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0 pointsacjohnson5512y ago0 comments

I'm interested! I think Dedekind cuts are reasonably understandable, but infinitesimals are on the surface of much of our calculus syntax, so I'd be glad to understand where they become so tricky formally.

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igravious12y ago

I wrote a short paper on the topic once upon a time[1] which you may find interesting. It's part history of math, part philosophy of math.

It's not a great paper and most of the insights in it come from others but here is some of the arithmetic of nilpotent[1] infinitesimals as shown in the appendix.

Imagine an entity which is not equal to zero but that when raised to the power of 2 or higher is equal to zero! Sounds odd, doesn't it, but it works! (ϵ is an infinitesimal)

ϵ != 0 but ϵ^n = 0 | n>1

ok? so we get:

(ϵ + 1)^n = 1 + nϵ thus: (ϵ + 1)^−1 = 1 − ϵ

e^ϵ = 1+ϵ

(ϵ + 1)(ϵ−1) = −1, or alternately (1 + ϵ)(1 − ϵ) = −1

and finally (for calculus): ϵf′(x) = f(x + ϵ)−f(x)

1: http://leto.electropoiesis.org/propaganda/The_Analyst_Revisi...

2: https://en.wikipedia.org/wiki/Nilpotent

edit: clarity, line breaks!

JadeNB12y ago

How do these differ from the [dual numbers](https://en.wikipedia.org/wiki/Dual_number)?

ColinWright12y ago

You appear to have an error. You write:

  (ϵ + 1)(ϵ−1) = −1, or alternately (1 + ϵ)(1 − ϵ) = −1

That alternative should surely be:

    (1 + ϵ)(1 − ϵ) = 1

Not least, in a commutative system (1+x)(1-x) = 1-x^2. Thus

    (1 + ϵ)(1 − ϵ) = 1 - ϵ^2 = 1

igravious12y ago

Thanks, well caught :)

j / k navigate · click thread line to collapse

0 comments

igravious12y ago

I wrote a short paper on the topic once upon a time[1] which you may find interesting. It's part history of math, part philosophy of math.

It's not a great paper and most of the insights in it come from others but here is some of the arithmetic of nilpotent[1] infinitesimals as shown in the appendix.

Imagine an entity which is not equal to zero but that when raised to the power of 2 or higher is equal to zero! Sounds odd, doesn't it, but it works! (ϵ is an infinitesimal)

ϵ != 0 but ϵ^n = 0 | n>1

ok? so we get:

(ϵ + 1)^n = 1 + nϵ thus: (ϵ + 1)^−1 = 1 − ϵ

e^ϵ = 1+ϵ

(ϵ + 1)(ϵ−1) = −1, or alternately (1 + ϵ)(1 − ϵ) = −1

and finally (for calculus): ϵf′(x) = f(x + ϵ)−f(x)

1: http://leto.electropoiesis.org/propaganda/The_Analyst_Revisi...

2: https://en.wikipedia.org/wiki/Nilpotent

edit: clarity, line breaks!

JadeNB12y ago

How do these differ from the [dual numbers](https://en.wikipedia.org/wiki/Dual_number)?

ColinWright12y ago

You appear to have an error. You write:

  (ϵ + 1)(ϵ−1) = −1, or alternately (1 + ϵ)(1 − ϵ) = −1

That alternative should surely be:

    (1 + ϵ)(1 − ϵ) = 1

Not least, in a commutative system (1+x)(1-x) = 1-x^2. Thus

    (1 + ϵ)(1 − ϵ) = 1 - ϵ^2 = 1

igravious12y ago

Thanks, well caught :)

j / k navigate · click thread line to collapse