Timeline for Was the early calculus inconsistent?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 20, 2013 at 21:44 | comment | added | Toby Bartels | It might have been more interesting if L'Hôpital had calculated that $dx\,dy + dy\,dx = 0$, then concluded that $dx\,dy = −dy\,dx$ rather than that $dx\,dy = 0$. | |
| Mar 20, 2013 at 17:26 | comment | added | Mikhail Katz | @François G. Dorais: you wrote above that "I can't imagine 17th century mathematicians writing $dx^2=0$". I would like to comment that 17th century mathematician Nieuwentijt did explicitly envision nilsquare infinitesimals. Bell mentions this in his piece "continuity and infinitesimals" at SEP. However, Leibniz proceeded differently. | |
| Mar 20, 2013 at 15:48 | comment | added | François G. Dorais | L'Hôpital saw that $dxdy = 0$ follows from his first postulate: "Grant that two quantities, whose difference is an infinitely small quantity, may be taken (or used) indifferently for each other: or (what is the same thing) that a quantity, which is increased or decreased only by an infinitely small quantity, may be considered as remaining the same." (Same source.) | |
| Mar 20, 2013 at 15:44 | comment | added | François G. Dorais | However, Leibniz had a different take: "He also assumed that the $n$th power $(dx)^n$ of a first-order differential was of the same order of magnitude as an $n$th-order differential $d^nx$, in the sense that the quotient $d^nx/(dx)^n$ is a finite quantity." (Same source.) | |
| Mar 20, 2013 at 15:38 | comment | added | François G. Dorais | Newton, for one, did exactly as I said. "Thus, for example, in the case of the fluent $z = x^n$, Newton first forms $\dot{z} + \dot{zo} = (\dot{x} + \dot{xo})^n$, expands the right-hand side using the binomial theorem, subtracts $z = x^n$, divides through by $o$, neglects all terms still containing $o$, and so obtains $\dot{z} = nx^{n−1}\dot{x}." (From section 4 of plato.stanford.edu/entries/continuity ) | |
| Mar 20, 2013 at 15:27 | comment | added | Andrej Bauer | But if they write $(x + dx)^2 = x^2 + 2 x dx$ then it follows immediately by basic algebra that $dx^2 = 0$. Why of why didn't they just follow their noses? | |
| Mar 20, 2013 at 15:24 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
added 289 characters in body; added 3 characters in body; added 1 characters in body; added 13 characters in body
|
| Mar 20, 2013 at 15:19 | comment | added | François G. Dorais | I don't know but it's an interesting question. I can't imagine 17th century mathematicians writing $dx^2 = 0$ but I can imagine them writing $(x+dx)^2 = x^2 + 2xdx$. | |
| Mar 20, 2013 at 15:13 | history | edited | Andrej Bauer | CC BY-SA 3.0 |
added 2621 characters in body; added 228 characters in body
|
| Mar 20, 2013 at 13:27 | history | answered | Andrej Bauer | CC BY-SA 3.0 |