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6:01 PM
@Astyx Zeros, in a field?
Oh, wait, nevermind :P
Brainfart
 
We have that A, B, S are collinear. If S is in $\vec{AB}$ then does it hold that $\vec{A}-\vec{S}$ is a multiple of $\vec{B}-\vec{S}$ ?
And does it hold that $|SA|^2|SB|^2=((\vec{A}-\vec{S})(\vec{B}-\vec{S}))^2 \Rightarrow |SA||SB|=(\vec{A}-\vec{S})(\vec{B}-\vec{S})$, or can we not take the square root?
 
@SteamyRoot i proved it the way you suggested but i dont feel comfortable with $p \ ^ {-1} (p (U) ) = U$. i get that $U$ is contained in the left set. but if we take $z \in p \ ^ {-1}(P(U))$ which is not in $U$, then $z \in V$ where is the contradiction?
$p \ ^ {-1}p(z) \subset V$
 
What are $U$ and $V$ ?
 
the separation of $X$
 
Recall that $p^{-1}(y)$ was either in $U$ or in $V$
 
6:12 PM
right so $z$ is in both
thanks again :)
wait no
$p \ ^ {-1} p(z) \subset V$
 
Just think of it as this: $U$ is the disjoint union of a bunch of $p^{-1}(y)$'s, and so is $V$.
 
alright ,i agree with this .
 
@SteamyRoot I know, right ?
:p
 
we also get $p(z) \in p(U) \cap p(V)$ and that can't happen.@SteamyRoot
 
My physics oral exam if anyone is interested
(@Semiclassical in particular)
 
6:28 PM
Hrmf.
 
oof, nice
 
Make a gif of it going through the loop
claim you invented a perpetuum mobile.
 
I should be able to explain this after I learn more EM
 
m8 do rep theory not EM
Lol jk
 
6:41 PM
So I went over to the black board, and then he told me : "We're going to watch a video"
 
@Daminark i dunwana do rep theory
 
I was like "wtf ?"
 
i wish i had more time for riemannian geometry though
@Astyx lol
 
Then we watched the first 45 seconds, and he told me : "Explain it, and give me the maximal speed"
It was very unusual but very fun
I enjoyed it a lot
 
hello chat
 
6:42 PM
Hi
 
@BalarkaSen Do geometric rep theory then
 
@Astyx nice
 
is anyone doing interesting things
 
Not really. Just relaxing right now
 
@Eric Benson probably is
 
6:46 PM
@TobiasK what does that entail
 
probably
 
@EricSilva I'm listening to Talking Heads
 
@BalarkaSen Not entirely sure actually
 
Hope that answers your question! sits in self-satisfaction
 
pretty interesting
 
6:47 PM
you mean the band or is their something else with that name @Balarka
 
the band
 
ah ok, my roommate plays them all the time
ive never sat down to listen to them though myself
 
they are brilliant
fargle put me on it
 
cool cool, yeah they seem like something id like if i were in the right mood
 
you could also listen to the super obscure albums of Byrne with Eno
like there are 2 of them but yeah
 
6:51 PM
im down for obscure music
 
@EricSilva Then try D.A.A.U (Die Anarchistische Abendunterhaltung)
 
Balarka why
 
Also recommended are Youngblood Brass Band who play hiphop on brass instruments
 
y u do dis
oh shit that sounds right up my alley @Tobias
 
6:54 PM
Does anyone know a reference with explicit examples of surfaces in euclidean 3-space parametrized by asymptotic lines?
 
cuz joyce eez rockin' it bruh
if you like hip hop on horns you could listen to Byrne and St Vincent's album
I like the first track so much
 
@Balarka this is a sick jam
 
right?
 
^^
 
m8s u gotta listen to this glorious music video called "Get rekt"
 
7:07 PM
oh yeah that is VERY good
definitely my favorite of all time
 
Yeah Balarka can confirm
Like that's what this video is referring to: youtube.com/watch?v=bRmIjBYToi0
 
@Daminark well, 7K The GOD is a better musician than NANDA tho
 
@Daminark That's gold
 
listen to the commentary here: youtube.com/watch?v=fbIyAOhNRGs
 
Yeah Bill Wurtz is just great
 
7:09 PM
Did he take a random recording or did he speak like that just for the sake of this video ?
 
I think he took the recording?
Doesn't sound like him, and he mentions Paul Mccartney at the end who I presume is the person who said that
 
That's so cool
 
I always forget the title of the original video
 
Show that $X$ is simply connected if and only if any two paths in $x$ with the same initial and terminal points are path-homotopic
I'm trying the prove the forward direction of the above, but I can't seem to construct a path homotopy
 
Do you define simply connected as path connected plus all loops are nullhomotopic?
 
7:15 PM
Yep, that's the standard definition I think
 
@Perturbative So the two paths give you a loop and you just need to turn the homotopy that contracts it into one from one path to the other
 
@Tobias That does not suffice, because you want to fix the endpoints
 
I was thinking of doing something like the above to $I \times I$
 
The nullhomotopy of the loop does not necessarily do it.
 
@Perturbative Well, you also need to still assume the space path connected
 
7:16 PM
Okay so, if all loops are nullhomotopic, they are homotopic to each other. So traverse one path and then another to get a loop, and homotope that to the path which just goes forwards and backwards
 
@BalarkaSen I said turn it into
 
Ah ok I misread.
Right, that's the whole point of the exercise
 
Woops I misread the exercise too, sorry about that everyone
Just ignore that graphic I posted
 
Womp womp
 
Dami, what do you mean by path that just goes forwards and backwards?
 
7:32 PM
Goes from x to y back to x along the given path
 
Ohh okay
 
If a and b are your paths from x to y, think about a * b^-1 * b maybe
draw some pictures
 
Or some long exact sequences
 
how about spectral sequences instead
 
That, I don't know about
 
7:36 PM
It's like a book with each page having an elaborate chain of long exact sequences
and a formula to move from one page to another
The first page is your classical, nice (co)homology long exact sequence >:)
 
@BalarkaSen Each page doesn't have a chain but a grid of sequences
But they are complexes, not exact sequences
 
Sounds lit. Why don't we teach it in schools?
 
And the next page has the (co)homology of the previous one
Or rather, it is not a grid, but you have a bunch of chain complexes at a certain angle, and that angle changes as you move up the pages
 
@TobiasKildetoft Sure you have a chain? It's just a bunch of chain complexes with arrows going between them
 
@BalarkaSen Well, you have a grid of objects with chain complexes at a certain angle
With the angle determined by which page you are on
 
7:42 PM
@Daminark @BalarkaSen So if $f$ and $g$ are two paths in $X$ with initial point $p \in X$ and terminal point $q \in X$, then $f \cdot \bar{g} \sim c_p$, that part I get but what I don't get is how to homotopy that to the path which goes backwards and forwards
 
And for some reason they almost always start on page 2
 
This angle thing confused me greatly
 
@TobiasKildetoft I mean it's a doubly indexed grid, sure, but the stair-step diagram gives you a (sequence of) exact sequence
 
Like the Grothendieck spectral sequence which has $R^iF(R^jG)$ on the second page at place $(i,j)$
And which abuts ("converges") to $R^{i+j}(F\circ G)$
 
Eg think about spectral sequence of homology of a CW complex filtered by it's CW decomposition
 
7:44 PM
(for nice enough functors $F$ and $G$)
 
I dunno the Grothendieck SS. I am almost always thinking of filtered things and basic examples of that (Serre SS, say)
@Daminark The amount of information one needs to process to compute with them is a massive pain in the ass
 
Nah I'm replacing the elementary school curriculum with spectral sequences
 
Usually one only uses them when they either abut quickly (like on page 3 or 4) or for the fact that in some nice cases they lead to a 5-term exact sequence
 
Or at least, upto scale. (like, the only interesting pages can be 2 and n (and infinity), say)
I derived the Gysin sequence out of Serre SS recently
@Daminark At least use these: math.cornell.edu/~hatcher/SSAT/SSATpage.html
other than some Peter May thing
 
@BalarkaSen Have you read "You could have invented spectral sequences"?
 
7:53 PM
yah.
Tim Chow writes nicely, but unfortunately it was all from an algebraic and dry point of view for me
i think i even emailed him about it, and he replied back with nice insights
 
Lol I mean, I had a very brief excursion with Hatcher since we did some stuff in difftop along those lines and I dunno, it didn't resonate well with me in my memory
Maybe I should see more of both it and Concise to really tell which is more my speed, and also check out the other one Eric mentioned, but as of now I just prefer Concise
Which means that I will make everyone use it because no one is allowed to have different preferences
 
Ahh, so if $G$ is the homotopy $f \cdot \bar{g} \sim c_p$, then a path homotopy between $f$ and $g$ would simply be $H(s, t) = G(s, t) \cdot g$
Since $H(s, 0) = f(s)$ and $H(s, 1) = g(s)$ and $H(0, t) = G(0, t) \cdot g =p$ and $H(1, t) = G(1, t) \cdot g =q$
 
8:16 PM
awake midnight, this night dream has this integral loat across in one scene:
$$\int \sin (\theta)\sin\left(\sin\left(\frac{\pi\tan \theta}{2}\right)\right)d\theta$$
Now I should go back to sleep cause I am tired
 
Hi Ted
 
peter pan
 
Hi Astyx :)
Hi, Forever.
 
What's up ?
 
Just back from 2 1/2 hours at the dentist. You done studying? :P
 
8:28 PM
I still have one math exam tomorrow, the importantest actually
 
I berembered.
 
I kinda messed up today's math exam though :/
 
Oh?
 
Like, I was going slow and stuff (and it was analysis, again)
 
I'm surprised they don't think linear algebra is important
 
8:30 PM
At the end the examiner wanted to speed things up so I could finish the exercise, but instead that confused me and I became stupid for 10 minutes
They do, I'm just unlucky
Most of my classmates have had an algebra exam
 
I would ask a mixture of stuff, personally.
 
Possibly tomorrow (hopefully) ?
Yeah, but an hour is a short amount of time
 
What was being discussed when you got confuzled?
 
So the problem was showing that if $P\in \Bbb R[X]$ and $(\cos(P(n)))_{n\in \Bbb N}$ has a finite number of limit points, then $P-P(0) = \pi Q$ where $Q\in \Bbb Q[X]$
So the way she wanted me to try was to introduce $\Delta : P\mapsto P(X+1)-P(X)$
 
Interesting ... I've never seen anything like that before.
 
8:33 PM
And at the end she obviously wanted me to say that we could "invert" $\Delta$
(I actually said that $\Delta$ established a isomorphism from the space of polynomials that cancel at 0 into the space of polynomials)
 
Well, if it makes you feel better, I don't see how to do this ...
cancel at 0?
 
I mean $P(0) = 0$
 
Oh, 0 constant term.
 
See that if $(\cos(P(n)))$ has a finite number of limit points, then so does $(\cos ((\Delta P)(n)))$
 
I don't see why it's invertible.
 
8:36 PM
(lots of parenthesis)
You restrain yourself to $\Bbb R_n[X]$
 
Oh, I guess you look at a formula for it.
 
Then $\Delta (X^k)$ is of degree $k-1$
So it's surjective
 
Right. It lowers degree.
Yeah, I see.
 
This means that $X, \dots, X^n$ get mapped to polynomials of degree $0, \dots, n-1$ QED
 
Yeah, I got it.
OK, now what?
 
8:38 PM
See that if $(\cos(P(n)))$ has a finite number of limit points, then so does $(\cos ((\Delta P)(n)))$
By expanding the $\cos$ and extracting five times (she asked me to be specific on that one ... ugh)
 
extracting?
 
Isn't that correct english terminology ?
 
One can extract teeth and one can extract roots.
 
Like, if you have a sequence $(u_{n})$, if $\phi : \Bbb N \to \Bbb N$ is strictly increasing, $(u_{\phi(n)})$ is said to be an extracted sequence in french (suite extraite)
 
oh, take a subsequence
we definitely don't say that in English
 
8:40 PM
Yeah, don't you have a word for that in english ? What a poor language :p
 
Pass to a subsequence :)
 
Right :)
So now we have the tools to do our induction
First, with degree $1$ write $P-P(0) = a_1X$
 
This is clearly a pet problem of hers and she expects to lead people through it.
BTW, I don't call this an analysis problem.
 
Neither do I
Well I did
But ... :p
 
More linear algebra than analysis.
Once you know definition of limit point, it's just trigonometry and linear algebra.
Oh, and proof by induction :P
So the point of $\Delta$ is that it's degree-lowering for the purposes of your induction. I get the main idea.
 
8:43 PM
At first I thought of proving $(Q(n)-\lfloor Q(n)\rfloor)$ (where $Q$ is defined as above) had a finite number of limit points and to conclude that $Q$ was in $\Bbb Q[X]$
And also that it's a bijection from $X\Bbb Q[X]$ into $\Bbb Q[X]$
 
It still seems weird to me that it must be rational polynomial times $\pi$ if you're getting finitely many limit points.
 
Degree $1$ is just proving $a_1$ and $\pi$ have to be commensurable
 
Oh, is there some stipulation that $\cos(n\theta)$ is dense in $[-1,1]$ if $\theta$ is an irrational multiple of $\pi$ ... or something like that?
 
Bob
What does it mean for a function to be dense?
 
I think she thinks I didn't get the main idea, because she was stressing me out, so I'm not too satisfied
 
8:46 PM
Values are dense.
 
Yes, that's the intialisation with degree 1
 
I said $\theta$ is a particular sort of number.
That itself takes a proof that's not a joke.
 
what's up guys?
 
That is interesting to beginning manifold students because you get the dense line on a torus from this fact.
 
Yeah, I just claimed it and she acted like it was trivial
 
8:47 PM
I've seen plenty of students who can't prove it :P
Interesting what she's interested in and what she isn't ...
 
I can but she didn't want me to :(
Yeah, I found her questions to be way too specific, preventing me to express what I actually understood
 
Well, as I said the other day, I suspect she had already decided you knew stuff before you all got into this question.
 
@TedShifrin what kind of number?
 
There are examining styles just as there are teaching styles.
 
I had only this question :p
 
8:48 PM
Well, scroll up and read, Typhon.
 
i did
 
That's probably true
 
What, Astyx? The whole hour was this question.
 
45 minutes, but yeah
 
Oh yuck.
 
8:49 PM
At first she let me try my approach (which was unsuccesful)
After one hour I would probably have finished it
 
I think the first question should be something relatively standard just to settle things down.
Again, different styles.
 
hmmm
 
But I guess they've seen your written exams?
 
so theta is an irrational multiple of pi?
:/
 
Probably not specifically
 
8:50 PM
Wow, you can read ...
 
But since I'm one among 200 to have passed the written exams, she probably knew I was not completely stupid
 
@TedShifrin it wasn't entirely clear if that was what you were claiming made cos(n*theta) dense
 
hopefully
Did you see my physics oral exam by the way ?
 
Excuse me? It was completely clear.
 
to me it wasnt
 
8:51 PM
No, I got here when you saw I got here, Astyx.
 
Bob
How did you do Astyx?
 
Typhon: Oh, is there some stipulation that $\cos(nθ)$ is dense in $[−1,1]$ if $θ$ is an irrational multiple of $\pi$ ... or something like that? ... WTF is unclear?
 
@Ted Here :)
 
i wasn't sure if you were talking about the same thetas
 
8:52 PM
Not as good as I would have hoped @Bob
 
or if you were saying that theta's which made the conjecture hold true were a broader set beyond just irrational numbers
 
Bob
are you joking?
 
I'm done with you, Typhon.
 
Hi Alessandro, what's up ?
 
Heya @Alessandro
 
8:52 PM
@Bob me ?
 
im not joking
 
Bob
are you joking @Astyx ?
 
why are you being rude?
 
@Astyx I had my probability exam today
 
No @Bob, why do you think I am ?
 
8:53 PM
It was much harder than expected, but I'll wait the results before commenting
 
Wise decision
 
Oh, this was your retake, @Alessandro?
 
Bob
The link I got was to a video of a simple electronic device. I guess I clicked on the wrong link.
 
Not really a retake since I never took the full exam, only a partial exam but yeah
 
Ohh.
 
8:54 PM
@Bob It was
 
Bob
that is why I thought you were joking.
 
Read the next messages
Yeah, it was amazing actually
Unluckily it won't get me any point for the ENS Paris
But hey, that's life
Anyway, big exam tomorrow, gotta sleep well
Bye chat
 
Night @Astyx
 
Hey there!
 
Bob
Good luck with the exam @Astyx
Bye Chat
 
9:03 PM
Hi @Dami
 
How's everything going?
 
Hi chat
 
I should have remembered today is nationnal day ... so fireworks, no sleep for me yet
Hi Semi, did you see my physics oral exam question ?
 
Oooh, right
I may be able to spot some fireworks outside :O
 
Are you that close to the border ?
 
9:14 PM
gg @Astyx
 
@Astyx Quite close, yeah.
Though, before my parents moved, we were literally 100m away from the border :P
 
I used to live in Strasbourg, I could go by foot to germany
 
Haven't, no
 
I pinged you
 
@Astyx I used to live here
 
9:18 PM
Indeed, quite close :p
Damn these fireworks just keep on going
 
@Astyx I was in Strasbourg a couple of times since I lived nearby for a year, it's a very nice city
(I was in Germany though)
 
Yeah, it's quite beautiful
I'll revisit it someday
 
Wasn't the storming of the Bastille on the 14th of July?
 
Yeah, but some cities celebrate it on the 13th cause we french like to make the most out of Bank holidays
 
Yup, in 1789
 
9:26 PM
Ah, I see
 
honestly at least like this they make the most of the fireworks
It's a bit of a waste to do all of them on the same night
This being said, it just finished, so I can finally sleep :)
Bye (for good this time)
 
Wait do you have an exam during a national holiday?
 
Well, g'night and good luck with the exam tomorrow, then.
 
I do, and I had on the weekend (saturday + sunday) too
They just don't give a damn
 
That's so weird
 
9:29 PM
Thanks, bye !
 
(I had exams on Saturday before, but Sundays and holidays is crazy)
 
It's not weird that they make students take exams during weekends and holidays.
It's weird that they find staff willing to assist in that.
 
Good luck for your exam then!
@SteamyRoot My university is just closed the whole day during national holiday
 
Mine is too, officially. I can still get in if I want to :D
 
It's an holiday for the people working in it as well so...
 
9:48 PM
My school's math building is never actually closed, which is very convenient
 
I am a perfectly articulate counter-monster
2
 
@ForeverMozart so random. much random. doge
 
10:25 PM
of the writing of many papers there is no end
- Ecclesiastes 12:12
 
-Yoda 3500
 
@HenningMakholm I recently have been having problems with the same user! Although the question did make sense in this case (with some clarification about some "bad english"). I am starting to think this guy is a troll (looking at the \sqrt 2 question or the ridiculous edits for the one you participated in) or growing into a crank and deeply affected by Dunning-Kruger.
 
@HenningMakholm I didn't know you came here... I feel I'm in greater company now
 
In greater company, now I feel.
 
10:41 PM
0
Q: Verifying $|F(r)| \geq \frac{1}{1-r}\log(\frac{1}{1-r}) ] \text{and} |F(re^{i \theta})| \geq c_{q/r}\frac{1}{1-r}\log({\log(\frac{1}{1-r})})$

ZophikelIn the text "Complex Analysis", by Elias M Stein and Rami M. Sharachki i'm attempting to take take a Taubrian Route to verifying the following properstion in $(1)$ Let, $F(z)$ be the following series: $$F(z) = \sum_{n=1}^{\infty}d(n)z^{n} \, \, \text{for} \, |z| < 1$$ $$\text{Remark}$$ One can...

^ Can anyone help or give any helpful hints plz :-) thanks
 
hey
 
11:41 PM
Hey @Ted!
 
11:59 PM
@ForeverMozart I don't often. Several years ago I used to hang out here a lot, but I had do cut down on the timesink.
 

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