Timeline for If d/dx is an operator, on what does it operate?
Current License: CC BY-SA 3.0
42 events
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| Oct 22, 2023 at 23:38 | comment | added | Gerald Edgar | At least we have given up 19th-century notations like $$\frac{d^n\,0^m}{d0^n}$$ | |
| Oct 22, 2023 at 22:38 | answer | added | NinjaDarth | timeline score: 0 | |
| Mar 25, 2023 at 13:59 | answer | added | Tom Copeland | timeline score: 1 | |
| Feb 23, 2023 at 15:37 | answer | added | Alexey Muranov | timeline score: 2 | |
| Aug 28, 2018 at 19:05 | review | Close votes | |||
| Aug 29, 2018 at 15:35 | |||||
| Aug 21, 2018 at 2:30 | answer | added | Terry Tao | timeline score: 35 | |
| Aug 12, 2018 at 16:33 | comment | added | Michael Bächtold | Btw: the OP is in very good company with his question: math.stackexchange.com/questions/1258923/… | |
| Aug 12, 2018 at 11:01 | history | edited | Wojowu |
Tag fixes
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| Aug 12, 2018 at 10:50 | answer | added | Michael Bächtold | timeline score: 12 | |
| Feb 15, 2017 at 20:51 | review | Suggested edits | |||
| Feb 15, 2017 at 21:07 | |||||
| Dec 7, 2012 at 16:09 | comment | added | user21349 | I'm surprised to see such a lengthy discussion of this notation with no mention of the history. To me, this is a case where historical changes have led to the reinterpretation of an elegant old notation in an awkward new way. The Leibniz notation $dy/dx$ was originally meant to represent the quotient of two infinitesimal changes. After Cauchy-Weierstrass, people wanted to make $d/dx$ an operator. It doesn't quite make sense because it's an attempt to retrofit an old notation. There's a similar awkwardness in NSA, where we'd like $dy/dx$ to mean the standard part of the quotient $dy/dx$. | |
| Dec 7, 2012 at 7:41 | answer | added | Dirk | timeline score: 2 | |
| Dec 6, 2012 at 19:35 | vote | accept | Jason Howald | ||
| Dec 6, 2012 at 7:01 | comment | added | Joel David Hamkins | Since the question opened up again, I've now posted here. Thanks! | |
| Dec 6, 2012 at 6:58 | answer | added | Joel David Hamkins | timeline score: 50 | |
| Dec 6, 2012 at 5:53 | history | reopened |
David E Speyer Todd Trimble Carl Mummert Kevin Walker François G. Dorais |
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| Dec 5, 2012 at 22:39 | comment | added | Joel David Hamkins | I posted my answer, such as it is, on my blog at jdh.hamkins.org/the-differential-operator-ddx-binds-variables. | |
| Dec 5, 2012 at 17:49 | comment | added | Peter Dalakov | I totally agree with Sam, though I would tend to say that we're working with the manifold $\mathbb{R}$, together with a volume form $vol=dx$, so one could write this operator as $d/vol$, if so desired. | |
| Dec 5, 2012 at 15:14 | comment | added | Joel David Hamkins | Well, I'm not satisfied by the current answers. I've written up an answer, which I will post if the question is reopened again. | |
| Dec 5, 2012 at 14:21 | comment | added | user9072 | @David Speyer: please just check the revision history, it was reopened and reclosed. (I had a comment to that extent but decided it was too unfriendly so deleted it after some rethinking but did not manage to think up a friendly one, so there is no trace of the reclosure.) The meta thread you link to was declared essentially obsolete a long time ago. But in any case those written votes are effective/technical votes so it would not even apply regardless. | |
| Dec 5, 2012 at 14:15 | comment | added | David E Speyer | I also would like to see this reopened. I'm a little confused though -- JDH and algori both say they have reopened, but the reopen counter is only at 1 (my click). Are we vote trading? tea.mathoverflow.net/discussion/506 I accept and follow the vote trading protocol, but it hasn't been clearly initiated. | |
| Dec 5, 2012 at 13:47 | history | closed |
Daniel Litt user9072 Chandan Singh Dalawat José Figueroa-O'Farrill Chris Gerig |
not a real question | |
| Dec 5, 2012 at 11:54 | answer | added | Goldstern | timeline score: 2 | |
| Dec 5, 2012 at 9:18 | answer | added | David Roberts♦ | timeline score: 2 | |
| Dec 5, 2012 at 7:22 | history | edited | Andrés E. Caicedo |
edited tags
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| Dec 5, 2012 at 7:14 | answer | added | Bazin | timeline score: 1 | |
| Dec 5, 2012 at 4:27 | history | reopened |
Joel David Hamkins algori Goldstern Frank Thorne MTS |
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| Dec 4, 2012 at 23:41 | comment | added | Sam Gunningham | I've always thought of $x$ as just a choice of coordinate on the $1$-manifold $\mathbb R$ (i.e. picking a diffeomorphism with a ``standard'' copy of $\mathbb R$). Without choosing coordinates, we have for each function $f:\mathbb R \to \mathbb R$ a linear operator $df$ on each tangent space. Picking a coordinate function $x$ allows us to express these as numbers, hence we get a function $df/dx$. | |
| Dec 4, 2012 at 20:50 | comment | added | Joel David Hamkins | Of course we all understand the basic issue---none of us is confused about any elementary matter---but I feel that the explanations given don't yet get at the conflation of syntax and semantics that the question is about. For example, Goldstern says "$d/dx$ and $d/dt$ mean the same thing", but I find it unlikely that he would write ${d/dt} (x^2)=2x$ in place of ${d/dx}(x^2)=2x$ for his calculus students, without further remarks. It is the precise nature of this particular confusion that the question is about. For example, the $\lambda$-calculus is quite insistent about avoiding this collision. | |
| Dec 4, 2012 at 20:26 | comment | added | Margaret Friedland | @Goldstern gives a serious answer, but (basically) the same argument can be stated in an entertaining way; see Darsh Ranjan's answer to this question:mathoverflow.net/questions/1083/do-good-math-jokes-exist-closed | |
| Dec 4, 2012 at 18:38 | comment | added | Joel David Hamkins | Thanks, algori, although I was hoping to read an answer, rather than write one. | |
| Dec 4, 2012 at 18:37 | comment | added | Goldstern | As long as we are only looking at functions in one variable, $d/dx$ and $d/dt$ mean the same thing, and one of them is redundant. As soon as we look at functions in, say, two variables, we implicitly introduce an (arbitrary) order of variables, say $x$ is the first and $t$ the second, and $d/dx$ is the derivative with respect to the first variable. This makes sense even for variable-agnostic sets of ordered pairs. True, it is an abuse of notation, but $(d/dx, d/dt)$ is sometimes just more readable than $(D_1, D_2)$. | |
| Dec 4, 2012 at 18:35 | comment | added | algori | I've voted to reopen as well, as I would be interested to hear what Joel Hamkins has to say about this question. | |
| Dec 4, 2012 at 18:10 | comment | added | Joel David Hamkins | I find this question to be both deeper and more interesting than it appears by the comments that some others here do. So I have voted to reopen, and would look forward to reading a thoughtful answer that takes the issue seriously. | |
| Dec 4, 2012 at 17:39 | comment | added | algori | Jason -- you could try asking this on math.stackexchange.com; there is a chance you'll get a detailed answer there. | |
| Dec 4, 2012 at 17:14 | comment | added | Qfwfq | Mathematics is a human activity, hence it often benefits from some en.wikipedia.org/wiki/Abuse_of_notation | |
| Dec 4, 2012 at 16:42 | comment | added | Jason Howald | Thank you for the criticisms. I have rephrased to clarify my intended meaning. | |
| Dec 4, 2012 at 16:41 | history | edited | Jason Howald | CC BY-SA 3.0 |
Rephrased, motivated by first two comments
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| Dec 4, 2012 at 16:34 | history | closed |
Ryan Budney Dan Petersen Michael Renardy Will Sawin George Lowther |
not a real question | |
| Dec 4, 2012 at 16:30 | comment | added | Thierry Zell | I don't think this is, at least as stated, a good question for MO (and apparently I am not the only one, because I did not go as far as to downvote). Also, I would challenge some of your assumptions. The notation $\frac{df}{dx}$ is used very frequently in the books I'm reading, for instance. Also, does $\frac{d}{dx}$ denote a single operator? As Poincare remarked: "Mathematics is the art of giving the same name to different things". | |
| Dec 4, 2012 at 16:27 | comment | added | Ryan Budney | What does your question have to do with logic? People do write $\frac{d}{dx} f$ sometimes. And $\frac{d}{dx}$ and $\frac{d}{dt}$ are frequently seen as essentially identical in these contexts. Many calculus textbooks where functions are not synonymous with formulas have this outlook. | |
| Dec 4, 2012 at 16:21 | history | asked | Jason Howald | CC BY-SA 3.0 |