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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 und
by acjohnson55 13y ago
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.
- igravious 13y agoI 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_Revisited.pdf http://leto.electropoiesis.org/propaganda/The_Analyst_Revisi... 2: https://en.wikipedia.org/wiki/Nilpotent https://en.wikipedia.org/wiki/Nilpotent edit: clarity, line breaks!
- ColinWright 13y agoYou 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
- igravious 13y agoThanks, well caught :)
- JadeNB 13y agoHow do these differ from the [dual numbers](https://en.wikipedia.org/wiki/Dual_number https://en.wikipedia.org/wiki/Dual_number)?