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Jul 11, 2019 at 13:23 comment added Lo Scrondo I deeply thank you!
Jul 11, 2019 at 13:12 comment added Willie Wong Because the derivatives agree @LoScrondo: if two functions have the same derivative at one point then there Taylor expansion agrees to first order.
Jul 11, 2019 at 11:38 comment added Lo Scrondo Thank you @WillieWong , I think yours is the only possible explanation. Just to verify if I've understood correctly...why your Taylor expansion lacks the linear term, i.e. why it is $\xi t \to \xi t + O(\xi^2t^2)$ and not $\xi t \to \xi t + O(\xi t)$?
Jul 9, 2019 at 16:08 answer added Bazin timeline score: 2
Jul 9, 2019 at 13:22 comment added Willie Wong I am pretty sure implicitly the coordinate systems defining $V$ and $V'$ are the "same" at $x$ (the transition map should have derivative that equals the identity there) (this is given the figure illustrating the situation on pg 191). Then it really should just be a change of variables and application of Taylor's theorem. Following Arnold's notation, for $\partial V'(\epsilon)$, instead of integrating over the four segments $\xi t, \xi t + \eta, \eta t, \eta t + \xi$, you would be integrating over $\xi t + O(\xi^2t^2)$ etc.
Jul 9, 2019 at 12:49 comment added Lo Scrondo I just followed the description given by Arnold, but in fact he treats $V$ as an ordinary parallepiped, and $V'$ as a curvilinear one
Jul 9, 2019 at 11:53 comment added Ben McKay I think you will have to define the term "curvilinear parallelipiped". If $V$ is just a parallelipiped, then $V'$ isn't. But if you allow arbitrary curvilinearity, it is not clear how it behaves with $\varepsilon$.
Jul 8, 2019 at 23:15 review First posts
Jul 9, 2019 at 3:05
Jul 8, 2019 at 23:13 history asked Lo Scrondo CC BY-SA 4.0