$ 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 $\overline{F}$ containing infinitesimal elements. We will also specify that $F$ is an ordered field, having a $<$ binary relation.

First note that if $\overline{F}$ 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 ${\overbrace{1 + 1 + \ldots + 1}}^{\text{N times}} < H$ for any positive integer $N$. So $\overline{F}$ is a non-Archimedean field.

If $P \left(x\right)$ is any polynomial with coefficients in $F$, then $P \left(H\right)$ must also be in $\overline{F}$. If $Q \left(x\right)$ is another non-zero polynomial with coefficients in $F$, then $\frac{P \left(H\right)}{Q \left(H\right)}$ must also be in $\overline{F}$.

Now notice something a little subtle: If $P \left(H\right) \ne 0$ then the multiplicative inverse of $\frac{P \left(H\right)}{Q \left(H\right)}$ must be $\frac{Q \left(H\right)}{P \left(H\right)}$, so we must have:

$\frac{P \left(H\right)}{Q \left(H\right)} \cdot \frac{Q \left(H\right)}{P \left(H\right)} = \frac{P \left(H\right) Q \left(H\right)}{Q \left(H\right) P \left(H\right)} = 1$.

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

As a result, we consider:

$\frac{P \left(H\right)}{Q \left(H\right)} = \frac{R \left(H\right)}{S \left(H\right)} \iff P \left(H\right) S \left(H\right) = Q \left(H\right) R \left(H\right)$

That's our field $\overline{F}$ or in particular $\overline{\mathbb{R}}$

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

$\frac{P \left(t\right)}{Q \left(t\right)} \equiv \frac{R \left(t\right)}{S \left(t\right)} \iff P \left(t\right) S \left(t\right) = Q \left(t\right) R \left(t\right)$

May 25, 2016

Here's a little discussion of application...

#### Explanation:

Let $\overline{\mathbb{R}}$ be the field of rational expressions $\frac{P \left(t\right)}{Q \left(t\right)}$ with Real coefficients modulo the equivalence relation:

$\frac{P \left(t\right)}{Q \left(t\right)} \equiv \frac{R \left(t\right)}{S \left(t\right)} \iff P \left(t\right) S \left(t\right) = Q \left(t\right) R \left(t\right)$

Let $\epsilon = \frac{1}{t}$ be our designated infinitesimal.

Is this field useful for simple Calculus?

Consider $f \left(x\right) = {x}^{3}$

If $a \in \mathbb{R}$ then:

$f \left(a\right) = {a}^{3}$

$f \left(a + \epsilon\right) = {\left(a + \epsilon\right)}^{3} = {a}^{3} + 3 {a}^{2} \epsilon + 3 a {\epsilon}^{2} + {\epsilon}^{3}$

So the average slope of $f \left(x\right)$ over the interval $\left[a , a + \epsilon\right]$ is:

$\frac{f \left(a + \epsilon\right) - f \left(a\right)}{\left(a + \epsilon\right) - a} = \frac{\left({a}^{3} + 3 {a}^{2} \epsilon + 3 a {\epsilon}^{2} + {\epsilon}^{3}\right) - {a}^{3}}{\epsilon} = 3 {a}^{2} + 3 a \epsilon + {\epsilon}^{2}$

We can introduce another concept to our field $\overline{\mathbb{R}}$, which is that of "standard part". The standard part of a finite number in $\overline{\mathbb{R}}$ is the part formed by throwing away any infinitesimal terms.

For example, the standard part of $3 {a}^{2} + 3 a \epsilon + {\epsilon}^{2}$ is $3 {a}^{2}$, which is the derivative of $f \left(x\right) = {x}^{3}$ 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 $\overline{\mathbb{R}}$ supports some interesting ideas:

If:

$f \left(x\right) = \left\{\begin{matrix}0 & x < 0 \\ \frac{1}{2} & x = 0 \\ 1 & x > 0\end{matrix}\right.$

Then we can 'approximate' the derivative of $f \left(x\right)$ with:

$f ' \left(x\right) = \left\{\begin{matrix}\frac{1}{\epsilon} & x \in \left[- \frac{\epsilon}{2} & \frac{\epsilon}{2}\right] \\ 0 & x \notin \left[- \frac{\epsilon}{2} & \frac{\epsilon}{2}\right]\end{matrix}\right.$