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8:00 PM
@Mike @Huy I tried to to help Huy and now you two are roasting me??
I am hurt
 
Huy
who's roasting you
 
I didn't see anything except for the phrase continuous fit
 
@Leaky: I just told you an elementary way in my book.
 
which what else was I going to do, not make a joke
 
hmm anyone knows good resource to learn about hessian matrix and more specifically it's eigenvalues ? maybe one with more intuitive explanation on how things work ?
 
8:01 PM
what things work, @Tuki?
 
i have some understanding on how eigenvalues/ eigenvectors work on geometric level
hessian matrix
you can use it to define critical points of vector valued function
 
My sleep cycle is a topologically mixing dynamical system, by the way
2
 
You can find stuff about it in my lectures on YouTube ... application to classifying critical points of functions, proved. @Tuki
 
@Daminark Weird!
 
Huy
just look at pictures of paraboloids
 
8:01 PM
Not vector-valued, @Tuki. Multivariable scalar function.
 
yes i mean the input is vector
for function
= multivariable ?
 
He didn't actually prove the fact that you can find a module such that the E infinity are graded pieces in either vertical or horizontal orientation, said to check a book by Vakil called "Foundations of Algebraic Geometry" for more details
 
What's the difference between vector and multivariable ?
 
But we used that to get the stuff we needed
 
8:03 PM
@Daminark I take that as blackbox too
 
@LeakyNun
 
@Daminark honestly, that sounds awfully sophisticated for a proof of the Snake lemma
 
I never did the calculation
 
Like that?
 
Looks right, DogAteMy.
 
Huy
8:04 PM
@AkivaWeinberger how ?
 
Geogebra, manually fiddling to get P and Q
 
Huy
ah ok
 
It's definitely overkill but it was quite a lot of fun
 
@Tuki: Look here‌​, at lectures 47-50.
 
@Daminark Learn spectral sequences from Hatcher with me
 
8:05 PM
so $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is only "multivariable" not vector function ?
 
Right.
 
Possibly relevant lines
 
Vector function normally means vector-valued.
 
And Rohit's explanation of the stuff was really helpful, the first time around I was like, what?
Now I'm happy
 
Some isosceles triangles
 
8:06 PM
but (x,y) is vector ?
 
@TedShifrin Just the basics, for while :)
 
Or happier at least, I'll need to work some stuff out
@Balarka never let go of an opportunity to preach for the church of Hatcher
 
@AndersonFelipeViveiros: I don't know if it's too elementary, but some of the stuff in my lectures (just linked up there for Tuki) might help. Lots of concrete computations, but also applications to differential topology.
 
Just go read a calculus book about taylor series, boom you're done
 
But yeah I may do so soon, for now I'll have to finish functional
 
8:07 PM
@Dami want to think about functional analysis?
 
@Tuki: That's true, but usually "vector function" refers to the values, not the inputs.
 
Sure
 
@TedShifrin now i understood thanks
 
Anyhow, @Tuki, if you're trying to get examples and understanding of some of the theory, my lectures might help you.
 
8:09 PM
Take the unit ball of $C^1(0,1)$, this is also a subset of $C^0(0,1)$, what does its closure in the latter space look like?
 
@TedShifrin Thank you!
 
Yes i am watching the lecture 47 right now @TedShifrin
 
LOL, OK. Maybe it'll help.
 
That should be a compact subset of $C^0(0,1)$ so it is somehow much smaller than a ball of this space
 
8:10 PM
@Tuki: Eigenvalues come later in the course ... but the signs of the entries of the $D$ you'll see are the same as the signs of the eigenvalues. That gets proved in the very last lecture in the second semester :P
Hi @Antonios.
 
ok
 
how's it going @TedShifrin
 
Wait you're taking the open interval? You'll need to be a bit careful then
 
My neck hurts like hell, but otherwise fine, @Antonios-AlexandrosRobotis. You healthy again?
 
healthy and back in school.
 
8:11 PM
@TedShifrin darn, what happened?
 
@Daminark oh, no, it's closed
 
yeah what happened lol
 
I always have back/neck/shoulder problems, Demonark, but I did something that made it worse.
 
Okay okay phew
 
Yippee, @Antonios.
 
8:12 PM
im pretty sure I did have the flu
it wasn't awful, just had a fever for like 5 days
 
There are people dying from the flu more than usual this time. Probably not kidlets your age.
 
yeah I heard it's not great this year
 
OK, lunchtime for me. Back later.
 
Bye @Ted
 
8:15 PM
See you @Ted!
Hope you feel better
 
Two pairs of similar triangles and a pair of congruent triangles
 
8:35 PM
are there integrals which cannot be done by u-substitution, but only by integration by parts?
 
@Trey I'm not sure how you would do $\int\ln x~\mathrm dx$ by substitution
Or $\int xe^x~\mathrm dx$
 
@BalarkaSen Is Agrawal a very popular name in India ?
 
@Akiva Substitute $x\log(x) - x = u$ :3
@Gabriel Hm, it is, yes
Agarwal, I think, you meant
 
Aborderwal, I think it's called
 
Abigbeautifulwall
Thats the right thing
 
9:18 PM
Hi! $2^x =x^3$ has 2 answers. as you plug in the values of the argument the difference of two functions gets big initially, which means the graphs are to be apart at a sizeable distance; but in the long run they again intersect but initially one may think there is 1 solution. So, is there a way/method to get to know that there is 2 solutions not 1 not thinking and observing much the graphs?
 
@AkivaWeinberger ?
 
Can there by a differentiable curve in some Euclidean space whose derivative at zero is in direction $\hat v$, yet such that every neighborhood of zero contains a point at which the derivative is orthogonal to $\hat v$?
 
What about the level set of something like $y^3 - x^4\cos(1/x)$ in $\Bbb R^2 \subset \Bbb R^3$?
Parametrize it near the origin by some $\alpha(t)$
Should do the trick
 
10:15 PM
hi
 
@BalarkaSen to make sure I understand, are you suggesting the graph of $x\mapsto x^{\frac 43}\cos(\frac 1x)$ about zero?
 
(I raised it to the 5th power because I am paranoid and the graph comes out nicer)
@Arrow Sure, I mean, the locus of points on the plane satisfying $y^3 = x^4 \cos(1/x)$
Actually yeah I guess it needs to be $x^5$ not $x^4$
Whatever, doesn't really matter. You can make it smooth near $0$ of arbitrary order
By raising $x^n$ for large $n$
 
Sorry, why is having 5 and not 4 important? Is only one of them differentiable?
 
Something like that.
I haven't done the computation by hand. But the more you raise the power, the smoother it becomes
I think you can work it out now.
 
10:25 PM
Thank you
 
@LeakyNun i formalized the limited numbers
 
Hi. I'm interested in the asymptotic expansion of $f(N,N)$ as $N\to\infty$ for some function $f$. I've found a paper giving a result for $f(N,z)$ as $N\to\infty$ uniformly in $z$. The question is: due to this uniformity, can I apply this result for my case (i.e., to take $z=N$)?
 
@BalarkaSen This is why I'm alive
@Daminark Not a book yet, but I do highly recommend
 
Wait it's not a book? How so?
 
@Daminark They're his course notes, but my adviser said that he thinks Ravi will publish them soonish
 
10:37 PM
Oh, I mean even if they're yet to be published it's practically a book
 
They're really often mentioned by people, since they're basically the best notes you can find that'll get you into scheme theory
They still change too frequently for me to call it a book
Of course, you need look nowhere else than the best version of his notes: math.stanford.edu/~vakil/216blog/FOAGjan2915ereader.pdf
 
Lol I just took the most recent version. What's good about those over the current one?
 
Oh, that ones a fun one
It's worse in most ways
 
Weird, I mean Vakil himself must've had some reason in mind for making the changes
 
Oh, I linked the ereader version
Which looks hilarious on a computer
(Which he himself notes in the list :P)
 
10:49 PM
Kek
 
@Daminark You know about measure theory I think I read. Do you know much about Fourier theory?
 
Nope, just how to prove that trig polynomials are dense in $L^2$
 
Pontryagin duality etc
 
Oh no I don't know any of the fancy business like that
I mean I know the statement of it, dual group being continuous homomorphisms to the circle, and that the double dual is isomorphic to the original group
But I dunno how to prove that
 
I think that's a special case maybe
The circle part I mean
 
10:53 PM
@user21820 sorry we got your room frozen
my fault
 
@Narcissus what's the general version?
 
I think one can lift the requirement that its abelian
 
Hi guys
 
@Daminark Nah, we good fam, I think that's the most general it is. I think it is called something else in the higher generality
 
We had derivatives of power series lately and I'm confused why we didn't make the index shift from $0$ to $1$ in the derivative, e.g. we have a power series $f(x) = \sum_{k=0}^\infty c_k x^k$ and for its derivative we specified $f'(x) = \sum_{k=0}^\infty k c_k x^{k-1}$ which starts again at $k=0$.
 
11:01 PM
In mathematics, Tannaka–Krein duality theory concerns the interaction of a compact topological group and its category of linear representations. It is a natural extension of Pontryagin duality, between compact and discrete commutative topological groups, to groups that are compact but noncommutative. The theory is named for two men, the Soviet mathematician Mark Grigorievich Krein, and the Japanese Tadao Tannaka. In contrast to the case of commutative groups considered by Lev Pontryagin, the notion dual to a noncommutative compact group is not a group, but a category of representations Π(G) with...
 
All other resources I found move the index up by one, e.g. proofwiki.org/wiki/Differentiation_of_Power_Series Is this a problem with our definition? Because then the first term is a fraction $\frac{1}{x}$ which isn't defined at zero....
 
But k=0 kills the term, so it doesn't matter in the end
 
Oh yeah totally looked over that
Thanks
 
@Dami do you want to think about another functional analysis thing?
 
11:27 PM
@Narcissusjewel Is this a generalization though? I think Pontryagin is a thing in locally compact abelian groups, so it seems like they generalize the compact abelian group case in two different ways
@Alessandro sure
 
@Daminark You're right, I'm too tired to type carefully
 
Nevermind, I shouldn't be doing functional analysis during the night, I write dumb things
 
Lol rip Alessandro
 
Ok let's try again
I want to construct a counterexample to the statement that the distance between a point $x$ and a closed set $C$ is attained for some $c\in C$ in a normed vector space
I think that $C=\{(1+\frac1n)e_n\}$ in $\ell^2$ should work, with $x=0$ I get $d(0,C)=1$ yet there is no point of norm $1$ in $C$
 
yes that works
 
11:43 PM
More generally if $u_n$ is any sequence on the closed unit ball with no convergent subsequence in some Banach space I can pick $(1+\frac1n)u_n$
So when we prove that the projection on a closed convex subset of an Hilbert space is well define, convexity is needed to prove the existence of a point minimizing the distance and not only to prove its uniqueness
 
Actually I want $||u_n||=1$ rather than them being in the closed unit ball
 

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