Skip to main content

[Community Question] Calculus: I need to prove a few vector identities using Cartesion Tensor Notation, and I can't figure out how!

One of our user asked:

I have been all over the internet, but I just can't make sense of this stuff. I have done my best to learn from my textbook and different websites, but this is confusing for me. I haven't taken any calculus in years, and I'm jumping in headfirst. If anyone can help me understand how to prove these using Cartesian Tensor Notation, I would really appreciate it!

First identity: x ( x a) = ∇(∇ . a) - (∇^2)a

Second identity: . (a b) = a . b + b( . a)

Third identity: . (f δ) = f

Fourth identity: δ : a = . a

Thanks everyone


Comments

Popular posts from this blog

[Community Question] Calculus: Manifold with boundary - finding the boundary

One of our user asked: I have the manifold with boundary $M:= \lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1\geq 0, x_1^2+x_2^2+x_3^2=1\rbrace \cup\lbrace (x_1,x_2,x_3) \in \mathbb R^3 : x_1= 0, x_1^2+x_2^2+x_3^2\leq1\rbrace$ and I need to find the boundary of this manifold. I think it is $\lbrace (x_1,x_2,x_3) \in \mathbb R^n : x_1= 0, x_2^2+x_3^2=1\rbrace$ , the other option is that the boundary is the empty set? I think the first is right? Am I wrong?

[Community Question] Linear-algebra: Are linear transformations between infinite dimensional vector spaces always differentiable?

One of our user asked: In class we saw that every linear transformation is differentiable (since there's always a linear approximation for them) and we also saw that a differentiable function must be continuous, so it must be true that all linear operators are continuous, however, I just read that between infinite dimensional vector spaces this is not necessarily true. I would like to know where's the flaw in my reasoning (I suspect that linear transformations between infinite dimensional vector spaces are not always differentiable).