ewwww

fake

@TedShifrin I don't think that can be true, because $\left(\frac ne\right)^n\cdot\frac1{n!}\sim\frac1{\sqrt{2\pi n}}$

@Pedro, nice, she's hot

@EnjoysMath really

@EnjoysMath That's not the point of the video. (?)

I click on the sexiest thumbnails in the sidebar too!

that's the point

out of all the videos you chose the one with the sexy face

:D

sex sex sex sex sex everywhere

please stop

My mom doesn't like that

o_o

wat

She's in the room

@EnjoysMath I should flag you.

But I won't.

Hi people

Hi @JasperLoy

@JasperLoy pls respond

You should ping @willhunting

Why

@JasperLoy how come you haven't pinned your location on the map

@user127001 I have changed my username, I won't get a signal unless you use @willhunting

ok @JasperLoy @willhunting

@user127001 Do you see my name as Will Hunting on the screen?

@JasperLoy no

@willhunting no

@user127001 I understand now. Please refresh your browser and tell me what you see

@JasperLoy @willhunting Nothing changes when I refresh

@user127001 Oh dear, there is something very wrong then. You should see Will Hunting as my user name, maybe refresh again?

@JasperLoy @willhunting it just says "Will" next to your blue box

@user127001 OK, that's right, that's what you see on a small screen!

Oh @Jasperloy @willhunting

@user127001 So you just ping me as will if that is what you see

People really love to downvote me

Why @JasperLoy @willhunting?

@user127001 Hehe, you don't get how to ping me. You can just use will without jasper to ping me!!!

Ok @will

When will you change back to JasperLoy @will

@user127001 I am not sure, just try to identify me in some way, and beware of imposters!

Ok @will

@willhunting: Even on my huge desktop, all I see is Will, unless you write a bunch of lines.

@TedShifrin Ah, that means my desktop is really huge then, or I am using a smaller font

I think the downvote was because I did not show the method, I just gave the answer, but it is so trivial

@robjohn: I know Stirling's formula quite well :) I don't think we have a contradiction. Certainly (at least for all but a few $n$), $e(n+1)>\sqrt{2\pi n}$, so $\dfrac1{\sqrt{2\pi n}}>\dfrac1{e(n+1)}$.

I downvoted someone who wrote garbage and gave an explanation of why it was garbage and what needed to be done. I still haven't had a response.

hi @user127001

Ugh, I guess I should go back to grading my horrid exams.

Hi @MickLH lol

@TedShifrin Well, user geometry hasn't been seen since three hours before your comment, so I'd say (s)he has a good excuse for showing no reaction so far.

hey mon

Good @WillHunting

I think downvotes should be abolished!

I should get ready and work today

@WillHunting Actually, I'd like the concept of downvotes to be extended to starred chat utterings -- a way to unstar some of them.

@EnjoysMath Who removed your message?

@WillHunting Hey there :)

Now that I have 3k on Math and 3k on Eng, I retire from answering questions, lol

I left steel cut oats in honey overnight, hoping to have made oatmeal that can satisfy my insatiable sweet tooth this morning for breakfast

no dice! the steel cut oats wont budge without heat

I did not get any email yesterday after leaving it in this chat, lol

Nowadays, I only get spam

that reminds me to check my email

Hi all! A quick question to everyone (or just to @DanielFischer ;) ). Could you give me some guidelines on how to compute multivariate limits? If such a thing is feasible. To be specific, I have a $n$-dimensional function, which envolves the $n\times n$ matrix $A$. I would like to find the limit of the function when the matrix ``tends to'' the zero matrix... Is such an approach correct? Any piece of info would be very helpful! Thanks a lot!

bleh, all spam except one work related email

yo yo yo how do I post to meta?

yep I'm one of those lazy internet users

@EnjoysMath meta.math.stackexchange.com/questions/ask

@nullgeppetto Is there anything stopping you from just filling the matrix with $x$ and taking $x$ to your limit?

Idling around... I would like to read a surprisingly elegant proof, something that should be in The Book (Erdős), but posted here on MSE. Suggestions?

All proofs are equally good to me

@nullgeppetto The matrix "tending to $0$" is no problem at all, that just means all entries tend to $0$. Finding the limit of the function may still be difficult, but without anything to go by, there's nothing specific I could say.

@WillHunting Can you proove that the "Proofiness" of any given set of correct proofs is equal?

@DanielFischer, hmm, yes, that would be correct I think! The only limitation is that my matrix needs to be symmetric positive definite, and thus I think I should present it as an appropriate function of $x$, but is there any way to do so?

Is this meta-proof still a proof or an axiom?

I can't sleep, so I should drink more coffee, lol

Hi @gabriel I admire the French universities, excellent math curriculum

@WillHunting Hi. If you say so ... What in particular strikes you in their curriculum ?

@GabrielR. Well, they cover many more things than what I did in my university, just too different

@nullgeppetto That the matrix shall be symmetric and positive definite is no problem, that set is a fine cone having $0$ as an adherent point, so you can have $\lim\limits_{\Sigma\to 0} f(\Sigma,\dotsc)$. Whether writing $\Sigma$ as a function of $x$ has any benefit depends on the situation.

Our "universités" are really not renowned though... Our engineering schools (Polytechnique and the Ecoles Normales Superieures) are the flagship of French secondary education. The curriculum (and mastery) required to get into those schools is quite interesting. Take a look at a part of the math exam (aimed at people between 18 and 21) sujets-de-concours.net/sujets/xens/2013/mp/maths2.pdf

French schools are not renowned?

This is news to me

Revisiting this question. Achille Hui has given all the hints, but somehow the OP won't compare coefficients? Or does she consider the question already answered, as suggested in the comments?

I kid you not : our high schools really suck, it has really been dumbing down. Only our greatest engineering schools that demand a competitive entrance exam are worth something

@GabrielR. Universitie Paris, Sud, Orsay is pretty renown.

And it

's not a school, but I.H.E.S.

Hi Alex

Bye Alex

@Mike yeahwhateverman

@GabrielR. Do you not think so?

@DanielFischer, many thanks once again! That really helped!

Thanks @DanielFischer

whoa, thanks all pointing at DF atm

he got a thx hurricane

@AlexYoucis Yeah Orsay is quite good for research... But nothing can replace getting into ENS-Ulm or ENS-Lyon

@GabrielR. It depends what you're interested in. If you wanted to do algebraic geometry, I don't even know if that's true. About generic undergraduate math programs though, I have no idea.

@Mike Did you go to that thing?

ah, @AlexYoucis I'm only considering undergraduate studies. But seriously, only the 50 brightest students of France can get into Ulm (math-physics majors). I guess that says a lot about the level of this ENS

@GabrielR. Yeah, but getting into the best school isn't everything, right? :)

get into the school w da hottest chix

@AlexYoucis yes, you're right

@GabrielR. Plus, you can always transfer, or go to a better school for graduate school (if you're planning on going) ;)

@EnjoysMath we don't have hot girls walking around with tight yoga pants here

@EnjoysMath Wolfram Alpha has some of the required features, and its code is certainly far more complex than could be spontaneously achieved via crowdsourcing.

@TedShifrin Do you mind if I ask you a geometry question?

:)

heya @Alex

et salut @Gabriel

Hi everyone

Depends if I can answer it, @Alex.

I'm pretty depressed grading my undergrad diff geo exams today :(

@Darksonn Hi

@TedShifrin How hard is it to show that X is Stein if and only if $H^1(X,\mathcal{F})$ vanishes for all coherent $\mathcal{F}$?

I don't remember, @Alex. Isn't it all higher $H^q$? I haven't thought about this since my days in Evans Hall ...

@TedShifrin Yeah, the above if and only if is equivalent if we replace 1 with q>0 :)

Oh well, I guess I'll just have to tap into Evans hall myself.

Sorry :P I'm not as smart as I once was.

We obviously need to produce enough holomorphic functions to embed in $\Bbb C^n$ ... but I don't remember how to use the cohomological condition ... obviously have to choose some clever sheaves.

It should be somewhat similar to the Kodaira Embedding proof in wanting to separate points and tangent vectors by tensoring with appropriate sheaves and getting sections.

@TedShifrin Yeah, it's something to that effect. It looks like it may be in a patper of Serre.

Yeah, that sounds right. It's also in Gunning and Rossi.

assuming x<0;y<0, then if sqrt(x) < y, is this then true? x < y^2

nope, @Darksonn ... Try an example.

derp x>0;y>0

oh, that was a derp.

Hi @TedShifrin

hi again @user127001

@Darksonn Suppose $f:\mathbb{R}_+\mapsto\mathbb{R}$ is strictly increasing, then $0<x<y$ implies $0<f(x)<f(y)$. Here, set $f(x)=x^2$.

--- Nonsense, forget the $0<$ before the $f$

But you get the idea

Hi, I'm facing an interesting problem: for which natural n is floor((3+sqrt(5))^n) odd? It looks like it might be a solution of some recurrence relation... tinyurl.com/ncfkx7f

@TedShifrin Hope you don't mind if I ask another :). Is there an obvious topological reason why $H_\text{sing}^i(X,\mathbb{C})$, should vanish if $X$ is stein of (complex) dimension $n$?

Hi @TedShifrin How long does it take you on average to grade one of these papers ? My teacher told our class he spends less than ten minutes each (for 4-hour exams). I think it's quite short

I write a fair number of comments, @Gabriel, as I do on weekly homework. So for 30 students, I tend to spend 6-8 hours grading, much as I tire of it. I'm almost done with the exams, and it's probably about 4-5 hours.

You mean for $i>n$ or something, @Alex?

@TedShifrin Yeah, for $i>n$ :)

Yeah, I am pretty sure recurrences is the way to tackle it. But how to go from a solution to the recurrence?

I don't remember this, either, of course. :P

@mirgee ... Did you see my various replies to your question earlier?

Were you asking about the sequence or the series, @mirgee?

@mirgee Yes, $\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^2$, and you know where the golden ratio comes from.

@TedShifrin Does this sound right? If you use the $d$-Poincare lemma, you can show that $0\to\underline{\mathbb{C}}\to\Omega^0_{\text{hol}}\to\cdots\to\Omega^n_{\text{h‌​ol}}\to0$ is a resolution. If $X$ is Stein, then the $\Omega^i_{\text{hol}}$ are acyclic, and so $H^k_{\text{sing}}(X,\mathbb{C}})=H^k(X,\underline{\mathbb{C}})= H_k(X,\Omega^i_\text{hol}(X))$. But, since $\Omega^i_{\text{hol}}=0$ for $i>n$, this proves the result. Yeah?

@TedShifrin I don't know, I was working on series today, but I think I asked about limits of sequences here

Well, @mirgee, my responses relate to the series, not to the sequence.

Your question was one of my favorite series questions. But it's somewhat subtle.

Where are you using $i>n$ for that, @Alex?

@TedShifrin Just the fact that the resolution by the $\Omega^i_{\text{hol}}$ terminates after $n$ :)

@TedShifrin: Ah, I see. Well, I figured I will ask my professor about that particular problem

So, acyclic tells us that $H^i(\mathcal S) \cong (\ker d\colon \Gamma(S_i)\to \Gamma(S_{i+1}))/(\text{im}\, d\colon\Gamma(S_{i-1})\to\Gamma(S_i))$.

Well, @mirgee, if it's series, I've solved it for you. :P

I see, @Alex, so of course you're right.

@TedShifrin Ok cool, thanks!

Thanks, @Alex, for reminding me of cool stuff I'd forgotten. :)

Sigh ... OK, back to finish grading.

@TedShifrin No problem. Thanks for listening to my ramblings :)

@TedShifrin: Yeah, you did :) I asked the question in relation to examining the endpoints of radius of convergence of $\sum_{n=1}^{\infty}{(-1)^n n^n \over e^n}x^n$. Thanks ;)

People, you are so smart! I admire you :) I feel so dumb all the time when studying math

@AlexYoucis Yes, I'm missing a talk right now

What's the title @Mike?

I dunno, I'm missing it because I'm dealing with something that came up, not because I don't want to attend

Romanians got talent - youtube.com/watch?v=E7voaVjhE-o

(that song inspires me while doing math)

@Mike D'ja Devadoss's talk?

If I have some fraction a/b, where a and b are some very large numbers, do anyone know of an efficient method for finding two numbers c, d which approximate a/b, but which are much smaller?

@Darksonn Take a convergent of the continued fraction expansion of $a/b$. The convergents are best rational approximations, if $c/d$ is a convergent (other than maybe the $0$-th), and $p/q$ is a fraction with $$\left\lvert \frac{p}{q}-\frac{a}{b}\right\rvert < \left\lvert \frac{c}{d}- \frac{a}{b}\right\rvert,$$ then $q > d$.

@Darksonn Exactly as Daniel Fischer said. I might add that you can think of $a/b$ as a single real number for that purpose.

ok, thanks I'll look into this

A little demo: 3.14. Store the 3 and reciprocate the 0.14, so you get 1/0.14 = 7.14... Store the 7 and reciprocate the 0.14... but no, let's stop here. We already got 3.14 = 3 + 1/(7+...), and neglecting the ... leads to 3.14 ~ 22/7.

ok thanks

@AlexYoucis Yeah. Great stuff.

That dude loves a good pentagon.

aww, not today I guess

How many digits of precision can I expect that method to get for x number of iterations?

@Darksonn Look at Theorem 5. Roughly, the accuracy is in the order of the square of 1/denominator.

Quick question: Any materials/books for studying counting problems?

What do people think of putting footnotes in a math thesis for comments too long to be parenthetical? My school has no policy, but I also cannot find a single person who gives asides in the footnotes.

Just did a whole bunch of edits, glad nobody complained

@Vladhagen At least in the papers I read, there are no parenthesized comments at all. The writer has to lay out a single thread of thought for the reader to follow. In the process of writing, it seems to me that side remarks are either moved to their own section where on-topic, or thrown away mercilessly. IMHO the reader's attention shall not be used up for "everything that can be said".

Hi @will

@WillHunting Hey, stop doing that!

is that better?