Mathematics - 2014-10-17 (page 2 of 3)

Balarka Sen

14:25

@DanielFischer

The Game

Ellu

Balarka Sen

Hello @TheGame.

Daniel Fischer

@BalarkaSen To prove Brouwer's separation theorem, also wait until you have learned singular homology.

Balarka Sen

@DanielFischer What made you think that I want to prove Brouwer's separation theorem? :P

Daniel Fischer

@BalarkaSen You wanted to prove the Jordan curve theorem yesterday, the separation theorem is the natural generalisation.

Balarka Sen

14:31

:P No, I just want a bit of help with point-set topology.

Daniel Fischer

@BalarkaSen Okay. Specifically?

The Game

@Chris'ssis

You're not the only one to slay titans :P @Chris'ssis

Balarka Sen

@DanielFischer Simmons wants me to prove that if X is a metric space, Y a subspace, A an open subset of Y, then A is intersection of Y with a set that is open in X.

Is this OK? : Take a point $x$ in $A$. The $Y$-ball $S_r(x)$ by openness is in $A$. Now the $Y$-ball $S_r(x)$ is $S_r^X(x) \cap Y$ for the $X$-ball $S_r^X(x)$ around $x$. Then taking unions like mad, one gets $A = Y \cap X'$ for some open $X' \subset X$.

Daniel Fischer

@BalarkaSen Well, strictly, you need to say that "There is an $r > 0$ such that ...". But I presume you just omitted that here due to laziness. The argument is the/a right one.

Balarka Sen

Yes, right, I shouldn't've skipped the $\exists$s. Thanks.

@Mike mystical greetings upto deformation retract.

Daniel Fischer

14:41

What's with all the mysticism in here?

Mike Miller

Deformation retract is not an equivalence relation.

Balarka Sen

@MikeMiller yes, ah. for example, wedge of two circles and a square with a cut on it are not equivalent upto retract, but both are deformation of a more general figure, right.

well that fail so hard.

@MikeMiller off-topic : can galois theory be defined for transcendental extensions?

Mike Miller

@Balarka you have yet to define what it means to be "equivalent up to retract"! One can't even define it for a pair of arbitrary spaces, since the property of being a deformation retract is one that a subspace of another space has.

Balarka Sen

for example, you can't treat Gal(C(z)/C) as really a "galois group" in a sense, as if C sits below E sits below C(z), there is no SES 1 --> Gal(C(z)/E) --> Gal(C(z)/C) --> Gal(E/C) --> 1 as E is isomorphic to C(z) by Luroth.

Mike Miller

@DanielF While a normal person would prove Jordan with winding numbers, my taste is to use Alexander duality...

@Balarka no idea

Balarka Sen

14:50

@MikeMiller hey no fair. you called me silly once for saying that C(z)/C can't be treated as a galois ext.

:P

Mike Miller

No, I didn't. I said that the group of automorphisms should still be called the galois group.

Balarka Sen

1 --> 1 --> PSL(2, C) --> 1 --> 1 is not really a SES is it?

oh noes wait. 1 --> 1 --> PSL(2, C) --> PSL(2, C) --> 1 is a SES =(

Mike Miller

I am claiming nothing mathematical. Just terminological.

Balarka Sen

@MikeMiller i don't care about names

Mike Miller

Ok :)

Balarka Sen

14:57

hmm

Mike Miller

@DanielF Oh, I just saw the comment. I once said "mystical greetings" in here because I thought the phrase sounded funny, and it had mass appeal with some of the crowd here.

Balarka Sen

@MikeMiller mass appeal? O_o

chat.stackexchange.com/search?q=mystical+greetings&room=36

i wouldn't call you, rehband and me as "mass" @Mike

Mike Miller

Ah, I seem to have said it more than once

Balarka Sen

15:18

@MikeMiller Zariski cancellation problem has been solved recently.

Though I barely understand what the author did, being unfamiliar with algebraic geometry.

Alec Teal

16:16

Hello again

Are the @Committingtoachallenge and @robjohn pair related?

Nick

@TheGame: You share so muc onion, that it makes my eyes tear up.

$\lceil testing \rfloor$

The Game

@Nick onion ?

@AlecTeal Nope, their pics are just similar

Nick

@TheGame: Those comics are from theonion, right?

The Game

No

theoatmeal.com

Nick

16:34

potato, tomahto.

Real oatmeal is barf though.

@TheGame: :D I'm actually beginning to enjoy these comics from theoatmeal. Thanks for reintroducing me. (who need stumbleupon when they have friends)

The Game

:-)

Ice Boy

c:

The Game

@IceBoy wat

Ice Boy

wuzzup?

The Game

The sky

Nick

16:51

Actually $\hat j$ conventionally denotes up.

Random Variable

17:38

@DanielFischer Yesterday you said that $$ 2\pi i\operatorname{Res}\left(\frac{z}{1+z^2} \sum_{n=1}^\infty a_n e^{inz}; i\right) = 2\pi i \sum_{n=1}^\infty a_n \operatorname{Res} \left(\frac{z}{1+z^2}e^{inz};i\right).$$ I assume that is because $ \sum_{n=1}^{\infty} a_{n} e^{inz}$ converges uniformly. Aren't we basically moving a limit inside of the infinite sum?

Daniel Fischer

17:55

@RandomVariable Yes, the uniform convergence is a sufficient condition for the equality. Remember that $$\operatorname{Res}(g,\zeta) = \frac{1}{2\pi i}\int_{\lvert z-\zeta\rvert = \rho} g(z)\,dz$$ for all small enough $\rho$. Since the convergence is uniform on that circle, we can interchange summation and integration.

Random Variable

@DanielFischer I was looking at it from the perspective that if $c$ is a simple pole, then $\text{Res}[f(z),c] = \lim_{z \to c} (z-c) f(z)$.

Daniel Fischer

Works too. Or you can use the Laurent expansions.

Sawarnik

@nick

Random Variable

18:16

@DanielFischer And it's because the convergence of $\sum_{n=1}^{\infty} \frac{x}{1+x^{2}} a_{n}e^{inx} $ is uniform on that circle, not just because the convergence of $\sum_{n=1}^{\infty} a_{n}e^{inx}$ is uniform on that circle, right?

Nick

@Sawarnik

Sawarnik

@Nick What are ya doing?

Nick

@Sawarnik: Um, readin webcomics, watching cinema sins piss over michael bay's latest transformers installment, trying to translate South Korean.. which I now realize is just Korean ... other than that, nothing much.

Sawarnik

Nice question that :D

And i am trying to re-learn some Canvas, to make a game of Tetris :| @Nick

Nick

@Sawarnik: O_O sounds like your intentionally chopping off your fingers... Good luck with that.

Sawarnik

18:23

@Nick Very much like that :|

Nick

@Sawarnik: Actually ignore me, I have no clue if whether Canvas is easy or difficult because of my reluctancy to learn... One day, I'll have the spark for it. Till then, Hakuna Matata.

user1257255

hi! can you please help me with some proof by induction. i'm stuck at what i actually need to prove because i only know how to correctly prove equations but not inequalities. the basic form is n! > 2^n and once i insert k+1 -> (k+1)! > 2^(k+1) i don't know what do i need to prove... thanks!

Nick

@Sawarnik: What's the best CMS you know. I need to create a database of questions that I can access and manipulate to automatically create worksheets.

Sawarnik

@Nick Its difficult :'(

@Nick I don't know any.

user1257255

@Nick if you're familiar with any programming language it shouldn't be too hard to do this :)

Nick

18:26

@Sawarnik: Nah, it's just like cooking. Just keep your workstation (code) clean and add the right ingredients. Unlike actual cooking, you can restart and scratch out the bitter parts.

Sawarnik

@Nick Unlike cooking, this never works though.

Nick

@user1257255: All you want to do is prove $n! > 2^n$ , right?

user1257255

@Nick that's true, but the problem is that I don't know how to do that for k+1

Sawarnik

because k+1>2 :|

Nick

@user1257255: Don't worry. Calm down. You're problem is simple. Do you remember that you're assuming $k! > 2^k$ , right? Use this fact. C'mon. Let your mind think. What do you know about factorials?

Sawarnik

18:32

"Don't worry. Calm down. Use this fact. C'mon. Let your mind think." Wow.

Nick

@Sawarnik: You will not believe how many times that kind of motivation helps.

Sawarnik

Yes, I won't. But you have a little brother...

user1257255

n! = 1 * 2 * ... * (n-1) * n

Nick

@user1257255: :D Yes, n! = n(n-1)! = n(n-1)(n-2)!

You're right

You have all the pieces. Solve the puzzle.

@user1257255 Hint: Multiply both sides of $k! > 2^k$ with (k+1) ... Now, do you see?

@Sawarnik: Do I suck at communication, bud? Sometimes I feel like I do. I need to work on my approach to things and how I explain things and express ideas.

user1257255

@Nick do i need to multiply them to get something out of it? :|

Sawarnik

18:41

@Nick No you don't :)

Nick

@user1257255: $k! \times (k+1)$ = ?

user1257255

k*k! + k! ?

Random Variable

@DanielFischer Ignore my last question. It was a silly question and the answer is obvious. Thanks for your help.

Nick

@user1257255: Um, why don't you try expanding $(k+1)!$ like you did $n!$

$$(k+1)! = (k+1)\cdot \underbrace{k\cdot (k-1)\cdot (k-2) \dots 3 \cdot 2 \cdot 1}_{\text{Does this not look like k! ?}}$$

user1257255

where should i put your code to see nice structure of it (is there any online editor or other program because i see only a lot of {}$$ and other characters)?

Nick

18:49

@user1257255: Do you not have $\LaTeX$ enabled ?

@Sawarnik: Where are the chat rules and guidelines! We have a new user!

Sawarnik

@Nick On the star board!

Nick

@Sawarnik: Not available on my end on my current connection. Just link the MathJax thing to @user1257255

Daniel Fischer

@user1257255 math.ucla.edu/~robjohn/math/mathjax.html

There you can get a bookmarklet (?) to have $\LaTeX$ rendered in chat.

Sawarnik

@Nick Is height-height-height a congruence criterion?

Nick

@user1257255: Use the start ChatJax thing on this page.

user1257255

18:53

thanks, it works

Nick

@Sawarnik: ... without context, it can also be criterion for kidney transplants. Congruence criterion for what? And height of what?

Sawarnik

@Nick Triangles!

Alec Teal

OVer 5 hours less than 10 views, gotta bump: math.stackexchange.com/questions/978181/…

Sawarnik

+1 view

Nick

@Sawarnik: SSS congruent triangles. That's a thing.

Sawarnik

18:55

@Nick But HHH?

Nick

@Sawarnik: ... In which dimension is your triangle. Your terminology is confuzzling me.

@user1257255: Ok, so did you prove it yet?

Sawarnik

@Nick 2D!

user1257255

@Nick i don't think so... i have (k+1)k! > 2^(k+1)

Nick

@user1257255: ... Are you sure that's what you have? In that case is the point that (k+1)k! = (k+1)! that difficult to comprehend?

You yourself defined a factorial to be this way.

25 mins ago, by user1257255

n! = 1 * 2 * ... * (n-1) * n

@user1257255: Look, I'm tired. here's the part of the proof that you need:

\begin{align*}
(k+1)! &= (k+1)k!\text{ (by the definition of factorial)}\\
&\ge (k+1)2^k\text{ (by the induction hypothesis)}\\
&> 2\cdot2^k\text{ (since }k\ge 4\text{)}\\
&= 2^{k+1}.
\end{align*}

@user1257255: Any doubts?

Sawarnik

@Nick Well done Nick!

user1257255

19:03

@Nick I understand, I need to learn how to prove inequalities. Thank you very much for your help! :)

Nick

@Sawarnik: I lost, you cheescake.

Sawarnik

Cheescake?

Nick

@user1257255: No, it's not about inequalities. All the induction questions are mostly the same. You use your first domino to knock over all the other dominoes.

@Sawarnik: =_= Yes, well, you see, my friends turn into edible delicacies once I am hungry enough.

Sawarnik

@Nick I would prefer to be mushroom soup!

Nick

@Sawarnik: Cream of mushroom or clear mushroom soup?

Sawarnik

19:07

Cream of mushroom.

What is clear mushroom? :O

Nick

@Sawarnik: ... boiled mushrooms in salty water.

Just to make it sound more dramatic: It's dirt covered white fungii of bare minimum edibility boiled in heavily chlorinated tap water and seasoned with iodized sodium chloride that was stored in some filthy man's smelly socks.

ccorn

This question has been edited by Community. Is that a bot? So ... in math formulae are automatically replaced by \ldots or \cdots? I'd like to see such features extended to the most frequently done math edits to new questions.

Nick

@ccorn: Ah yes, @Community, the monster that keeps testing me with strange edits in the review centre. If he is not a bot, then he is a madman. Either way, I despise his shmuck attitude.

Sawarnik

@Nick :D

Nick

@Sawarnik: So, how's tetris going?

Sawarnik

19:18

@Nick Bad. Its tough [in terms of motivation] to learn Canvas again :(

Nick

@ccorn: I looked into it. This is the claim that the User Bio is making:
Hi, I'm not really a person.
I'm a background process that helps keep this site clean!
I do things like
- Randomly poke old unanswered questions every hour so they get some attention
- Own community questions and answers so nobody gets unnecessary reputation from them
- Own downvotes on spam/evil posts that get permanently deleted
- Own suggested edits from anonymous users

Random Variable

It says that I've been a member for 3 years and 1 month. But I noticed that my first activity of any kind on MSE was on March 16, 2012. What was I doing for all those months? Could the system be wrong?

Sawarnik

Hmm.

I despise him too anyway.

ccorn

@Nick Thanks. So it's probably an edit by an anonymous user. Alas, I hoped to see a few of those standard editing tasks automated :-)

@RandomVariable Could that have something to to with your login method? Perhaps your authentication provider has an older account creation date.

Or they measure membership time in Venus years.

2

Random Variable

@ccorn I'm going with that they measure time in Venus years since my Google account is much older than that.

Care Bear

19:32

@ccorn Edits are attributed to community when they are made by an anonymous visitor from the Internet. You can get the story by clicking "edit approved" in the revision history.

It was approved by the author of the question, by the way.

here's a duplicate to close, for those who have close votes left.

ccorn

@CareBear Thanks!

Jasper Loy

20:20

Hello @alizter, lol. I am bored.

Balarka Sen

@Daniel

Jasper Loy

@BalarkaSen Do you like Munkres so far?

Balarka Sen

I'm not using Munkres, @Jasper.

Jasper Loy

@BalarkaSen Oh, then what?

Balarka Sen

G. F. Simmons.

Jasper Loy

20:26

OK. I think Munkres is better than Simmons, though I dislike both, lol.

Balarka Sen

Kids : If Jasper says "A", be sure that "¬A" is true.

@Jasper Jokes apart, let me have a look at Munkres.

Alyosha

If $\Bbb{R}=S \times \Bbb{Q},$ prove that $S$ is uncountable. I'm not sure if this is actually true.

Balarka Sen

Hello @Alyosha!

Alyosha

I mean, it probably is.

Hello.

Balarka Sen

@Alyosha product of countable sets is countable

R is uncountable, thus so is S.

Alyosha

20:29

Yes, but you're assuming the continuum hypothesis.

Daniel Fischer

@BalarkaSen Yes?

Balarka Sen

Huh, @Alyosha?

Alyosha

That is, if $|\Bbb{R}|>|S|>|\Bbb{Q}|$, prove that $|\Bbb{R}|>|S \times \Bbb{Q}|.$

Balarka Sen

that was not your original problem.

Alyosha

It's the nontrivial bit.

Balarka Sen

20:31

as Q is countable, if S is countable then so is Q \times S.

It's a trivial fact.

Alyosha

Yes, that is the trivial bit.

Balarka Sen

And R is countable requires no CH at all.

Alyosha

But that leaves the case of $|\Bbb{R}|>|S|>|\Bbb{Q}|$.

Balarka Sen

no.

Alyosha

I don't see why not.

Well, if such a set exists it's not obvious to me that $|\Bbb{R}|>|S \times \Bbb{Q}|$

Balarka Sen

20:34

oh right

sorry i am not diving into CH :P

Alyosha

I suppose that in general the theorem is for infinite sets that $|A \times B|= \text{max}(|A|,|B|)$

Balarka Sen

@Alyosha what are you studying these days, apart from sets?

Jasper Loy

Sets and cats.

Balarka Sen

And Tsetse fly?

Alyosha

My university time (apart from basic lectures) is made up of combinatorics/graph theory/algebraic topology and cats.

Balarka Sen

20:36

and no general topology?

Alyosha

I've read some of munkres. General topology is later in the year (AT is for a different year).

Balarka Sen

I'm studying point-set at the moment ;)

Also doing commutative algebra hard.

Alyosha

I really like the look of the book 'Galois Groups and Fundamental Groups' by
Szamuely, but I have less time to read it than usual.

Oh, excellent.

Oh, by the way, do you know how to prove Rogers-Ramanujan with modular forms?

Balarka Sen

Was gone for a while.

Alyosha

I encountered them over the summer and didn't see a proof.

Balarka Sen

20:41

You need some algebraic topology and commutative algebra before reading up Grothendiek's galois theory, @Alyosha

Did you read up Alekseev?

@Alyosha What, Rogers-Ramanujan identities?

Jasper Loy

Well, I am only going to read Calculus I for now. You are light years ahead of me @bal, lol.

Alyosha

Yes.

That book doesn't seem to be very CA heavy, and initially it's only got basic AT stuff in it.

Balarka Sen

As far as I recall, RR identities doesn't actually require modforms, do they, @Alyosha?

@Alyosha I have skimmed through it. It's not very good.

Alyosha

No, but I think the person who showed them to me claimed they did.

Perhaps I didn't remember correctly.

In what way does it look not good?

Balarka Sen

@Alyosha There are combinatorial proofs. But I believe it can be proved through some algebraic number theory on $\Bbb P^5$. Observe that the rogers ramanujan functions are actually shifted generators of the functions field of $X(5)$.

Care Bear

20:45

@MikeMiller I stand corrected: the all-caps warning was added on Programmers two days earlier. Programmers and Math are the two SE sites that have it so far.

Balarka Sen

i.e., hauptmoduls.

Jasper Loy

Hello @carebear, do you like the cartoon?

Balarka Sen

@Alyosha It's way too intuitive. And too theory-based.

Alekseev is intuitive too but it rigorously makes up the whole idea through just excercises

Hello @TedShifrin

Alyosha

You mean it appeals to intuition rather than rigour?

Balarka Sen

Right @Alyosha. And more than that, it describes the theory rather than building it up from foundations.

It's a survey.

I'd rather study hard and read up SGAs rather than studying surveys.

Alizter

20:48

@JasperLoy hello

Alyosha

SGA?

Jasper Loy

@Alizter Have you been working on math problems with Sarah?

Balarka Sen

Séminaire de Géométrie Algébrique du Bois Marie

In mathematics, the Séminaire de Géométrie Algébrique du Bois Marie was an influential seminar run by Alexander Grothendieck. It was a unique phenomenon of research and publication outside of the main mathematical journals that ran from 1960 to 1969 at the IHÉS near Paris. (The name came from the small wood on the estate in Bures-sur-Yvette where the IHÉS was located from 1962.) The seminar notes were eventually published in twelve volumes, all except one in the Springer Lecture Notes in Mathematics series. == Style == The material has a reputation of being hard to read for a number of reasons...

Alyosha

I assume there exist English translations.

Balarka Sen

I believe so.

Alyosha

20:51

Well, also I gather that I probably wouldn't understand SGA at the moment.

Balarka Sen

@Alyosha It'd take a lot of knowledge in comm. alg., alg. geo. and alg. top. to even grasp it.

Alyosha

So you essentially recommend reading CA etc.?

Balarka Sen

Fundamentally it builds up the unifying connection between galois theory, covering spaces and number theory.

Right @Alyosha. In fact I am studying them. My temporary goal is to understand Grothendiek's works.

And it'd probably take me years to get there.

Jasper Loy

@BalarkaSen His works on what?

Alyosha

OK, fair enough. I suppose it's good that I hadn't started that book then.

Balarka Sen

20:55

@JasperLoy Galois theory/Covering spaces/Number theory. =P

@Mike mystical greetings

Jasper Loy

I see Care Bear doesn't want to talk to me.

Alyosha

Both commutative algebra and algebraic geometry lectures clash with my other lectures, sadly. I suppose I could use Lang for CA.

Jasper Loy

I won't ping him anymore then.

Balarka Sen

@Alyosha I am using Atiyah-McDonald.

Mike Miller

Hello

Balarka Sen

20:57

Again with the universal helloes @Mike

Alyosha

In your notation for a function field, was $X$ an arbitrary field?

Balarka Sen

@Alyosha huh?

oh you mean $X(5)$?

Alyosha

$X(5)$

Balarka Sen

no, that was the compactified modular curve of level $5$.

Alyosha

Oh, OK. Evidently I don't know enough MF.

Balarka Sen

20:58

me neither

read some of them during my days of quintic-obsessions.

i don't really "know" about them. i am only familiar with what they are.

Alyosha

Ah. I don't know the definition of a compactified modular curve.

Balarka Sen

@Alyosha quotient the upper half plane by $\Gamma(5)$ and compactify

Alyosha

What does the $5$ here signify?

Balarka Sen

$\Gamma(N)$ is called the congruence subgroup of level $N$. defined as the group of 2x2 matrics with elts from F_N such that diagonal elts are $\pm 1$ mod N and the rest are 0 mod N

if I recall correctly

Alyosha

Oh yes, of course.

Jasper Loy

21:05

Prof @bal is uttering profound stuff.

Balarka Sen

@JasperLoy no, I am mumbling them

Jasper Loy

Whereas I am still at Calculus I, LOL.

Balarka Sen

I only have a super-intuitive super-nonrigorous idea about modular forms =(

Pedro Tamaroff

@DanielFischer

Balarka Sen

Hello @Pedro

Pedro Tamaroff

21:07

Hello.

Jasper Loy

However, it is possible that the solution to Riemann Hypothesis comes from calculus...

So that's where I am putting the money, LOL.

Alizter

@JasperLoy She has not replied atall.

Daniel Fischer

@PedroTamaroff Evening, @Pedro.

Balarka Sen

@Pedro $A, B \subset X$. I want to prove that $int(A) \cup int(B) \subseteq int(A \cup B)$. does it not suffice to take an elt $x$ of $int(A) \cup int(B)$ and say that it is either in $int(A)$ or in $int(B)$ which automatically implies it is in $int(A \cup B)$?

Jasper Loy

@Alizter OK, I guess she is usually very busy. Might be the next female winner of the Fields medal, LOL.

Alizter

21:09

@JasperLoy Do you know anybody on MSE that has a fields?

Jasper Loy

@Alizter No. I had some Mrs Fields cookies though.

Balarka Sen

@Alizter not a field medalist, but en.wikipedia.org/wiki/Jonathan_Lubin

Jasper Loy

@BalarkaSen Ah, another J L, LOL.

Balarka Sen

Lubin, Minnesota

13.3k 1 17 33

Jasper Loy

I am very sad that my favourite John Lee does not have a Wikipedia page. Maybe I should start one for him...

Pedro Tamaroff

21:11

@DanielFischer Today I had my complex analysis test.

Jasper Loy

@PedroTamaroff My analysis test sure was complex.

Daniel Fischer

@PedroTamaroff I hope you did well.

Pedro Tamaroff

@BalarkaSen Use the characterization of the interior as the largest subset of $A$ that is open.

Alizter

I reckon that the RH will be solved with new mathematics.

Pedro Tamaroff

Note that $A,B\subseteq A\cup B$

So certainly the interior of $A$ and of $B$ is contained in $A\cup B$.

Balarka Sen

21:13

@Alizter yes, google [Field with one element]

@Pedro OK

Jasper Loy

I hope John Lee does not retire for another decade or so. I might want to do my thesis under him if possible.

Pedro Tamaroff

In particular, ${\rm int}\,A\cup{\rm int}\,B$ is an open subset of $A\cup B$.

Jasper Loy

@Alizter It might just involve calculus, you know, like what Chris's Sis does.

Balarka Sen

right, as union of open subsets is open

Pedro Tamaroff

@DanielFischer I had to prove that if for $f$ entire ${\Im }f(z)>0$ for all $z$, $f$ is constant.

I used Liouville's theorem.

Alizter

21:14

@JasperLoy It seems to deep of a result to just use calculus

Pedro Tamaroff

I simply showed that an unbounded entire function has a zero, @DanielFischer

Balarka Sen

Picard, @Pedro

Pedro Tamaroff

@BalarkaSen We can only use results proven in class, smartpants.

Daniel Fischer

@PedroTamaroff Umm, $e^z$?

Alizter

@BalarkaSen what is your favorite group?

Balarka Sen

21:15

@Alizter $\mathbf{Gal}(\overline{\Bbb Q}/\Bbb Q)$

Alizter

@BalarkaSen what about finite group?

Balarka Sen

$A_5$.

Pedro Tamaroff

@DanielFischer Sorry, I mean like polynomials.

Where $|f(z)|\to \infty$ as $|z|\to\infty$.

Daniel Fischer

@PedroTamaroff Those are polynomials.

Balarka Sen

@Alizter did you google field with one element?

Alizter

21:18

@BalarkaSen yes

Balarka Sen

did that blow your mind? did it? did it? did it?

Pedro Tamaroff

@DanielFischer I might have screwed up then. I cannot remember what I wrote now. How would you prove it, Daniel?

Jasper Loy

@BalarkaSen Nothing can blow my mind because it is already blown.

Jayesh Badwaik

I proved a result which said that if any disc is not in the range, then the entire function is constant.

Daniel Fischer

@PedroTamaroff That an entire function with positive imaginary part is constant? Liouville $\frac{f(z)-i}{f(z)+i}$.

Or $\frac{1}{f(z)+i}$.

Jasper Loy

21:20

Very interesting that @dan puts his hand on his mouth and @jay puts his hand on his cheek. Where should I put mine?

Alizter

oof

Pedro Tamaroff

@DanielFischer Why cannot I Liouville $1/f(z)$?

Oh, right.

It might get close to $0$.

Jasper Loy

@Alizter Is that a woof from your doggie?

Daniel Fischer

Right, @Pedro. You don't know a priori that it doesn't.

Jayesh Badwaik

@JasperLoy That interesting place?

Pedro Tamaroff

21:22

@DanielFischer OK. Then I had to find all branches of the logarithm such that $$\int_{|z-i|=1/2}\frac{z^{4/3}}{(z^2+1)^2}dz=-\frac {\pi}6$$

Jasper Loy

@JayeshBadwaik Flag!

Pedro Tamaroff

Another one was finding all entire functions for which $f(x+iy)=f(x)+if(y)$.

Balarka Sen

@PedroTamaroff Residue it?

@PedroTamaroff C-R equations?

Pedro Tamaroff

@BalarkaSen You use Cauchy's formula. Then it boils down to finding the branches for which $i^{1/3}=-i$.

Balarka Sen

Bah I am just raving names.

zips lip

Pedro Tamaroff

21:25

@DanielFischer How would you solve the second one?

Daniel Fischer

@PedroTamaroff $f(0)= 0$, and $f((1+i)t) = (1+i)f(t)$ for all real $t$, hence for all complex $t$. Now look at the power series expansion to see that $f(z) = c\cdot z$.

Pedro Tamaroff

@DanielFischer What if I showed that $f'(z)$ is constant?

@DanielFischer Yes, $f(z)=f'(0)z$.

I proved that.

I showed that $f'(z)$ is constant on $\Bbb R$.

Daniel Fischer

@PedroTamaroff Yes, showing that the derivative is constant also works. You know that.

Pedro Tamaroff

@DanielFischer I am missing one more problem.

Oh. It was some convergence stuff of series.

Not worth it.

Daniel Fischer

Trivial?

Pedro Tamaroff

21:33

It was $$\sum_{n\geqslant 0}\frac{w^n}n$$ with $w=i\frac{1-z}{1+z}$.

I showed the thing converged only for $w\in \bar B(0,1)-\{1\}$ using Dirichlet's criterion.

Then I pulled back with the Cayley map.

I got that $z$ must be in $\{\Re z\geqslant 0\}-\{i\}$.

Balarka Sen

Cayley map, @Pedro?

Pedro Tamaroff

@DanielFischer What I did not do (there where five exercises) was finding some Möbius transformation that sent two points to two others and a line to a circle.

I'm a wuss like that.

Daniel Fischer

@PedroTamaroff Cross ratio.

Pedro Tamaroff

Yes, it is a biholomorphic mapping $\Bbb E\to \Bbb H$.

Daniel Fischer

@PedroTamaroff $\mathbb{E}$? The standard is $\mathbb{D}$, for disk.

Pedro Tamaroff

21:37

@DanielFischer You mean that $$\frac{z_3-z_4}{z_3-z_2}\frac{z-z_2}{z-z_4}$$ thingy?

I don't really know how to use it @DanielFischer

@DanielFischer Dunno, Remmert uses $\Bbb E$.

Balarka Sen

@PedroTamaroff I thought that was called Riemann mapping theorem? Or am I wrong?

Daniel Fischer

@PedroTamaroff Ah, "Einheitskreis", German.

Pedro Tamaroff

@BalarkaSen The Riemann mapping theorem says every simply connected domain of $\Bbb C$ is biholomorphic to $\Bbb E$.

@DanielFischer Cool beans =)

Balarka Sen

Ah ah.

Daniel Fischer

@PedroTamaroff It maps $(z_2,z_3,z_4)\mapsto (0,1,\infty)$.

Mike Miller

21:38

@Pedro Slightly incorrect. Be careful.

Pedro Tamaroff

@DanielFischer What's the word breakdown there?

@DanielFischer Yes, I know.

@MikeMiller Damn it, Mike.

NONEMPTY.

There.

Happy?

Mike Miller

Still wrong.

Pedro Tamaroff

Proper, nonempty.

Daniel Fischer

@PedroTamaroff Einheit: unit; Kreis: disk (but it can also mean circle, use Kreisscheibe or Kreislinie to avoid ambiguity)

Mike Miller

Good.

Balarka Sen

21:39

Does Remmert require prerequisites on differential forms?

Pedro Tamaroff

@BalarkaSen No.

The book starts with baby steps.

Balarka Sen

Cool.

Pedro Tamaroff

@DanielFischer So what to do with it, Daniel?

What's the general MO?

Balarka Sen

Not sure if I should cut off something from topology, commutative algebra and sieve theory to include complex analysis. =P

Daniel Fischer

@PedroTamaroff Choose third points, and compose the cross ratio mapping the first three to $0,1,\infty$ with the inverse of the CR mapping the second three to $0,1,\infty$. Of course if you see something more direct, use that.

Pedro Tamaroff

21:42

@DanielFischer Right, yes. But I have to map a line to a circle.

Apart from two points to two points.

Daniel Fischer

For pairs of points in the plane, you can of course immediately write down an affine transformation.

Pedro Tamaroff

Yes.

@DanielFischer So I guess I am passing this one, but I screwed up in the $\Im f(z)>0$ part.

I mean, I guess I got the jist of it, but I think my solution is incorrect.

Daniel Fischer

@PedroTamaroff A line is a circle passing through $\infty$. So you can just map $z \mapsto \frac{1}{z}$, scale, and translate.

Mike Miller

@Balarka If you are a normal human, you will not absorb things properly when trying to learn huge swaths of vastly different subjects all at once. Take your time.

Balarka Sen

@MikeMiller that was meant as a joke.

i am not really studying anything other than topology atm.

Pedro Tamaroff

21:45

@DanielFischer Yes, so I started doing it and did use $z\to z^{-1}$, using that if I have a circle with center $w$ and radius $|w|$, it gets mapped to a line through $1/2w$ perpendicular to the segment from $0$ to $1/2w$. But then I was like "Nah, I'll get stuck."

So I just did four out of five.

Mike Miller

Ok.

Balarka Sen

@Pedro except you screwed up at the first, so that's 3 outta 5.

Pedro Tamaroff

@BalarkaSen Yeah, it's still a pass though.

Care Bear

Getting tired of programming contests dumps. Where is this from? I suspect HackerRank. Both questions of this users are of this sort.

In any case, off-topic -- contest or not.

(And what's with contest organizers who don't display the problems publicly? They'll be posted around anyway.)

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$Mathematics$ Mathematics