DogAteMy: Quite seriously. I was trying to have a serious conversation with a Mexican guy who spoke only Spanish (and I know virtually none). I was using an app (not Google translate) to do consecutive translation from Spanish to English to Spanish. It kept saying "If ..." in contexts that made no sense, and I finally realized it couldn't distinguished between "si" and "si." It had the same problem when I used "too" in English, and it translated it as "to." AGH.

Si j'aurais la possibilité de parler d'espagne je ferrais.

@TedShifrin Haha. (Also, sí and si.)

or sth. like that.

Yikes, @Maks. If $\sum a_n$ converges, then $a_n\to 0$, so for large enough $n$ we'll have $|a_n|<1$. So if $\sum \frac 1{b_n}$ converges, so will $\sum \frac{|a_n|}{b_n}$, which means that $\sum \frac{a_n}{b_n}$ will as well.

@Physics, that French ain't so good.

@TedShifrin You know better ?

Yup.

And we have a bunch of Frenchmen who inhabit this chatroom, as well.

Je comprends.

Félicitations!

Peut-etre ils me peuvent dire ce qu'était mal.

Was that correct ?

Nope.

No

Salut, @Astyx.

Hello

Quoi de neuf ?

@Astyx Are you from France ?

Yup

What was wrong ?

What were you trying to say ?

Pas grand'chose. Je viens d'écrire une longue réponse à une question d'algèbre linéaire (ou géometrie complexe).

I recognize some of these words

"Maybe they could tell me what was wrong/bad"

LOL, @Astyx. Ne sois pas méchant :P

I meant Ted's comment

So, DogAteMy, last I saw, you almost had a solution to that convex curve question?

@PhysicsGuy No but I mean were you trying to say "If I had the chance to speak Spanish, I would" ?

Hey, you're goofing off in school again today?

Ah, yeah

No

School lets out early on Fridays

Ah ... the bonuses of religion :D

@Astyx Yes

@TedShifrin Yeah, the post-doc running the seminar was happy.

@AkivaWeinberger Not at all.

@Danu: Did your fellow students ask questions and pay rapt attention?

More like Steel-tjuhs

Then it's "Si j'avais la possibilité de parler Espagnol, je le ferais"

@TedShifrin Some. But some others are lazy---they already weren't paying attention when I was doing the easy stuff at the start.

I don't know the whole Phrases conditionelles thing.

Accent on the first sylláble, though, right? @Danu

@AkivaWeinberger Yeah, on Stiel

I hate this language anyway.....

@TedShifrin Consider the two points with horizontal tangents

Sigh. Oh well, you did your best, @Danu. (One of the reasons I've always preferred teaching undergraduates to graduate students. ... Although I've had some great students in grad classes, always some disinterested ones.)

That's just mean ... :(

I didn't say that I hate you

Fair enough

$T=d\gamma/ds$ where $\gamma$ parametrizes the curve, so $\gamma=\int Tds$

The thing with Dutch is that the pronunciation of vowels is very "pure", so the "ie" sound for instance is very... "ie" :D

I just had very much difficulties "learning" that language.

Oh, that makes it all clear, @Danu.

@TedShifrin And undergrads are better? Lol, that's opposite to my expectations.

@TedShifrin I know, right?

@PhysicsGuy: It seems you have had some difficulties learning English, too. What is your native language?

And so the difference in $y$-coordinate between the two points of horizontal tangents is $\int_{\theta=0}^{\theta=\pi}\sin\theta ds$ @TedShifrin

@Ted Shifrin German.

Das stimmt, vielen Dank, @PhysicsGuy

Was hat das damit zu tun ?

So now I'm back to focusing on my own shit, I guess @Ted. Though this week a PhD student took over the symplectic geometry lectures and did a bad job, so typing that is shit this week.

You're doing very cleverly, DogAteMy. :)

since $\sin\theta$ is the horizontal component of $T$

no, DogAteMy, vertical

I still have to complete it---he was trying to prove this theorem by Poincare-Birkhoff on fixed points of symplectomorphisms of annuli that twist the boundary components in opposite directions

And then that's $\int_0^\pi\sin\theta(ds/d\theta)d\theta$

@Danu: Oh yeah, cool stuff.

But he messed a lot of stuff up so I couldn't keep up live-TeXing :(

@TedShifrin Yes, typo, sorry

He did the proof by Brown-Neumann (referenced on wiki)

In any case, that's $\int_0^\pi\sin\theta/\kappa~d\theta$

@TedShifrin Can't say I found it very inspiring.

We also don't know any proper motivation. He very very briefly mentioned some stuff about Poincare return map invariant tori something something

Bound with $\kappa_{max}$, pull it out, evaluate $\int\sin$, conclude

What is the name of the theorem that states that if $f:X\to Y$ is surjective, then the cardinality of $Y$ is smaller than the one of $X$ ?

@Danu: This is one reason it's often disappointing to have students lecture instead of the faculty member, who actually can provide insight. You learn a lot doing your own lecture, but often very little from the others'.

Super, DogAteMy. Pretty cool, eh?

Yup

@TedShifrin Yeah, lol. This was really bad. But the faculty member gave him material to lecture on---he just decided to do something esle instead...

@TedShifrin I haven't ever had much difficulties learning english, I just don't speak like a professional. And I make mistakes.

I only recently "made up" that question, DogAteMy. :)

@TedShifrin Much like the weather outside, to which I now must go.

Bye

Bye, DogAteMy

That's a very poetic way of saying you're going

Hi chat

Hi @Semi

@PhysicsGuy: That's OK. I just don't think there's any need to say deprecating things about other people's languages.

HI @Semiclassic

heyo Semi

Only deprecating thing I can say about languages is that in choir I enjoyed singing in Latin a heck of a lot more than French

LOL ... Well, I have enjoyed every foreign language I've studied. I'm more proficient at some than others, but ...

h

Big difference between singing and speaking, of course

How's the freezing north, @MikeM?

Man, why are you hating me so much for not liking french? It's just terrible to learn, that's it. I never made any statement about people talking french.

Different Mike, but it's Minnesota cold here

It's not that cold. It is dark. It looks like midnight and it's 6pm.

Another two hours on the road.

Good thing you don't know how to drive, @MikeM.

Currently 15F here. Weeee

@PhysicsGuy: I am not hating you. :) I just hate physics. Let's move on. (And, no, I don't really hate physics.)

How dare you, ser.

We spent 20 minutes trying to figure out how to open the gas valve.

gas valve on what?

Outrageous, simply outrageous.

@TedShifrin Today, I spent a few hours looking at (abstracts of) papers by Hitchin (and Atiyah and Donaldson...). Hitchin did a lot of awesome stuff on the border between math and physics.

Yup, smart dude :)

The car.

@Ted Shifrin Two topologists talk with each other: "Hey, could you please bring me a cup of coffee ?" The other goes away. After a short time he comes back with a donut, the other says: "Thanks".

72km to Vik.

Those jokes are awful.

That must be one of the worst joke I've heard in a while

@PhysicsGuy 2/10---the cup of coffee with donut thing is way overused

Lol, the pile-on :P

And also one of the most hated joke apparently

@MikeMiller "Vik" lol

Local slang confirmed

@Danu No. That's literally the name.

@Danu What do you get if you cross an elephant and a chicken ? The trivial elephant bundle over the chicken.

It's not Reykjavik.

@MikeMiller :D :D Damn. I assumed you meant Reykjavik

@PhysicsGuy not nice either.

The one about crossing a citrus fruit* with a bull is much better.

Here's a decent joke

what's purple and commutes? An abelian grape

@Danu Old Macdonald had a form; ei /\ ei = 0 ?

Anyways, I think we all get these jokes from the same MO thread.

$\displaystyle \oint_{\text{Western Europe}} \operatorname{Europe}\left({z}\right) \, \mathrm dz$?

We're in the southeast right now. Vik is a small city on the way back to where we're staying. Vik, of course, just means bay.

The same MO thread which isn't funny

0, because the poles are in eastern europe

I like the citrus fruit with a bull one. It is at least sufficiently absurd.

I will leave you all to your humorless humor :P

PleSw don't leave me

PleSw

@MikeMiller Say I want to understand the cobordism rings a little bit. Is it worthwhile to go through Thom's original paper (I have an English translation as well as the French)

Witten is such a terrible lecturer. It's funny :P

@Danu Witten is a very good mathematical physicist.

No. Read Milnor.

@PhysicsGuy Duh :P

@PhysicsGuy We are aware

@MikeMiller Which book/text/paper?

Oh, he has "notes on cobordism"

well... h-cobordism theorem

close enough?

Characteristic classes

I' m going to leave.

Bye.

Bye @PhysicsGuy

@MikeMiller That's like... Just 10 pages or so though.

So?

I would think that there would be more important stuff to know about cobordism than what Milnor/Stasheff in about 10 pages.

Nah.

Ok... Fine.

hey @Danu you know about de rham cohomology right ?

I think there is a error in my notes

Okay...

If $f : X \rightarrow Y$ differential map of (real) manifolds, show that there is a ring homomorphism $f^{} : H_{DR}^{\times}(X,R) \rightarrow H_{DR}^{\times}(X,R)$ induced by $f^{}[w] = [f^{*}(w)]$

the last line should be $f^{*}[w] = [f(w)]$ right ?

I can't read your TeX.

ok let me just take a pict

Just tex it properly

What is $\{\}$ and what is $\omega$?

w is a class

$\{w\}$ is a class of omega

you mean a form

So how do you propose to define $f(\omega)$?

I mean we don't have definition for $f^{*}$ how do we define it we are trying to induce a map on it.

@Adeek I cannot parse that sentence, unfortunately. Can you rephrase?

morning math guys! and girl(s)!

probably $f^{*}[w] = [f(w)]$ but this isn't really a ring homomorphism though.

Again, you propose to use "$f(\omega)$". But what do you propose $f(\omega)$ means?

oh I see yes that doesn't make sense.

But $f^{*}(w) = [f^{*}(w)]$ doesn't make sense to me as well we because we are trying to define $f^{*}$.

You're supposed to know what $f^*\omega$ is for a form $\omega$, probably.

oh

So who's got a cool weird commutative ring that I haven't heard of today??

$\Bbb Z$

I guess if you want to get pedantic... sure.... nobody has explicitly brought up integers to me today

oh @Danu he is using different definition on both ends. On one end it is the regular function $f^{*}$ and on other end it is the algebra homomorphism $f^{*}$ on the forms.

depends on your definition of cool and weird, what about the powerset of a set with symmetric difference as sum and intersection as multiplication?

yus

mmm tell me more

I wanna know how it's represented, let me have it

I'm not sure what you mean

lol sorry

I get excited about maths

Do you know everything about $\Bbb Z[X]$ ?

nope, not too much at all

It's more interresting than it seems

My only real excursion from the integers / reals / complex has been the galois fields

actually even the complex I'm pretty weak on

Work on that then

Complex numbers are primordial in most math

I'm really tripping on extension fields lately

I still don't quite see why reducing modulo an irreducible polynomial works

I mean complex domain is nice but in practice I only need a weak knowledge of it because in engineering applications I've always been able to get by with simple things or characterize the system in the reals if it's not so simple

What kind of engineering applications @MickLH

mostly electronics and drafting

im not really qualified for much else lol

I still advise you to study it if you intend to do anything in analysis

Make what you want of it

Is Westworld any good ?

I'm not actually sure where to continue with complex analysis, I've got past all the simple stuff with euler's identity and polar coordinates

@MickLH Usually I would not actually out those under complex analysis but calculus. I would not call it complex analysis until it is the more abstract study of holomorphic functions

Is that identical to analytic functions?

@MickLH Complex analytic, yes (which is not the same as real analytic)

feel free to give homework :P

I simply don't get enough examples to work through in complex analysis

Most textbooks should contain plenty of those

I had to leave school very early and don't really even know what would be a good textbook

What language do you speak ?

actually everything I know is because I was forced to self study to solve a task

english

math.ku.dk/noter/filer/koman-12.pdf these are the ones I learnt from

thank you, downloading immediately lol

How I wish I had time to read that ..

math.ku.dk/noter/filer/matematik.htm has a decent collection of free lecture notes, though some of them are only in Danish

my danish is super weak lol

I've only stepped foot in denmark twice

Actually, most seem to be only in Danish (a few in Norwegian as well)

this is some good stuff, but trying to boot up my brain just smacked me with an Insufficient Energy error lol, be back soon, and much appreciated

Hullo there

If you didn't have enough languages, I'll throw in Spanish

Let $$f: x \mapsto \sum_{n=1}^{+\infty} {\sqrt{x}\ln n\over {1+x n^2}}$$. How can I find an equivalent of $f$ at 0 ?

what is n?

@Sophie My bad

$f(0) = 0$, no?

Yeah

Not sure what's meant by an equivalent of $f$ at 0.

So what's the problem

^ (scratching head)

(Also, you could start the sum at n=2 since ln(1)=0. but that's a detail)

A (simple) function $g$ such that $$\lim_{x\to 0}{f(x)\over g(x)} = 1$$

Ah, so you want some function which is asymptotic to $f(x)$ as $x\to 0$.

Exactly

it has a simple-ish power series?

So I'm trying to illustrate examples of contours and I'm not sure whether to put the parameterization of the contour on the graph itself or as a caption. Thoughts please?

@MickLH What do you mean ?

Two convergent and one divergent

In this example

it's an aesthetic question not a math question

The issue I see is that if you just do a leading-order approximation $\sqrt{x}\ln n$, then you get the divergent series $\sum_n \ln n$.

So that bodes ill.

Did you check the Pade approximants?

@MickLH One issue: The radius of convergence of each geometric series will be $|-1/n^2|$.

I don't know those

the Pade approximant might not be divergent though, if you pick whichever order for numerator and denominator match the asymptotic behavior properly

So if $x$ is small but nonzero, then eventually $n$ will be so big that $x$ falls outside of the radius of convergence

Hence it's only going to work for $x$ being exactly zero.

^

what's the domain of $f$ anyways?

You could possibly focus on the high-order terms, though, since there one instead can expand in powers of $1/(x n^2)$.

$\Bbb R_+$

But I think something more clever is needed.

@Semi What do you mean ?

$$\frac{1}{n^2x+1}=\frac{1}{n^2 x}\frac{1}{1+1/(n^2x)}=\frac{1}{n^2 x}-\frac{1}{n^4 x^2}+\frac{1}{n^6 x^3}-\cdots$$

For sufficiently large $n$.

Right