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Jun
9
comment Prove that $(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$ using Gauss's theorem
@Samuel 1) just a fancy way to say "cube of side length $\epsilon$ around $p$ 2) the cube has $V_n = \epsilon^n$ 3) the face has area $dS = \epsilon^{n-1}$. 4). $f(p+\epsilon_k)$ is slightly wrong notation. Since it's $n$ dimensions we can move $\epsilon$ units in any direction. I just mean $\pm \epsilon$ units in the $k$ direction.
Jun
9
revised Functional analysis: Why $|[x,y]|\leq [x,x]^{1/2}[y,y]^{1/2}$
added 304 characters in body
Jun
8
comment Prove that $(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$ using Gauss's theorem
@Samuel If you use a sphere instead of a cube, it changes how small $r$ has to be in order to approximate the divergence, but the limit is the same. We could rescale $r \mapsto \frac{1}{ \sqrt[n]{V_n}} \, r $
Jun
8
comment Functional analysis: Why $|[x,y]|\leq [x,x]^{1/2}[y,y]^{1/2}$
it's just bra-ket notation. It's just another way to write your definition emphasizing that $A$ is symmetric.
Jun
8
revised Functional analysis: Why $|[x,y]|\leq [x,x]^{1/2}[y,y]^{1/2}$
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Jun
8
answered Functional analysis: Why $|[x,y]|\leq [x,x]^{1/2}[y,y]^{1/2}$
Jun
8
comment Functional analysis: Why $|[x,y]|\leq [x,x]^{1/2}[y,y]^{1/2}$
I can't tell... why is $[x,x] = \langle \epsilon x + Ax, x \rangle$ but $[y,y] = \langle \epsilon y + Ay, \color{green}{\mathbf{\epsilon}} y \rangle$ ? Using your definition I got $[x,x] = \langle x + \epsilon + Ax , x \rangle $. So I am even more confused.
Jun
8
comment Prove that $(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$ using Gauss's theorem
why don't you show your proof in 3 dimensions? Here is a rather abstract version of your question, though it does not give a proof.
Jun
8
comment Prove that $(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$ using Gauss's theorem
@Samuel I am sorry it's that $r \to 0$ so I always think of it as $\epsilon$. They are the same
Jun
8
revised Prove that $(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$ using Gauss's theorem
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Jun
8
comment $T^2=I$ implies $T$ is diagonalizable
If we use $\mathbb{Z}$ instead of $\mathbb{R}$ we get a $\mathbb{Z}$-module. In the other case, $\mathbb{R}_{\geq 0}$ is not even a group, so non-negative matrices are a world of their own.
Jun
8
comment $T^2=I$ implies $T$ is diagonalizable
Which field? Non-negative reals $\mathbb{R}_{\geq 0}$ are not a field, since $1$ does not have an additive inverse. $\mathbb{Z}$ is not a field since $\frac{1}{2} \notin \mathbb{Z}$.
Jun
8
comment $T^2=I$ implies $T$ is diagonalizable
Aren't $\{(1,1), (1,-1)\}$ eigenvectors?
Jun
8
answered Prove that $(div \ w)(p) = \lim \limits_{r\rightarrow 0} \frac{1}{V_n r^n}\int_{S_r(p)} \langle w,\nu \rangle dS$ using Gauss's theorem
Jun
8
revised $T^2=I$ implies $T$ is diagonalizable
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Jun
8
comment Limits and Series in Smooth Infinitesimal Analysis
@StefanPerko judging from your question you know nothing about infinitesimal analysis. so why don't you make your question more specific?
Jun
8
answered $T^2=I$ implies $T$ is diagonalizable
Jun
8
revised Limits and Series in Smooth Infinitesimal Analysis
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Jun
8
comment Limits and Series in Smooth Infinitesimal Analysis
@Ian I am not fimiliar with this version of non-standard analysis, so I am outlining some of the things that are going wrong when I try to help him.
Jun
8
revised Limits and Series in Smooth Infinitesimal Analysis
added 694 characters in body