Timeline for Was the early calculus inconsistent?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2013 at 16:57 | review | Suggested edits | |||
| Aug 21, 2013 at 17:14 | |||||
| Apr 18, 2013 at 8:56 | comment | added | Mikhail Katz | Good point. Did you ever try seeing what Bos has to say about this? See ams.org/mathscinet-getitem?mr=469624 | |
| Apr 17, 2013 at 17:56 | comment | added | Toby Bartels | Yes, the correct chain rule for second derivatives is $d^2y/dx^2 = (d^2y/du^2)(du/dx)^2 + (dy/du)(d^2u/dx^2)$. But unlike the chain rule for first derivatives, you can't derive this by treating $d^2y$ and the rest as if they were elements of an infinitesimal-enriched continuum obeying the ordinary rules of algebra. At the very least, this is annoying (and I have found it so since since high school); but more than that, it suggests that the $dx^2$ that appears in $d^2y/dx^2$ is not really the square of $dx$ in such a continuum. | |
| Apr 16, 2013 at 16:11 | comment | added | Mikhail Katz | @Toby Bartels: I've thought about this for a few days but I am still not sure what point you are trying to make. Surely there IS a correct formula for the second derivative of a composite function, and there is a coherent theory of higher order differentials. | |
| Apr 15, 2013 at 3:28 | comment | added | Toby Bartels | @Vladimir: This does not contradict your point (or indeed any of the text of your latest comment), but one must be careful with $d^2y$, $dx^2$, and the like. Second derivatives in Leibniz's notation don't work as well as first derivatives, because the chain rule $d^2y/dx^2 = (d^2y/du^2) (du/dx)^2$ is false. I would prefer to write $\partial^2y/\partial{x}^2$ myself, on the grounds this is the coefficient on $dx^2$ in an expansion of $d^2y$ (in $d^2x$ and $dx^2$), analogous to the coefficients that are partial derivatives, rather than the ratio of $d^2y$ to $dx^2$. | |
| Apr 9, 2013 at 20:21 | comment | added | Mikhail Katz | I hope Andrej or somebody else knowledgeable about SDG comments on this, but awaiting their comment, it seems to me that they can arrange for nilpotency of any finite order one wishes. I have the impression they can accomodate higher order differentials this way. | |
| Apr 9, 2013 at 19:10 | comment | added | Vladimir Kanovei | And those denoted as dx, dy etc, what sort are them of? Hardly nilpotent since second derivatives involve $dx^2=(dx)^2$. - Vladimir | |
| Apr 9, 2013 at 8:53 | comment | added | Mikhail Katz | @Vladimir: In SDG they have both kinds of infinitesimals if needed: the nilpotent ones and the invertible ones. About 1/i it is certainly the case that it is positive and not merely that it is not the case that it is or isn't distinct from zero. | |
| Apr 8, 2013 at 19:29 | comment | added | Vladimir Kanovei | >>given any infinitesimal, it is not the case that it is distinct from zero. This is so funny! I mean, you claim in particular that any argument starting from "let e be a small positive infinitesimal number (hence non-0)". is wrong from the beginning? What then about the Euler factorization of sin which starts from an inf. large number i and then involves infinitesimals like 1/i - is it "not the case" that 1/i is definitely "distinct from zero"? | |
| Apr 8, 2013 at 13:31 | comment | added | Andrej Bauer | Old habits die hard ;-) | |
| Apr 8, 2013 at 12:58 | comment | added | Andrej Bauer | This is not the only sound way to view the controversy. Another sensible way is to go intuitionistic. The controversy is then resolved by realizing that (a) not all infinitesimals are zero and (b) given any infinitesimal, it is not the case that it is distinct from zero. There is no valuation map, or a sharp distinction between "standard" reals and infinitesimals, only weird (but sound!) intuitionistic sort of half-existence of infinitesimals. | |
| Apr 8, 2013 at 12:58 | comment | added | Mikhail Katz | There is a more down-to-earth construction of "nonstandard" reals with nilpotent infinitesimals by Paolo Giordano. I believe in his construction there is a valuation map. Perhaps Paolo could comment. | |
| Apr 8, 2013 at 12:53 | comment | added | Andrej Bauer | There can be no such valuation map in Synthethic Differential geometry, and neither is it the case that the smooth real line is an extension of "true reals" (whatever those are supposed to be), at least I do not see how that could be, since the smooth real line is not even Cauchy complete. | |
| Apr 8, 2013 at 12:49 | comment | added | Mikhail Katz | Thanks, Vladimir. Indeed, Leibniz provided consistent rules of inference in terms of his Transcendental law of homogeneity ("discard the negligible term"), without of course providing any interpretation of the number system itself (which had to await Hewitt, Los, and Robinson). | |
| Apr 8, 2013 at 12:45 | vote | accept | Mikhail Katz | ||
| Apr 8, 2013 at 12:29 | history | answered | Vladimir Kanovei | CC BY-SA 3.0 |