Timeline for Why $h_t$ maps into $\mathbb{R}^{\nu}$?
Current License: CC BY-SA 4.0
13 events
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| Jul 31, 2020 at 15:59 | comment | added | Zeno cosini | Got it. Really appreciate it. | |
| Jul 31, 2020 at 14:26 | comment | added | LSpice | I'm not sure what you mean by "the product". Do you mean the symbol $\langle{-}, {-}\rangle$? If so, then I have said what it means: evaluation of a linear map. If you want to think in coördinates, it is $(\partial h_1/\partial t, \dotsc, \partial h_\nu/\partial t) = (\nabla h_1\cdot(1, 0), \dotsc, \nabla h_\nu\cdot(1, 0))$, where $h_i \colon A \times U \to \mathbb R$ are the component functions. | |
| Jul 31, 2020 at 13:52 | comment | added | Zeno cosini | Yes, you're right @LSpice . I edited the title, thanks for that. About the definition of that product so, how can we define it? I didn't find any definitions and since I'm really new to this topic, I am really confused. | |
| Jul 31, 2020 at 13:52 | history | edited | Zeno cosini | CC BY-SA 4.0 |
improved formatting
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| Jul 31, 2020 at 13:32 | comment | added | LSpice | Incidentally, your title asks why $h$ maps into $\mathbb R^\nu$, but that's given; I think your question seems to be rather why $h_t$ maps into $\mathbb R^\nu$. | |
| Jul 31, 2020 at 13:27 | comment | added | LSpice | I'm not sure why you would expect the partial derivative of an $\mathbb R^\nu$-valued function to be $\mathbb R^{\nu - 1}$-valued; partial derivatives in some sense "reduce input variables", not "reduce output variables". Anyway, the 'definition' of $\langle{-}, {-}\rangle$ in terms of the gradient that you mention I guess is the one on p. 209, where it is specified only for $Y = \mathbb R$, so that evaluation is the inner product. For more general $Y$, one can only think of evaluation. | |
| Jul 31, 2020 at 12:49 | history | edited | Zeno cosini | CC BY-SA 4.0 |
improved formatting
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| Jul 31, 2020 at 12:48 | comment | added | Zeno cosini | Thank you so much @LSpice. Well, as I saw in the book, the definition of the $\langle -,-\rangle$ is exactly as I wrote and it'll be the inner product of the vector $v=(0,1) \in \mathbb{R} \times \mathbb{R}^n$ and $\nabla h(t,x)$. What I don't understand is that why this inner product lies in $\mathbb{R}^\nu$! I mean, in the definition, it is said that $h$ lies in $\mathbb{R}^\nu$. So, as $h_t = \frac{\partial h}{\partial t}$, shouldn't it lie in $\mathbb{R}^{\nu -1}$? | |
| Jul 31, 2020 at 12:30 | history | edited | LSpice | CC BY-SA 4.0 |
Oops, missed a few more small things
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| Jul 31, 2020 at 12:29 | comment | added | LSpice | Perhaps the point is that the notation is unclear, and looks like an inner product, whereas I think the significance is that $Dh(t, x)$ is a map $\mathbb R \times \mathbb R^n \to \mathbb R^\nu$, and all we are doing is applying it to the vector $(1, 0) \in \mathbb R \times \mathbb R^n$. (One could think of this as matrix-vector multiplication.) | |
| Jul 31, 2020 at 12:21 | comment | added | LSpice | The partial derivative at $t \in A$ of a map $h : A \times U \to V \subseteq \mathbb R^\nu$ is a map $U \to \mathbb R^\nu$; what is the question? (What I don't get is what it means to define $h_t$ twice, on the same domain, in two completely different ways ….) | |
| Jul 31, 2020 at 12:20 | history | edited | LSpice | CC BY-SA 4.0 |
Link to book, and specific page and section; proofreading
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| Jul 31, 2020 at 11:58 | history | asked | Zeno cosini | CC BY-SA 4.0 |