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8:00 PM
You can't "solve" in general, @Astyx. Do you know things about $u$?
 
u is $C_c^{\infty}(\Bbb R^d)$
I feel like this should be equal to $||\nabla u||^2_{L^2}$ modulo some constant
 
Just spitballing but you could try splitting $\mathbb{R}^{n}$ into some kinda $\mathbb{B}_{r}(0) \cup \left( \mathbb{R}^{n}\setminus \mathbb{B}_{r}(0) \right)$ decomposition if you're trying to bound bits of it, @Astyx?
 
So try to integrate by parts. Note that $u\nabla u$ is the gradient of $\frac12u^2$, and the other stuff is like gradient of $1/|x|$.
 
@TedShifrin hey ted!
 
hi Stan
 
8:03 PM
my professor for my machine learning class is good at applying ideas but kind of a sloppy mathematician :')
 
You just like to complain.
 
Top quality on my resume
if anyone's hiring complainers, let me know
my main skill
read anything interesting lately?
 
I read that Ted uses $dx$ instead of $\text{d}x$
Does that count as interesting?
xD
 
I have to go for now.
 
8:09 PM
Bye ted
Thank you for your help
And Khallil too
 
Would someone mind checking this?
 
that follows immediately from the definition of matrix-multiplication
 
@Masterphile Yeah, I know. I'm still working on making sure my proofs make sense. Sometimes they seem correct, but then I realized I made some logic error. This one was pretty intuitive though.
Plus, this is my first LA proof lol
 
errors are because of some misunderstanding of the "material" (if exercise is not too hard)
 
When I first started proofs, I was doing little more than regurgitating the form I saw others write proofs in. Now I go through multiple examples and make sure I know exactly what the definition is stating. The proofs just kind of flow naturally after that.
 
8:21 PM
yes
 
It's exciting :D
 
you´re first year of college?
 
In the fall. I'm finishing up my senior year of HS in a couple of months
 
gonna study math or?
 
8:25 PM
i think we almost do not mention LA here in high schools
 
My brain is fried right now, is this right ? $$\nabla^2{1\over|x|} = {d\over|x|^2} + {2\over|x|}$$
 
I'm self-studying. Ted suggested I start now
 
i am also self-studying, but "things" such as general topology, real analysis, elementary number theory, set theories, and such, ah, i guess you´ll get to know these on the college
 
My high school only goes up to Calc 2 and a statistics course
 
that´s far up, not bad at all
 
8:29 PM
@Masterphile Yeah
 
What does the $d$ stand for, @Astyx?
 
The dimension we're in
 
Oh right!
 
8:40 PM
does anybody knows to explain easily integrals in the complex plane, residues, and all that "stuff"? and is willing to?
i know i ask too much
but this is more fun than textbook
 
I'm really rusty but wouldn't $\nabla (1/|x|) = -x/|x|^{3}$, @Astyx?
Then you'd be taking the gradient of a vector field which would be something that isn't only a scalar
 
Indeed ...
 
OH
Do you mean the Laplacian when you write $\nabla^{2}$?
 
Yes
Thank you, I think I got it
at last
 
I was trying to work out the gradient of the gradient because I forgot that $\nabla^2$ denotes $\Delta$ which is just the divergence of the gradient
 
8:52 PM
You made me realize my mistake anyways :)
 
Did you also get $(3-n)/(|x|^{3})$, @Astyx?
(because that looks wrong and I'm so out of practice lol)
 
Give me 5 minutes and I'll tell you :)
 
9:15 PM
Oh I'm pretty sure it's correct
I was just tripping because in $\mathbb{R}^{3}$, it's equal to $0$ but that's consistent because $3-3=0$ lol
 
Seems right to me
 
I wonder what happens for $|x| = 0$
Probably some distribution stuff
 
Bellman's lost-in-a-forest problem is an unsolved minimization problem in geometry, originating in 1955 by the American applied mathematician Richard E. Bellman. The problem is often stated as follows: "A hiker is lost in a forest whose shape and dimensions are precisely known to him. What is the best path for him to follow to escape from the forest?" It is usually assumed that the hiker does not know the starting point or direction he is facing. The best path is taken to be the one that minimizes the worst-case distance to travel before reaching the edge of the forest. Other variations of the...
 
Do you know much about complex differentiation, @Masterphile?
 
9:30 PM
well, the definition of the derivative is formally the same as for real functions of a real variable, so conceptually i see no serious difficulties
 
Complex differentiation is a lot more restrictive
 
in what sense?
 
Do you know the limit definition of the derivative of a real valued function?
 
yes
 
So you understand that (on the real line/x-axis) both the left-sided and right-sided limits must agree for the limit itself to exist?
 
9:34 PM
yes, and in the complex plane it must be from all directions that they agree,and even more general requirements must be true
 
Oh sweet
So can you see why it's more restrictive a condition to be complex differentiable than real differentiable?
Also I wouldn't say the requirements are more general
I'd just say that there are more requirements
 
it´s not more restrictive actually, if it is it is not much, since on the line (real line) there are only two directions
 
It certainly is more restrictive
 
you ask for exactly the same conditions
just on the real line there are two directions from a point
 
Yea and you can approach in many more ways on $\mathbb{C}$ and all the limits must agree
Less functions satisfy more conditions
e.g. if I'm looking for numbers between 0 and 10, that's more restrictive than looking for numbers between -10 and 20
i.e. $[0,10] \subset [-10,20]$
In the same way, the more conditions you place on a class of objects, the less (not necessarily strictly) objects that satisfy all your conditions
e.g. $\{ \text{metric spaces} \} \subset \{ \text{topological spaces} \}$
 
9:43 PM
i understand you well, but the definitions are formally the same, the other reasons are why something is different when functions are from C to C
 
What do you mean by 'formally the same'?
(I might be misunderstanding)
 
they are exactly the same, just x is replaced by z
i think that differences in some theorems arise because C cannot be ordered as R
 
$\lim_{h \to 0 \text{ in } \mathbb{R}}$ and $\lim_{h \to 0 \text{ in } \mathbb{C}}$ are pretty different but I guess you mean that if you don't write 'in $\mathbb{K}$' then the definitions look identical
 
i understand why they are different, but formally, they are the same
 
You're not using the word 'formally' correctly
 
9:48 PM
i am using it formally :D
 
You actually aren't but I understand what you're trying to say so I guess that's all that matters
 
yes
 
A nice example of a function that isn't complex differentiable but is real differentiable is $f(x,y) = x - iy$
 
where it´s not complex differentiable?
 
Try approaching via the coordinate axes
That's an exercise for you
 
9:54 PM
when x tends to zero that tends to -iy/y=-i and when y tends to zero that tends to x/x=1
-yi/yi=-1
 
Anyone here hear of an embedding in the context of spectral clustering?
 
Sounds like machine learning
 
yeah im taking an intro class on it
 
10:10 PM
@Khallil what´s your expertise (college, fields of studies)?
 
 
2 hours later…
11:46 PM
okay, so metric spaces generalize euclidean spaces and topological spaces generalize metric spaces
again, i really like this theorem
Invariance of domain is a theorem in topology about homeomorphic subsets of Euclidean space Rn. It states: If U is an open subset of Rn and f : U → Rn is an injective continuous map, then V = f(U) is open and f is a homeomorphism between U and V.The theorem and its proof are due to L. E. J. Brouwer, published in 1912. The proof uses tools of algebraic topology, notably the Brouwer fixed point theorem. == Notes == The conclusion of the theorem can equivalently be formulated as: "f is an open map". Normally, to check that f is a homeomorphism, one would have to verify that both f and its inverse...
 
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