About Nabla and index notation

Remixex
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Homework Statement


Can I, for all purposes, say that Nabla, on index notation, is $$\partial_i e_i$$ and treat it like a vector when calculating curl, divergence or gradient?
For example, saying that $$\nabla \times \vec{V} = \partial_i \hat{e}_i \times V_j \hat{e}_j = \partial_i V_j (\hat{e}_i \times \hat{e}_j) = \partial_i V_j \epsilon_{ijk} \hat{e}_k$$
I have a feeling that is wrong, I've found all kinds of variations of this notation on the internet, almost no one seems to use the unit vectors, and that confuses me, being a total beginner on this kind of notation.

Homework Equations


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The Attempt at a Solution


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It's an 'abuse of notation', but it can be useful as a mnemonic. If it helps you remember the formulas, that's fine. But try not to lose sight of the fact that it is not strictly correct. For instance ##\nabla f## is not strictly a vector (it's a covector, aka one-form or dual vector). You probably don't need to understand nuances like that yet if you're just starting but it's good to remember that the notation is just a mnemonic, to avoid confusion later on.
 
Yes i have very clear what do gradients or divergences do...in a practical way (direction of max change, and flux per EDIT:volume unit :) i believe?) but I'm still struggling setting up the bridge between my vector calculus knowledge, physics knowledge, and this notation... But the course is only starting, and I'm trying to prove vector identities.

Thanks for the response :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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