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9:07 PM
Hmm
 
I think it's coming back a little bit
Say $f : \Bbb C \to \Bbb C$ is holomorphic and $U$ is open in the domain; pick $z \in U$ and look at $w \in f(U)$. I want a small disk around $w$ that's contained in $f(U)$.
 
True that
 
So, uh, for any $v$ near $w$ let $F(u) = f(u) - v$. Notice that $F(u) = (f(u) - w) + (w- v)$; I think the idea is if the $|w - v|$ bit is small enough then $f(u) - w$ would have the same number of zeroes as $F(u)$, but then $f(u) - w$ would have at least one since so does $F$ (dude, $F(z) = 0$)
That'd say for all $v \in f(U)$ with $|w - v| < \epsilon$, $v$ is in the image of $f$.
I think that bit is accomplished by Rouch\'e
Working with the linear homotopy $\{(f(u) - w) + t(w - v)\}_{t \in [0, 1]}$
 
9:26 PM
@Fargle: Do not abuse proofs by contradiction!
 
@TedShifrin :(
 
But hi :)
 
Hi @Ted
 
Hi @Balarka
You can prove a stronger statement than open mapping. You can prove that if $f-w_0$ has a zero of order $k$ at $z_0$, then near $z_0$ it takes on every value $k$ times.
 
hello handsome ppl
@TedShifrin Hello :D
 
9:33 PM
Of course, I personally think this is easier just from the power series expansion, but I do love Rouché.
hi @Kasmir
 
@TedShifrin I got something to ask ._. feel free to say no
 
@TedShifrin Right, I think the proof I remember the best is to understand that upto change of coordinates $f$ locally looks like $z \mapsto z^k$. That's a branched covering, hence open map. Your stronger statement is a corollary of this, of course.
 
Yes, but Rouché will prove the statement quite easily.
@Kasmir: Since when do you sit on ceremony?
 
doesnt roché has that theorem about finding roots?
 
More or less, yes.
 
9:36 PM
@TedShifrin sit on ceremony ? ><
dont know what that mean
 
The English phrase is "stand on ceremony." :) I modified it for you.
 
Rouché is black magic of a very high level
 
still dont get it :D
 
Google it
 
and why do you need to modify it for me
 
9:37 PM
@Alessandro It's topologically very apparent
 
Nah, Alessandro, there's a nice generalization of it and the argument principle in the differential topology setting.
 
cuz am black ?
 
we proved this fact using Rouché in my complex analysis class, it was pretty neat
 
LOL, @Kasmir. I had absolutely no idea.
 
@BalarkaSen what do you mean?
 
9:37 PM
I modified it for humor.
This obviously worked well. If I can't teach you math, I try to teach you humor. :P
 
haha
am not black but just like saying that
Its funny because many black ppl say that if they feel rejected by a girl =p
 
You're in Sweden, you told me. So what is your ethnic origin?
 
or by anything , am not rasist ofc ,just find that perticual thing funny in movies
french and tunisian
that is why I can speak french =p
and kasmir khan is not my real name
 
Ah, very cool. And you landed in Sweden. Très intéressant.
Well, I knew that much.
 
oui oui =p
 
9:40 PM
Anyhow, so what is your question of the moment?
And you should google "stand on ceremony." It's a nice phrase in English.
 
Verb: stand on ceremony (third-person singular simple present stands on ceremony, present participle standing on ceremony, simple past and past participle stood on ceremony)
  1. (idiomatic) To act in a formal, ceremonious or overly polite manner.
  2. Please make yourself at home - there's no need to stand on ceremony.
 
well its not a question really , I got 5 exercices that iv done
i want you to look at them and say if they are ok or not
 
I mean it's not that magical after you see the argument principle
 
ill figure out the "not" alone
oh i got it now :D
Ted did not expect me to be so polite?
grrrrrr
 
No, I'm not going to grade your homework for you. That truly is beyond what I'm willing to do on MSE.
 
9:41 PM
@TedShifrin So may I email it to you or you busy ?
 
If you have a focused question, of course, I'll try to help.
 
Okay =p
 
@Alessandro The argument principle says the winding number of a curve is the number of zeroes of the function. Rouch\'e just says that winding number is a homotopy-invariant. (The homotopy of the corresponding functions needs to be carefully done correspondingly)
 
this homework is graded
 
It was bad enough grading topology assignments when I voluntarily taught a topology course to 2 of my former students after I retired.
 
9:42 PM
I get from it 0.5 points
I understand that :D
 
Ted once gave me an exercise which proves a version of the argument principle in the smooth category.
 
I wanted to try my chances >< am happy either way
 
@Balarka: I think that is not a correct representation of Rouché.
Yes, and you never did it!
 
Which university are you reading at, @Kasmir?
 
@Ted False. I did and sent it to you :P
 
9:43 PM
It's needed for Hopf Degree Theorem, among others.
 
*studying
 
Oh, well, in that case I'm old and forgetful.
 
I quite liked the exercise.
 
@AndreasAlmgren Stockholm university
 
I see @Andreas is also a denizen of Sweden.
 
9:44 PM
@AndreasAlmgren är du svensk ? =p
 
haha najs vad pluggar du?
Ted snipped me again with that comment
 
snipe ≠ snip :P
 
Inget ännu, läser sista året på naturvetenskap. Funderar starkt på ren matematik. Hur är undervisningen på Stockholms universitet, om du känner för att berätta? @Ted, yes
 
Lots of Swedish people invading the room. Maybe not the best time to post Swedish curses.
 
9:46 PM
Probably not, Balarka. Stick to Bengali.
Someone is having fun downvoting me ... sort of at random. In one case, it was merited because the OP completely changed the post after reading my answer ... and then my answer was no longer appropriate.
 
Why care?
 
Reasonable question.
 
But I can understand your feelings on one side.
 
I have been the victim of a vendetta in the past, but I doubt that's it this time.
 
9:48 PM
@AndreasAlmgren helt generellt är det bra, fast man får inte mycket under föresningarna , man ska plugga mycket själv =p
 
Ok, tack!
 
I recently encountered someone whom I know well from answers in geometry. I have a feeling he and I disagree on style of appropriate answers (like he writes very complete ones and I often give hints). This system isn't perfect.
 
@AndreasAlmgren Lund och SU är bäst inom matte ( tror jag )
Np :)
 
So, @Andreas, what level of math person are you? :)
 
Ultimate noob
 
9:49 PM
Ah ... is there such a thing? :P
 
atm at least
 
Kasmir used to be one of those. :)
 
haha =p
 
I haven't finished high school yet
 
Ah, cool. I'm currently teaching some high school kids.
@Balarka: This is quite neat.
 
9:51 PM
You like it?
 
Andreas I recomand that you take "förberedande kurs i matematik"
 
Well, I spent 40 years or so teaching university, so it's different. But so far I think it'll be fun. It's a special school mostly for math geeks.
 
it is a good course before learning higher math
because they wont do anything basic from the start
 
Kasmir doesn't know what language(s) to speak in now.
 
9:52 PM
15 lectures then you take an exam
well
I figured its not nice for you
not understanding what we saying
 
Well, I speak French to the French folks and it's not nice to others. :P
 
it took me a while to get that , but I got there
Well Ted because you are very nice guy also , that too
 
Well, that sounds like something @TedShifrin. @KasmirKhaan I will look it up
 
So, if you're here, @Andreas, presumably you're significantly interested in mathematics?
 
@TedShifrin Ah, the deRham cohomology hint helps.
That is a nice fact
 
9:54 PM
We have a number of high school students in this chat ... Balarka, Akiva, MeowMix, Heather (is she high school yet? I think so) ...
 
yeah she's in high school now
i think
 
Yes I certainly am! It's one of the few things in school that is not boring.
 
My high school kids pretty much all claim they're bored in school. Makes sense, given that they're doing more advanced stuff outside of school. At least one of them may be bored even in this more advanced course; but I'm prepared to throw him harder stuff.
@Andreas: Here's a (not unusual) question I like to give. Maybe you know it; maybe you don't. If I choose $n+1$ integers from $1$ to $2n$, must it be that one of them divides another?
 
@TedShifrin you gonna be here for a while ? =p am planning on doing some exercices and it would be good if i could ask you about what i dont get =p
 
I'll be sorta around for a few more hours, then disappearing.
 
10:01 PM
okay :D dirichet is not something that they teach in sweden =p
at least not in highschool
 
Dirichlet what?
 
In my case I find school boring in large part because I don't like most of the things I am forced to read. I think that I should intuitively know the answer to your question to be able to answer it - the way you are asking - but I do not. I can tell you one thing and that is that I am certainly not one of the geeks you are teaching. My math interest is quite recent.
 
@TedShifrin pigeonhole principle i think its called too
 
LOL ... I like the question because there are different ways to do it, and because, like a lot of basic number questions, you can experiment and try lots of examples to develop intuition.
Oh, @Kasmir, that's only one of the ways to do it.
 
thats the first that come to mind for me =p
 
10:03 PM
It wasn't for me. :)
 
if i see that kind of Q's
haha =p
 
hey guys, can somebody help me out please
 
@Andreas: I don't need to give you questions. I'll shut up :)
That's a bit vague, @Sylent.
 
We all have to begin somewhere.
 
i need to get from $2(cos(x)^2 - sin(y)^2) - (cos(y)^2 - sin(x)^2)$ to ... $0.5*(cos(2x) + cos(2y))$
 
10:06 PM
I never liked the number theory type stuff that so many young people love. I developed a respect for elementary number theory for teaching purposes at the undergraduate level.
 
If somebody can point me in a direction that'd be greatly appreciated
im a bit stuck
 
Have you started with the double angle formula, Sylent?
 
Just started
 
What does it say?
 
erm, ive deviated from the question quite a lot
purely because im exploring something
 
10:07 PM
BTW, I don't believe it's correct.
 
wolframalpha says its true
 
Let's try $x=0$ and $y=0$.
What's $2\cos^2x + \sin^2 x$ when $x=0$?
 
the original question, prove the identity $cos(x-y)cos(x+y) = cos(y)^2 - sin(x)^2$
 
Aha ....
 
but i substituted using the identity $sin^2 + cos^2 = 1$
 
10:09 PM
You messed up somewhere.
 
and got to basically an identity that goes 2=2=2
so I thought, lets use the second two 2's to get the first 2
if that makes sense
its a mini adventure tbh
albeit not actually answering the question
 
So this question is correct. :)
You need to use the formula for $\cos(x-y)$ and for $\cos(x+y)$. Did you do that?
 
i can answer the question if i really wanted to
i just wanna complete my mini adventure
 
Well, did you try $x=y=0$ as I asked you to in your "mini-adventure" formula?
 
hold on
 
10:12 PM
Hey @Ted!
 
sub x = y = 0, got 1
cant be righjt
right*
 
Hi Demonark.
 
what are the guidelines to posting links?
 
@TedShifrin I wonder if there's a messy hands on way to do that problem by triangulating $M$ and homotoping $f : N \to M$ to inclusion of a subcomplex of dimension less than $\ell$. If $df$ has rank less than $\ell$, locally it's like the inclusion $\Bbb R^{< \ell} \subset \Bbb R^n$ after all.
 
Hmm, how do you get from local to global, Balarka, but, yeah, probably.
 
10:15 PM
I mean you can produce triangulations on $M$ and $N$ and homotopy $f$ to be a simplicial map by simplicial approximation theorem.
 
[start/end point](https://www.wolframalpha.com/input/?i=cos(x-y)cos(x%2By))

[stuck-point](https://www.wolframalpha.com/input/?i=simplify+2(cosx-siny)(cosx%2Bsiny)-(cosy-sinx)(cosy%2Bsinx))
@TedShifrin ^
 
It then remains to show that actually maps to a subcomplex of dimension less than $\ell$, which is not entirely obvious but I am definite that has to be true.
 
@Sylent: So your original thing simplifies easily to $\cos^2 x - \sin^2 y$, and I highly doubt that is $\frac12(\cos 2x + \cos 2y)$, but let me think.
 
Check the links, both equations simplify to that, I just dont see how
 
It's still amazing how it takes a one-line argument using forms. increase in RESPECC
 
10:17 PM
Well, let's see. $\cos 2x = 2\cos^2 x - 1$ and $\cos 2y = 1-2\sin^2y$. Does that give it?
Yup, it does.
You were right. I was wrong.
 
OOOHHHH
I completely forgot that identity
D'oh
 
I asked you about double angle formulas :P
@Balarka: It's about damn time!
 
I was completely over-looking those ones XD, thanks man
 
Sure.
 
@Balarka finally you've joined the dark side
 
10:22 PM
Nifty.
 
tfw you have to watch a 30 second ad before watching a 20 second meme
 
@Daminark I have sided the dark joint, rather
 
@SimplyBeautifulArt Have you thought of what to teach tomorrow?
 
@Balarka: It's actually not so astonishing that a differential topological condition plays nicely with forms.
Hi @Jasper.
 
10:24 PM
@Jasper Idk, I kinda have to see where people are at
 
@TedShifrin Hello, I should take some cooking lessons lol.
 
Ted you should write "Differential forms: A homotopic approach"
 
I'm done writing, but thanks.
 
do Carmo has a book Differential Forms, not well known.
 
:P
 
10:25 PM
He's not a natural differential forms user, so it's weird.
 
Yeah I've heard, I intend to look at that at some point
 
Henri Cartan has a very nice book on forms.
 
Differential forms using model categories
 
That's more like Griffiths/Friedlander/Morgan (rational homotopy theory).
 
Ah yes I have heard of that
 
10:26 PM
@TedShifrin That book is actually volume 2 of a course. I can't find volume 1 anywhere.
 
I still have a handwritten version of that book.
 
Maybe he wrote the forms book to learn about them?
 
Hm, I wonder what nlab has to say about differential forms
 
I never saw another volume, Jasper. I owned Cartan's book when I still had books. What was in volume 1?
Demonark, he had the same adviser I did, so I'm sure he knew about them. Just never naturally used them in his books or research, that I'm aware of.
 
@TedShifrin It's like differential calculus. I think what was translated into English got out of print.
 
10:27 PM
"One way to exhibit this statement nicely is:

A differential n-form on X is a smooth n-functor $P_n(X) \to \mathbf{B}^n \mathbb{R}$ from the path n-groupoid of X to the n n-fold delooping of the additive Lie group of real numbers."
@TedShifrin What is your opinion on this
 
puts Balarka on permanent ignore
 
When we ignore someone, we still can see the person's icon in small.
And when we navigate the rooms, we still see the icon in big.
There should be a way to make it totally unseen.
 
@Ted I didn't write that stuff!
 
You asked my opinion?
 
Also Urs Schreiber has discussed extensively below that statement in a small wikibox
 
10:30 PM
@BalarkaSen Is he a famous mathematician?
 
He's the founder of nlab
 
I have never visited nlab and don't intend to.
 
@TedShifrin a is equv to be mod H , where H is a subgroup of G
@TedShifrin can you clearify what this is saying ?
h in H , a= bh
 
Ted: Do you recommend Cartan over DoCarmo for forms?
 
@Daminark By the way, I was just listing the books. I didn't recommend anything for forms, lol.
 
10:35 PM
I don't remember details, Demonark. In general, I recommend a book that has true content and not just formal stuff ... so differential geometry with forms would be good.
@Kasmir, $a\equiv b\pmod H$?
 
@TedShifrin yes
 
I thought that Demonark was DogAteMy until I realised that DogAteMy is Akiva, lol.
 
@TedShifrin am doing cosets and quotient groups =p now I understand the notation for modulos, Z/nZ make sense now :D
 
Z/nZ = n
 
Typically, @Kasmir, that should mean $a^{-1}b\in H$, so, yeah, $b=ah$ for some $h\in H$. Notice you can solve for $a=bh^{-1}$.
 
10:38 PM
the idea of coset
we take a subgroup H
the elements of the coset are those of the from bh right?
I dont really get the process
 
A coset is just $aH = \{ah: h\in H\}$ for some $a\in G$.
 
do we fix one element of G
 
I think doCarmo does diffgeo of curves/surfaces in that book using forms. I'll check out Cartan and let you know
 
@TedShifrin so we do fix one element a in G right ?
 
Also @Balarka lol at nlab stuff
 
10:40 PM
Read my sentence, Kasmir.
 
@TedShifrin we multiply a by elements of H , either on the left or the right and we get the coset
 
morning ted
 
Tes Dhifrin @Faust
 
Well, you work with only one side, @Kasmir, most of the time.
glares at Demonark
 
@Faust sup faust7
 
10:41 PM
there the same if H is normal
Lol morning everyone
 
Yup, normality is equivalent to every right coset's being a left coset and vice versa.
 
okay but what is not clear to me
do we have to fix one element of G
or we do this for all elements?
like fix a , aH is the coset
 
one coset is generated by a
 
I dont see the use of this ><
we picked a subset of G
H is also a subgroup
we took element of G and we did the product
 
You will write $G$ as a union of these cosets when $H$ is a subgroup. Equivalence relation/partition.
 
10:43 PM
aH
I get that the union of these give us G
and the intersection is either the whole set or phi
Am very eager to see what is the use of such structure
-.-
 
Well, keep learning.
 
let me keep thinking and reading and ill come back to ya :D
 
Heisenberg group
is it commutative?
 
@Faust like the meth cook ?
 
Nope, not when $n>2$.
 
10:45 PM
why cant i find it
 
When $n=2$, it's pretty boring.
 
@Faust it is not =p matrices are not in general
 
i know but when i tried to show that it wasnt it didnt go well
 
Do XY , and YX
the entry 1.3
wont be the same
 
i did but in the example i picked it worked out
 
10:47 PM
Put in general letters, Faust.
 
one last stupid question
i did
 
And make sure it's not $2\times 2$. In that case the group is isomorphic to $(\Bbb R,+)$.
 
i made a stupid assumtion
$S_4$
it has 24 elements
the biggest is order 4 right?
 
i cant seem to think of how to list them
 
10:48 PM
well
easy way is to it by type
(123) , (134) , (124) ,(234) , (132) ,(143) ,(142) ,(243)
(12)(34) , (13)(24) ,(14)(23)
 
thee only one four cycle?
there should be 4 right?
 
no those are transpostions
1 maped to 2 and 3 mapped 4
 
@Kasmir: He knows. He's asking about 4-cycles.
 
no im asking about the four cycles lol
 
How many are there, Faust?
 
10:51 PM
oh ><
 
4? + identity?
 
I do it systematicly
identity has order 1
written (1)
 
ah ok think i can list them all now
thansk @KasmirKhaan
 
@Faust :)
 
Count carefully, Faust.
 
10:52 PM
if you get number that is not divisible by 24
then you did something wrong
possble types are 1 , 2,3,4,6,8,12
 
everything is order 4 3 2 or 1
in $S_4$
 
we talking different things =p
i means the number of elements of each type
 
ah
 
3 cycle you find 8 of them
 
@TedShifrin whats a chair of a department do/mean
 
10:55 PM
In charge of the department.
 
hello
 
Manages everything, makes some tough decisions.
Hi, Nate.
 
wierd ok
 
Most organizations need bosses, right?
Or else it's total anarchy.
 
yeah i guess
 
10:59 PM
Sorry, coming in halfway through a conversation. But would a department chair be listed as something like "Professor, Chair"
 
Something like that.
 
I've been trying to figure out who runs the math department around here.
 
aH is the left coset or the right?
 
I was listed as Professor, Associate Head of Mathematics Department for years.
I call it a left coset, Kasmir. Read your damn book!
 
wow.
 
10:59 PM
i didnt actually find out until today who the chair of our math department was
 
@TedShifrin okay sorry =p
 

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