Timeline for Why is differentiating mechanics and integration art?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2013 at 0:15 | comment | added | O.R. | If I understand correctly, this answer 'explains' why differentiation is simple. Explains is in ''s marks because what is really done is illustrating that the definition of derivative can be easily extended to a definition of a functor. On the other hand, it still leaves open the question of why integration is hard. | |
| Oct 2, 2012 at 19:02 | comment | added | Qfwfq | Something that puzzles me a bit, when seeing the differential of a function $f:\mathbb{R}\to\mathbb{R}$ as a map of vector bundles $df:T\mathbb{R}\to T\mathbb{R}$ on the real line manifold, is that it already contains all the information about its "antiderivative": indeed it's defined by $(x,v)\mapsto d_x f(v):=(f(x),f'(x)\cdot v)$. That is, if somebody gives you the derivative of a function as a map of vector bundles on the real line, then you are trivially able to reconstruct the function. | |
| May 31, 2011 at 1:50 | comment | added | Todd Trimble | Very true, Terry. | |
| May 31, 2011 at 1:21 | comment | added | Terry Tao | A related point is that differentiation can be defined in much more general settings than real-variable ones (e.g. formal differentiation of algebraic maps on varieties over an arbitrary field), whereas integration is mostly restricted to real-variable (and complex-variable) settings. So whereas integration needs both analysis and algebra, differentiation needs only algebra, which helps explain why it is simpler. | |
| May 30, 2011 at 23:10 | comment | added | Todd Trimble | @Mariano: I'm not sure what to make of your comment. Is there anything unreasonable in my answer that you'd like to point out? (Same to whoever voted my answer down.) | |
| May 30, 2011 at 22:59 | comment | added | Sridhar Ramesh | No, no, no; its with $(0,∞)$-groupoids that we are $\mathrm{Set}$ | |
| May 30, 2011 at 22:30 | comment | added | Mariano Suárez-Álvarez | Now we need to involve $(1,\infty)$-groupoids somehow and we are set :) | |
| May 30, 2011 at 20:20 | history | answered | Todd Trimble | CC BY-SA 3.0 |