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20:00
I'm like decent, but I have actual professional friends who blow me out of the water
Well it's all about building foundations and moving up and if they're professional I'm sure they took the time to make sure they had those solid foundations and work much longer at it than you do.
But that doesn't demote any of the progress you've made.
You should be proud of yourself.
@Zee I'm more of a jay-z fan. I also like Earl Sweatshirt.
I mean I am proud of what I can do musically :P I just wouldn't say I'm talented or very good at piano
Zee
Zee
@EricSilva that's ok , you just gotta be good enough to impress people, after all that's what life is about
besides I don't really play music for other people, for me it's a way to connect with heritage so I don't really think in terms of how presentable it is to others
My dad has always thought that I was a good musical person because when I played guitar I could always play decent music. (pink floyds "is there anybody out there?" for one) but I'm really musically incompetent... I guess it's all dependent on opinion.
Zee
Zee
20:03
We do everything for others , let's not fool ourselves
Well, I am not looking to perform at all. I play to learn something new and to escape for a little bit.
@Zee I am 23 years old. I have no friends, only a girlfriend. everything I do is mostly for my own enjoyment. I go fishing to catch fish for me, not to post it on facebook (which I don't have) I play piano to get better and to relax. I do math because it's beautiful.
We don't do everything for others.
Though you might.
Welcome back Ted.
Zee
Zee
You do all these things for society, fishing is for relaxing
oh hi again Ted
Long time no see Ted
20:06
meh what i do is ridiculously immaterial anyway so i try not to ponder whether i do it for society or myself
I am not usually in direct contact with society.
Zee
Zee
I agree
:P
Hi again, Eric, @Astyx. BTW, Astyx, the piece of music I detest most in the entire world (and the next world) is Bolero.
Does anyone have societies phone number? I haven't seen them in a while.
20:07
@Ted Lol, I can understand why. I personnaly really dislike playing it and can't stand to listen to it from the beginning to the end :p
@Ted the Ravel piece?
@TedShifrin Who is your favourite composer?
$\circ$
proceeds to leave the chat
Yes, Eric. It's worse than Phillip Glass, whom I abhor.
AHAHA I live in Phillip Glass's dorm from when he went to UChicago
he lived next door to where I do now
Zee
Zee
20:08
How can you hate glass? He's a genius
I don't have a favorite. But I listen mostly to chamber music (typically with piano in it) from 18th century through 20th.
Nice.
Monotony and totally uninteresting.
I also dislike Glass a bit, not a fan of the minimalism personally
My dad was actually a composer.
20:09
I don't mind the Ravel piece but I'm not excited by it
Woah cool Ted!
Zee, your idea of genius and mine clearly will never agree.
I totally mind it. So f***ing repetitious and uninspired.
What topic or branch of mathematics, or type of manipulation does this question & answer involve?: http://i.imgur.com/VAuiKH4.png

I'm being taught a specific syllabus and have exam for this specific component, this question supposedly being a part of it that we've never learnt, hmm.
Maybe we should do math before I get banned.
I think that's fair
20:10
Tchaicovsky (am I spelling that right?) is a great composer.
yeah I like Tchaikovsky
He's very low on my list, Dodsy.
That's Nate to you, Ted...
Despite my Russian heritage.
LOL
20:10
@Dodsy he's Russian so I guess it depends how you convert the spelling
I have a feeling you're a Bach fan, @TedShifrin
Bach is good.
Whoa @Ted you have Russian heritage? That's cool!
@EricSilva ah I see.
Somewhat, Lozansky, but I really listen to classical, romantic, and more modern stuff.
20:11
Bach was an acquired taste for me
Hya @Dami
Demonark, yes, 75% Russian, 25% Polish.
@TedShifrin Claude Debussy?
at first I didn't get it, but at some point I started to really get why Bach is good
@EricSilva Air on the G string is amazing.
I'll probably get banned for saying that?
20:12
Hey @Astyx, and everyone!
because someone will think I'm not talking about music.
@EricSilva I started to appreciate Bach more after functional analysis
How was class ?
Zee
Zee
@TedShifrin check out Max Richter , he's good
I should listen to music more systematically than I do but meh
20:13
But @Ted my house throws a phillip glass festival every year and we invite him to come. He never does though
@Zee On the nature of daylight is one of my favourites
I should do pretty much literally whatever I plan to do but don't
but people always give me crap cause I do not like his music
@TedShifrin how's it going?
It was good, Laci had a few fun quotes
20:13
I'll pass, Eric.
good choice :P
Ludovici Einaudi is a piano god.
"This is my result from way before you were born"
"Johnson is Schurian... Well that sounds like a conspiracy theory."
Zee
Zee
@Daminark did you appreciate Chopin after complex analysis?
@TedShifrin I solved that parametric surface problem. Don't need fancy approximations and stuff. The solution is actually superior that what I would've had without it.
20:14
@Zee Haven't had complex analysis :P
Well I should probably stop talking about music with you guys.
Zee
Zee
@Daminark if you don't like Chopin then we can test the hypothesis
And go do something productive with my life.
we should return to math
Zee
Zee
Yes I gotta go study
20:16
@Daminark is it because of Banach
idgi
Haven't heard any of that. I'll get back to you once I do it
@Daminark Hi. Bye.
Yeah @EricSilva
20:16
@EricSilva Quick, tell me, why are junctures of C^2 foliations compactly supported?
@amWhy how's it going?
[on getting back to math]
My favorite is, what's yellow, normed, and complete?
@TheGreatDuck ;-) $\uparrow$
Zee
Zee
@EricSilva I think couse Bach was big and topology vector spaces are big
20:17
puts Daminark on ignore
@Balarka I know $-\infty$ about foliations
@Daminark a banana
No, but I really must be going.
@TedShifrin Don't be a stranger, friend.
@amWhy I didn't say what's up.
@EricSilva we need to find Calegari
@Dodsy More like a Bananach space
@Daminark I was right then.
Bye, Nate.
20:18
@Daminark why Calegari...?
@Dodsy you ever do surface geometry?
I think he knows about foliations?
Danny, not Frank
OH the Thurston student
gotcha
I like his notes
@TheGreatDuck hehehe, clever... Actually I meant to smile and used $\uparrow$ to try to express being "up"-beat!
20:20
He's also a great lecturer @Balarka
Yeah, I've heard he's great
gave a talk on hyperbolic groups that I attended
was very cool
Fun!
I love hyperbolic groups but i know nothing about them
In the absence of grad analysis I may do his undergrad complex class, since he apparently covered things the TA didn't know
(Grad student TA)
This was after I did a reading with one of his grad students on hyperbolic groups and his talk tied everything together really nicely
20:21
@TedShifrin How are you these days....seems you're kept pretty busy! :-)
Like modular forms and elliptic integrals, he seems to actually be one of the professors who teaches a worthwhile undergrad class
@amWhy ah cool.
Hi @amWhy. Doing pretty well, thanks. Heading off to Europe for a month in a week.
Oh shit that's cool
I'll be taking his algebraic topology course in fall @Daminark
Along with Schlag and maybe Nori?
Ah, nice
Wait I just realized he's double teaching
20:22
@TedShifrin I might fit in a big suitcase, to tag along! That sounds awesome!
@TedShifrin Hello! Mr. Shifrin, is there any way to find out the algebraic multiplicity of an eigenvalue, if I know the geometric one?
no love for souganidis?
LOL ... sounds like a good idea.
Oh Soug's pretty great too
@Balarka why would I love souganidis
20:22
@Kirill: Usually its the algebraic that's easy. Just look at the characteristic polynomial.
I'll be back later. I need to reboot. A bunch of stuff isn't working on my computer.
It'll be a nice new source of memes (which we haven't been making nearly as many of with Schlag and Marianna, unfortunately)
Ah true Souganidis will be teaching Daminark
I forgot about this
@TedShifrin The other way round - my polynomial has to have factors $(x-1)$ and $(x+1)$, but I do not know how many
Inb4 75 problem psets
20:25
@Daminark inb4 playing basketball is on the final
Final
Question 1 - Fill in the blank:
"Math is not a ____________"
@TedShifrin ok
@Daminark fish
spectator sport
@Daminark Bath
20:26
hello, if $d(x_n,x_{n-1})>\varepsilon, \forall n$ why this means that we can't find a convergent subsequence ?
But yeah I won't be doing both of them. Basically, I'm doing algebra for sure, logic if I'm later gonna do grad geotop, core bio, and either grad Sougi or Calegariplex
" -I mean, I love watching basketball, but I don't play, so I can never be good ehhhh"
Aaaand Eric is the winner!
You mean math is a fish ?
@Astyx as it turns out, yes
20:27
(where ehhh is that sound he makes like the fonzie)
good life choice
@Astyx hello
Bonsoir @Vrouvrou
Wait hold on this isn't the first time I've done this, with grad Sougi
One time in problem session I was reminding people of a problem we did and I said "Soug chapter 2" instead of Rudin
@EricSilva lol
20:28
"It's calculus, you just gotta play the game ehhh"
I can't remember if I've heard that phrase before
I have heard him say it like minimum 40 times
@Astyx s'il te plait, si $d(x_n,x_{n-1})>\varepsilon, \forall n$ pourquoi cela veut dire que l'oon ne peut pas extrere une sous suite convergente ?
god this is 2much meme
@Balarka I showed you a few pictures but there are at least 10 or 15 more floating around
20:31
Ok I'm done making souganidis jokes
Schlag time? Lol jk
@Vrouvrou on peut, par exemple $(-1)^n$
Enfin je veux dire que ce n'est pas exclu
@Astyx j'essaye de montrer que "si de toute suite on peut extraire une sous suite convergente alors on peut recouvrir l'espace par un nombre finie de boules de rayon $\varepsilon$"
Ah ça c'est différent
Tu veux $d(x_n, x_k) \ge \epsilon$ pour tout $n\ne k$
j'ai construit une suite $x_n\in E$ tel que $d(x_n,x_{n-1})>\varepsilon , \forall n$
20:39
C'est pas suffisant
I want a diffeomorphism $f : [0, 1] \to [0, 1]$ such that it preserves particular strictly decreasing sequence $\{x_n\}$ with $x_n \to 0$, and $f(x_k) = x_{k+1}$. I can produce such an $f$ which is $C^2$, right?
Meh, sure, why not. Like $x/2$ with $1, 1/2, 1/4, 1/8, \cdots$ works, modulo bumping stuff up to make it compactly supported away from $0$ :P
@Astyx $d(x_n,x_0)>\varepsilon, d(x_n,x_1)\geq \varepsilon,...,d(x_n,x_{n-1})<\varepsilon$
Tu veux dire $\ge$ à la fin, mais oui
oui
et donc pourquoi on ne peut pas extrere une sous suite convergente ?
s'il vous plait
Procède par l'absurde, suppose que l'on peut en extraire une sous suite convergente, alors à partir d'un certain rang tous les termes sont à moins de $\epsilon/2$ de la limite
Essaie de voir pourquoi cela est absurde
20:46
$\forall \varepsilon\exists k_0>0, \foall k\in \mathbb{N}, k\geq k_0\Rightarrow d(x_{n_k},l)<\frac{\varepsilon}{2}$
@Astyx?
Oui, sauf que $\epsilon$ n'est pas quelconque ici
c'est le rayon de la boule ?
Oui
et donc je fait l'inégalité triangulaire ?
Essaie par toi même voir si ça marche, il ne faut pas que je fasse tout à ta place
Et une fois que tu auras trouvé essaie de comprendre l'intuition derrière mes indications
20:53
$d(x_{n_k},x_{n_{k_0}})\leq d(x_{n_k},l)+d(x_{n_{k_0}},l)\varepsilon$ contradiction
@Astyx
Oui
c'est juste ?
Oui
merci
Pas de quoi, retravaille cet exercice, c'est un classique et il entraine bien à avoir une intuition sur le sujet
Enfin ce n'est que mes conseils
Sur ce, bonne nuit
21:02
merci bonne nuit
@EricSilva It really depends on what you mean by computation. The question is really whether I could do the same computation in my head with enough time, and that's often not true.
Some tensor calculation no way I could understand out loud
Some spectral sequence computation maybe
maybe that says something bad about me
Hello, I am looking for a book/problem set on lattices and universal algebra with a lot of "is this a complete lattice", "is this a homomorphism of algebras", "is this class closed under H, S, P?" problems. Can anyone please recommend?
I'm I applying Ferynman's Integration trick correctly:
$$\frac{\partial{I}}{\partial{\theta}}=\int_{0}^{\pi}(-1)^{er}-1\frac{(-1)^\frac{{eir \theta - 2 \theta}}{\pi}i}{r}$$
$$\int_{0}^{\pi}\frac{\partial{}}{\partial{\theta}}(-1)^{er}-1\frac{(-1)^\frac{{eir \theta - 2 \theta}}{\pi}i}{r}$$
21:17
@ShaVuklia oh hey, separation of variables
My plan is to take the partial derative with respect to theta
That looks almost entirely incomprehensible.
@Semiclassical what's not clear
pretty much everything.
21:21
As a very obvious one, what is $cir\theta$ supposed to be?
And what's $(-1)^{er}$ supposed to be?
Plus, you haven't actually indicated what you're integrating with respect to
@Semiclassical it's that came form apply IBP for integrands in the form of $P(X)/Q(X)$, i'm integrating with respect to $\theta$
That really doesn't explain what the above is supposed to represent.
@Semiclassical they are just arbitrary values
That really doesn't help.
21:28
oh hey, dipoles.
Does anyone see why $r_2-r_1 \approx a \cos \theta$
Hey @Semiclassical
Also given is $r>>a$
@Semiclassical the integral i'm evaluating is part of bound i'm trying to construct
My guess is that what they mean is that the $z$-component of $\mathbf{r}_2-\mathbf{r}_1$ is approximately $a\cos\theta$. @Lozansky
But it seems like they've done some manipulations prior to that
I don't think so
Hmm.
I haven't done a dipole calculation in a while, so I may be wrong.
21:31
They've placed a point charge $+q$ and $-q$ at $K$ and $S$ respectively
Actually, looking at the geometry, $r_2-r_1\approx a\cos \theta$ looks plausible.
Yeah but how?
namely, draw a right triangle with KS as hypotenuse
Then $a\cos\theta$ will (approximately) be the base length and will (approximately) be $r_2-r_1$.
That said, this is handwavey
No no no, where did you get $r_2-r_1$ from?
That's way too handwavey for me
Well, if $r\gg a$, then the lines from $K,S$ to $P$ are more-or-less parallel
What I have in mind is the same approximation one makes when doing a double-slit experiment: hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html
21:35
Err don't agree with that
Isosceles, yes
Namely, the implication from that diagram that $\delta \approx d\sin \theta$.
But we don't actually have to be handwavey about this.
We know the locations of the points as such:
$O=(0,0)$, $K=(0,a/2),$ $S=(0,-a/2)$, and $P=(r \sin\theta,r\cos\theta)$
Agreed?
@Lozansky Is this Swedish?
@AkivaWeinberger Japp
@Semiclassical Yes, agreed
21:39
Mmkay. Then $r_1=|PK|=\sqrt{r^2\sin^2\theta+(r\cos\theta-a/2)^2}$.
And $r_2=|PS|$ which is the same but with $+a/2$.
Mhm yeah
Okay. Let's define $\delta = r_2-r_1$.
Ted showed that $\{(x,x^2,x^3):x\in\Bbb R\}$ has no three colinear points, right?
Do you remember how that went?
First, let's simplify: $r_2^2=(r\sin \theta)^2+(r\cos\theta+a/2)^2=r^2+ar\cos\theta+a^2/4$ and $r_1^2=r^2-ar\cos\theta+a^2/4$.
@Mike that's a very point. I was thinking something along the lines of a tensor calculation. It's weird for me because I actually usually can do those calculations in my head but I absolutely cannot verbalize my process until the end or i get lost.
21:42
@AkivaWeinberger Not sure if that's addressed for me?
To anyone, really
@AkivaWeinberger The message before the edit confused the hell out of me.
But $r_2^2=(r_1+\delta)^2=r_1^2+2r_1\delta+\delta^2$, or $$\delta^2+2r_1\delta=r_2^2-r_1^2=2ar\cos\theta$$
Deepest apologies
21:43
@Semiclassical Let me check
@Semiclassical Agreed.
Mmkay.
I'm going to have to head out, so I'll only justify the last steps briefly.
First, since $r\gg a$, we should expect that $\delta$ is a small quantity. In particular, it's a lot smaller than $r_1$.
So if the line is parametrized by $t$, the equation of the line is $f(t) = c + vt$. For $t = t_i$ that takes values $(x_i, x_i^2, x_i^3)$ where $i = 1, 2, 3$?
@BalarkaSen I actually was thinking about problem 12, and I feel like $(x,x^2,\dots,x^n)$ might work
For that reason, we can neglect the $\delta^2$ term.
That actually wants linear independence, which is stronger than not-colinear-ness
21:47
Second, we can also expect that $r_1$ and $r$ are not so different when we're far away.
Noncolinearity
Consequently, we can approximate the LHS of the last statement as just $2r\delta$.
In which case we can divide by $2r$ to get $\delta \approx a\cos\theta$.
Now, if that still seems handwavey...I pretty much agree.
Wait what
@AkivaWeinberger Can't you like, just consider the vandermonde determinant?
21:49
@SteamyRoot Don't know what that is, but the usual determinant would work for linear independence
yeah, but this is probably easier (I think)
I am blanking out and I don't have pen and paper in front of me so I back out
Probably the right way to do this is not the tricky way I just outlined, but rather to start from $$r_1=\sqrt{r^2-ar\cos\theta+a^2/4}=r\sqrt{1-\frac{a}{r}\cos\theta+(a/2r)^2}$$
I mean, I kinda get that we neglect $\delta^2$ but that invokes some problem with the $2r\delta$ term imo
the area of a triangle formed by three points $(x_i,y_i,z_i)$ is $$\det \begin{pmatrix}x_1 & y_1 & 1\\
x_2 & y_2 & 1\\
x_3 & y_3 & 1 \end{pmatrix}$$
area of triangle is zero iff collinear
21:51
And do a binomial expansion in powers of $a/r$
(actually, it's half that matrix, up to sign, but whatever)
@SteamyRoot Ahh, right, I see
And then binomial approximation?
in which case it's actually a lot easier, lol: $r_1\approx r(1-\frac{a}{2r}\cos\theta)$, and similalry $r_2\approx r(1+\frac{a}{2r}\cos\theta)$
Now, if $y_i = x_i^2$, it's a vandermonde matrix, with determinant $\prod (x_i - x_j)$
21:51
Right.
ooh neat idea
@Semiclassical Yeah I see how that might work
Oh that worked out pretty neatly
Like that a lot more
Yeah, I was really making things too complicated
okay, now a i reeeally need to go
21:53
Alright
Thanks for the help @Semiclassical
Hi all i'm sorry to ask you in this chat, but is there someone that understand about moebius strip here ?
Depends on what about the Möbius strip you want to know exactly.
The relationship between area and the diagonal of the inner polygon in it upper view..
Well, I don't really know what an inner polygon of a Möbius Strip is, so I'm afraid I'm of no help here.
21:59
Haha wait is bc I didn't explain it right..
@EricSilva I need to see it
Wait… $(x,x^2,\dots,x^n)$ contains the origin
So tensor calculations I need to see the symbol manipulation :P
I guess I want $(1,x,\dots,x^{n-1})$
.. actually I ask it a while ago but didn't had an answer

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