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Geometric Foundations of Calculus: Infinitesimals as Oriented Zeros

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Jun 6, 20265 min read

mooth Infinitesimal Analysis posits an infinitesimal such that the following statements are both true:

ε0ε ≠ 0
ε2=0ε^2 = 0

This is a very promising idea, but the way they pose it leans on ignoring the law of excluded middle, which is a little unsettling. I propose a different interpretation of the same definitions. In geometric algebra, squaring erases direction, as seen in how the basis vectors square:

ei2=1e_i^2 = 1

In light of this, I think there should be a different infinitesimal oriented along each basis vector, such that:

εi0ε_i ≠ 0
εi2=0ε_i^2 = 0

Now, one can see clearly why the infinitesimal is not zero. It's because of the direction, not the magnitude. Squaring erases the direction, giving us the pristine scalar 0. A result of this is all higher power are also 0, because they all contain a square that eats the rest:

εin= εi2εin2=0εin2=0ε_i^n = ε_i^2 * ε_i^{n-2} = 0 * ε_i^{n-2} = 0

This property will prove most useful later.

Addendum: In the rest of this piece, even though geometric algebra literature typically uses numerical subindices, we will use the subindices x and y to indicate orientation along the x and y axes, respectively.

Directional Mathematics

The math we are all most used to is the mathematics of scalars; quantities with magnitudes but no direction. Calculus, understood properly, requires us to consider the mathematics of direction without magnitude.

The first thing about pure directions that jumps out is equivalence over scaling. We can represent that with the equation:

nεx= εxnε_x = ε_x

Even though scaling is inert, addition is not. We can construct:

εx+y=εx+ εyε_{x+y} = ε_x + ε_y

As a statement about directions, this tells us about the directional sum of the x and y axes. We can also subtract them like so:

εxy=εxεyε_{x-y} = ε_x - ε_y

In fact, any linear combination is expressible:

εax+by=aεx+bεyε_{ax+by} = aε_x + bε_y

However, do not be fooled. Since this is only the mathematics of direction, equivalent ratios of a and b represent the same direction.

2εx+εy=4εx+2εy2ε_x + ε_y = 4ε_x + 2ε_y

As such, it is generally more useful to express directions relative to a reference direction. Let's choose the x axis.

aεx+bεy=εx+baεyaε_x + bε_y = ε_x + \frac{b}{a}ε_y

As a shorthand, we can then express ε_y with respect to ε_x:

εx+baεybaεy/xε_x + \frac{b}{a}ε_y \rightarrow \frac{b}{a}ε_{y/x}

This the form we will expect of our derivatives going forward.

We can recover ε_y by composing the relative direction with the reference direction like so:

εy=εy/x εxε_y = ε_{y/x} * ε_x

This will prove useful later.


Derivatives

The goal of a derivative is to find the direction of motion at a specific point. We represent this as:

x+ εxx + ε_x

To see how this construction makes derivatives easy, let's try taking a simple derivative with this construction.

f(x)=x2f(x)=x^2

By inserting ε_x, we can probe our function, and find our derivative by isolating ε_x on the other side.

f(x+εx)=(x+εx)2=x2+2xεx+εx2f(x+ε_x) = (x+ε_x)^2 = x^2 + 2xε_x + ε_x^2

We can delete the squared infinitesimal because it squares to zero. Defining the infinitesimal along f and subtracting f(x) from both sides:

εf=f(x+εx)f(x)=2xεxε_f = f(x+ε_x) - f(x) = 2xε_x

And the final trick:

εf/x=2xε_{f/x} = 2x

There we have it! The derivative, achieved with only our oriented zero. Note that this is not scalar division. This is a linear ratio of directions, expressing direction as a function of x. Just what we were looking for!

Another one

Let's try a cubic.

f(x)=x3f(x)=x^3
f(x+εx)=(x+εx)3=x3+3x2εx+3xεx2+εx3f(x+ε_x) = (x + ε_x)^3 = x^3 + 3x^2ε_x + 3xε_x^2 + ε_x^3

This time, two terms are deleted for being zero.

εf=f(x+εx)f(x)=3x2εxε_f = f(x+ε_x) - f(x) = 3x^2ε_x
εf/x=3x2ε_{f/x} = 3x^2

Power Rule

Indeed, this will work for any power of x.

f(x)=xnf(x)=x^n
f(x+εx)=(x+εx)n=xn+nxn1εx+...f(x+ε_x)=(x+ε_x)^n = x^n + nx^{n-1}ε_x + ...

Lucky for us, the remaining terms of the binomial expansion delete themselves because they all have higher powers of ε_x, no matter how big n is. Then:

εf=f(x+ εx)f(x)= nxn1εxε_f = f(x + ε_x) - f(x) = nx^{n-1}ε_x
εf/x=nxn1ε_{f/x} = nx^{n-1}

This is our power rule at full generality.

Sum Rule

The sum of two arbitrary functions, coming right up.

f(x)=g(x)+h(x)f(x)=g(x)+h(x)
f(x+εx)=g(x+εx)+h(x+εx)f(x+ε_x)=g(x+ε_x)+h(x+ε_x)

Assuming g and h are differentiable:

εf=f(x+εx)f(x)=g(x+εx)+h(x+εx)g(x)h(x)ε_f = f(x+ε_x) - f(x) =g(x+ε_x)+h(x+ε_x) -g(x)-h(x)
εf= g(x+εx)g(x)+h(x+εx)h(x)ε_f = g(x+ε_x)-g(x) + h(x+ε_x)-h(x)
εf= εg+ εhε_f = ε_g + ε_h
εf/x=εg/x+εh/xε_{f/x} = ε_{g/x} + ε_{h/x}

Simple, right?

Product Rule

This one is a little more interesting.

f(x)=g(x)h(x)f(x) = g(x) * h(x)
f(x+εx)=g(x+εx)h(x+εx)f(x+ε_x)=g(x+ε_x)*h(x+ε_x)

Here is the fun part. We can rearrange the following core identity for a shortcut:

f(x+εx)=f(x)+εff(x+ε_x) = f(x) + ε_f

Substituting in our product function:

f(x+εx)=(g(x)+εg)(h(x)+εh)f(x+ε_x)=(g(x)+ε_g)*(h(x)+ε_h)

Via binomial expansion:

f(x+εx)=g(x)h(x)+εgh(x)+g(x)εh+ εgεhf(x+ε_x)= g(x)*h(x) + ε_g*h(x) + g(x)*ε_h + ε_g*ε_h

The two infinitesimals multiply to zero via direction erasure. Subtracting the original product:

εf=f(x+εx)f(x)=g(x)εh+h(x)εgε_f= f(x+ε_x) - f(x) = g(x)ε_h + h(x)ε_g
εf/x=g(x)εh/x+h(x)εg/xε_{f/x} = g(x)ε_{h/x} + h(x)ε_{g/x}

Eh voila!

Chain Rule

Last but not least, the one for compositions!

f(x)=g(h(x))f(x)=g(h(x))

We have to be careful with this one. The infinitesimal only enters h directly:

f(x+εx)=g(h(x+εx))f(x+ε_x)=g(h(x+ε_x))

Using our nifty shortcut again:

f(x+εx)=g(h(x)+εh)f(x+ε_x)=g(h(x)+ε_h)

We now have an interesting problem. How to we handle ε_h inside of g? Well, let's pretend h is not a function for a moment. Our shortcut applies again!

g(h+εh)=g(h)+εgg(h+ε_h)=g(h)+ε_g

Of course, h is still a function, and we have to add the dependence on x back in. The infinitesimal on g is with respect to h, not x, so we make the ε_h explicit:

f(x+εx)=g(h(x))+εg/hεhf(x+ε_x)=g(h(x)) + ε_{g/h}*ε_h

Subtracting the original function as usual:

εf=f(x+εx)f(x)=εg/hεhε_f = f(x+ε_x)-f(x) = ε_{g/h}*ε_h

Finally,

εf/x=εg/hεh/xε_{f/x} = ε_{g/h}*ε_{h/x}

We arrive at our derivative, which all together says a change in x causes a change in h causes a change in g. This works multiplicatively because the local change is linear.


Coming Soon

That's about it for the foundations! The next installments to look forward to will cover integration, Taylor series, and deriving the trigonometric functions via Euler's formula as a definition.

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