Good day!

in Russia we say "to smoke" something in sense "to learn". I like it.

Hi.

@ParthKohli hi

Hey again @Nimza. How are you?

@ParthKohli don't know. I have just woke up. How are you?

I'm tired studying a subject for exam that has no application in today's world.

Sanskrit.

that's normal. I had the same problem on 1 course studying 16 bit assembler

@ParthKohli Nations should teach only english.

hi

@JonasTeuwen Are you able to generate previews for multi-file documents?

@N3buchadnezzar hi

Actually, Hindi(in India) is fine too, but Sanskrit? Really... that's a crazy idea

@ParthKohli You have to study three languages?

@ParthKohli It is better than to study something you do not understand.

@N3buchadnezzar: There is no need to learn Sanskrit in today's world.

@Jayesh: Yes, from 6th to 8th grade.

@ParthKohli Nationalistic cultural maintenanance and bullshit.

You can maintain your culture by learning the native language, which is Hindi.

Sanskrit, on the other hand, is like too ancient.

Idiot educationalists in India. sigh

@ParthKohli Was there not any option of other? You have English by default and then you get to choose the other language right?

English and Hindi by default and another language.

Help me please, why $\Gamma = \left\{ (z,w) \in \mathbb{C}^2 \colon w^2 -z^4 + 1 =0 \right\}$ has 2 points at infinity? I compactify it: $\overline{\Gamma} = \left\{ (z,w,\xi) \in \mathbb{C}P^2 \colon \xi^2 w^2 - z^4 + \xi^4 = 0 \right\}$. Then if $\xi = 0$ we have $z = 0$ and then there is only one point?

@ParthKohli Hmm. I had chosen marathi then.

Anyone keen on helping me figure out a few questions regarding functional analysis?

Let $A,B \in M_2(\mathbb{C})$ and suppose that $A^2=B^3=I$, $ABA=B^{-1}$, but $A,B \ne I$. If $C \in M_2(\mathbb{C})$ commutes with both $A$ and $B$, then $C=rI$ for some scalar $r \in \mathbb{C}$.

can anyone give me a hint on how to even start this?

reminds me of a group presentation and the center of a group, but that's probably not related

@JayeshBadwaik Yes.

@JayeshBadwaik You can set up master document.

@JonasTeuwen Okay. Found (setq-default TeX-master <master doc>). Would search around before asking you now. Have got some good hang of Auctex now.

@JayeshBadwaik Just ask, I probably don't know either but might know where to find it! 8-).

@JonasTeuwen Okay. 8-)

What's up with these goggle guys? 8-), B|, 8|, B)

We are nerds, we need glasses.

@JonasTeuwen Sunglasses to look cool bro!

@N3buchadnezzar Yeah, in winter!

Hey its still above 0 Celsius here in Norway, that means we can still wear shorts, have barbeques, and get a tan.

Yep.

@JonasTeuwen Do you know what it takes for a space to be closed?

@N3buchadnezzar It has to be isomorphic to a copy of me.

It is a metric space, hence every convergent sequence must have the limit in the space.

That is, take a sequence of $f_n$ in your space and show the limit is bounded, continuous and has the end-point thingie.

I was just thinking, hey dude all functions in BC are complete, so a subset is also complete since it is cloed.

Which is just uniform convergence.

Yes.

And it is closed, because it is finite and has a supremum right?

No holes, or other nasty thingies

No.

Let $(f_n)$ in $X$ with limit $f$ in the $BC$ thing.

Still duplicate. Call for votes.

Then $f$ is bounded and continuous because of uniform convergence.

Because every cauchy sequence in BC converges?

@JayeshBadwaik I know MobileOrg. It is pretty shitty...

@N3buchadnezzar Use that the whole space BC is complete. $X$ is closed, as is shown here, hence it follows that it is also complete.

@N3buchadnezzar Na.

BC stands for the space containing all bounded and contionous functions, but I guess you figured out that much.

Closed subsets of complete spaces are complete.

I think their proofs are stupid.

You show directly that it is closed, not that the complement is open.

Closed in metric spaces is easy.

@Matt Yeah, I was just trying to formalize why the subspace was closed =)

@N3buchadnezzar You can drop the $B$ since a continuous function maps a compact set to a compact set and is hence automatically bounded.

Anyway. Just came here for dupe votes.

I came here for destruction.

@JonasTeuwen Pissst

Takes from a bottle labeled 60% pain and destruction

Hands bottle to Jonas

Meur pain!

There

one out of 6 questions answered

What's on here...

@robjohn are you there?

@Matt yes

(Since you can read deleted stuff...)

@Matt are there 2?

@Matt Hey

Hi

@Matt You organised a firing squad above to close the question :D

in the words of BD

Oh good, it worked : )

@Matt I closed it

I swear to god I've seen that question asked like 5 million times

@Matt meta.math.stackexchange.com/questions/5030/…

@BenjaLim No, I did never close any of Miguel's questions.

@Matt No I'm just referring to the phrase of BD

the phrase that in general

firing squads are organised here on chat

Besides I mostly close dupes. I think Bill is talking about the NARQ votes and the "too localised".

NARQ?

not a real question

ah

@robjohn It's ok, never mind. I'll bbl!

Later!

@robjohn hey

@Matt bye

:6220028 which is the other?

@robjohn I saw you deleted some guys answer

@Matt see you

@robjohn I now have 10k goggles :D :D :D :D :D

@BenjaLim was it an answer?

@robjohn well

@JonasTeuwen

@JonasTeuwen hey

@JonasTeuwen analysis?

@robjohn

No. Beer.

Analysis later.

@JonasTeuwen I have lately been into representation theory

@BenjaLim Very good. You must teach me.

Must go now!

@JonasTeuwen Have you ever studied representation theory

Yes.

like group representation theory?

lie algebras?

@JonasTeuwen Do you know what is an ideal?

Duh...

You have plenty of those in analysis as well.

@JonasTeuwen really?

Yes.

@JonasTeuwen I heard from pierre he never really went heavily into algebra

Even dear Pierre, your supervisor is good at that!

Yeah.

But he does operator theory.

yes I know.

Surely he has representations, ideals and algebras.

I agree.

And you should take "heavily" with a grain of salt. 8-).

He probably does know it, just not at research level (maybe).

@JonasTeuwen What kind of rep theory did you do?

For quantum theories.

?

So representations of classes of operators.

ok but what about purely algebraic rep theory? @JonasTeuwen

No.

Well, yes, of course I needed to know that before I could go into this...

so what kind of algebraic rep theory did you do?

@JonasTeuwen By the way

I need to go. Man!

Do you know if someone wants to be an algebraist

Talk to you later.

how much analysis should they study in undergrad?

@JonasTeuwen WHY ARE YOU LEAVING ME

They should just do all available analysis.

@JonasTeuwen

Otherwise they become idiots.

@JonasTeuwen HOw can that be

Certainly functional analysis.

@JonasTeuwen do you know lenny taelman?

Because then you might do something the analysts have done way before you.

Yes, of course I do.

I took his courses :-).

@JonasTeuwen ah how?

And he is Belgian!

same uni?

Also, Lenny knows plenty of analysis!

@JonasTeuwen you are not an algebraist?

No. Closeby.

@BenjaLim No, but still I took algebra courses.

But he does algebraic stuff how did you take his courses?

It is important to at least know some of the other fields.

@JonasTeuwen what kind of algebra courses?

Galois theory, algebraic topology.

@JonasTeuwen you took galois theory?

Homological algebra, but was not his.

oh wow

Of course...

and algebraic topology?

even homological algebra?

It is not wow, I am much older than you. You can do the same. My point is: do some analysis too!

I'm stunned

Even pierre never did algebraic topology

If there would have been a good course, I am sure he would have taken it.

It is certainly not a matter of intelligent with your or Pierre.

huhuhuh?

It is a matter of not being a narrow idiot.

@JonasTeuwen I think he told me his uni never had algebraic topology courses

So, go take some functional analysis too, you might like it. It can be quite algebraic.

I think he went deep into analysis quickly :D

Yes, that is why.

@JonasTeuwen I have some friends doing analysis III now

Pierre knows quite a lot about other mathematics as well, he is humble.

Ok I have to go

You take it too. Bye.

@JonasTeuwen apart from analysis?

Yes.

and they tell me analysis 1,2,3 are all the same

The basic analysis courses are not so hard, but the knowledge is very useful.

Even when not doing analysis.

like?

Will tell you later. Will take more time!

ok

bye jonas

@BenjaLim I deleted two posts yesterday. To which were you referring?

I wish I was in university.

@WillHunting At least they let you choose your course :-)

@WillHunting: I have no problem doing Computer Science with Math.

No problem doing Math history as well (at least it's not like THAT history).

hello again. Is there anybody from France?

Bonjour, and nope.

:(

@ParthKohli If you stay in India, your freedom might be quiet limited!!

@JayeshBadwaik: Don't you too?

Once my HOD said to my class, if we provide flexibility in courses, there will be a certain drop in excellence.

@ParthKohli I do and I am telling you from experience.

Oh, no. I don't agree. People India have too much of freedom, IMO.

For example, no fines on stuff.

@ParthKohli What stuff?

I wouldn't like to name "stuff" here.

Maybe somebody knows, what may mean "fonction d'exhaustion"?

@ParthKohli There are no fines for "stuff" generally anywhere. In fact, in other places, it is not even frowned upon on.

@WillHunting yeah

@WillHunting No, he was specifically referring to the choice of courses you have and the pattern of evaluation. 10% assignment/class test, 2x15% mid sems, 60% end sem.

I don't know If there is a notion of exhaustive function in English

I don't know too. So I think that translation is incorrect

Ah, good. I've found

planetmath.org/encyclopedia/BoundedExhaustionFunction.html

@WillHunting I'm working with it, but I'm interested mostly in inverse problems

when youre doing trig substitution to solve an integral such as youtube.com/watch?v=3lC5AuCFK4c (around 12:00) is where I will be talking about. Can you simply just solve for theta and plug it in for x? The answer may not be completely simplified in this manner but wouldn't it still be an answer?

Hello guys

$2x^{\frac{1}{2}}-x^{\frac{1}{4}}-1=0$

@DantheMan What?

How do I solve that?

I guess you can use the quadratic formula.

What about factoring?

Let's test something - I'm gonna try it with mathematica.

ok

Opening it

k

@WillHunting Really?

@WillHunting Isn't it a quadratic equation in $x^{\frac{1}{4}}$

($x^{\frac{1}{2}})^{2} = x$

@WillHunting Or probably just lack of concentration after a tiring day?. :-)

@WillHunting Yeah. I have actually done this mistake on a particularly tiring night when I was 10.

Thanks @WillHunting

@WillHunting Your name should be changed to FlyBy - II. After a song similarly named by the pop group "Blue". The original FlyBy guy is Matt.

Avatars come and just fly by.

Wait

now how do I factor it?

@DantheMan Well, that is the teaching part of it. No one learns factorization without hit and trial.

Try various numbers until you get it right.

alright

But what do I do since there is a 2 constant on the first term?

Why does Wolfram|Alpha calculate it as $(y-1)(2y+1)$

@DantheMan Hint: $ax^{2} + bx + c = (px + r)(qx+s)$

@DantheMan This is a factored form

My above post was for the previous question. As for Wolfram Alpha, is that not correct?

What are $p,r,q,s$

What does mean "Martineau compact set"?

@DantheMan assuming the factors will have integer coefficients, which is what typically happens in these sorts of problems, you can immediately see the coefficients of $y$ in the two linear factors will be $1$ and $2$ (for they cannot be anything else). thus you look for $a,b$ such that $(2y+a)(y+b)=2y^2-y-1$. furthermore you see that $ab=-1$, so either $a=+1,b=-1$ or $a=-1,b=+1$. at this point there is only two candidate solutions to check, namely, (2y+1)(y-1) versus (2y-1)(y+1).

@JayeshBadwaik Right.... brain fart

@anon Aha...

@DantheMan Does that word mean: I'm more lost than I was before?

@GustavoBandeira Na... I got it factored but I don't really understand anon's answer

Then you should say so.

Every quadratic polynomial in the variable $x$ factors into a linear factor times another linear factor. A linear factor is of the form $rx+s$, where $r$ and $s$ are constant coefficients. So $2y^2-y-1=(rx+s)(ux+v)$, where $r,s,u,v$ are constants. Not only are they constants, they are integers, because the problem-makers always make these things integers. (This is not a mathematical deduction, it is an observation.) So in particular, expanding out the right side we would know the coefficient

of $x^2$ is $ru$, so $ru=2$. The only integer solutions to this are $r=\pm2,s=\pm1$ or vice-versa. Without loss of generality (I will explain that upon request), we may assume that $r=2$ and $s=1$, so now we have $2y^2-y-1=(2x+s)(x+v)$. Again, expanding out the right side we would see that $sv=-1$, and the only integer solutions to this are $s=\pm1,v=\mp1$, and so the only two possible factorizations are (2y+1)(y-1) or (2y-1)(y+1).

The idea behind the without-loss-of-generality assumption is somewhat expanded on here math.stackexchange.com/questions/123542/factor-14x2-17x-5

@anon hey

Taylor Series approximation is a pretty cool thing

Could a orange, some semi-old bread, and fresh noodles make a good combination (tastewise)?

@ParthKohli Try instead with fourier bro.

Maybe somebody knows, what is the "domain $\Omega$ dual to the compact set $K$"? I found a definition $\Omega = \left\{ w \in \mathbb{C}^n \colon w \cdot z + 1 \neq 0 \;\;\; \forall z \in K \right\}$

@WillHunting Nah.

Wow, I found a cool identicon generator.

finalboss.org/g

Good Afternoon!Good night!

@GustavoBandeira Are you there?

@Meand sim.

@GustavoBandeira Yay!How are you?

Good!

I see that you are sneaking my FB?what you think?

Sneaking?

yeah,taking a look

Yep.

so?

I wanted to see if you post interesting things.

and?

Yep. Some nice things. =)

:D

I see you've sneaked my sneakings.

What you think?

hilarious

Why?

it's funny

comment post that no one else does

Sim.

O povo só gosta de piada de peido. xD

é

e memes escrotos

tô cansada disso

Um amigo passou uns dias aqui.

Ele passou uns 20 dias.

Todo santo dia! Ele SÓ via isso!

uma chateação

de vez em quando,até vai...mais semppre perde a graça

Gustavo?

Fala.

@WillHunting Hi,Jas!

hmm

Tô ouvindo o raanerg.

normal

kraanerg

Do Xenakis

@GustavoBandeira cheu perguntar um negócio

@WillHunting How are you?

Pergunta.