Mathematics - 2017-04-07 (page 1 of 4)

Danu

12:00 AM

I need to wait for my laundry, which was delayed by 2 hours cause broken f*cking machine

Ted Shifrin

Ah yes, I remember my early life in laundromats. I'm hoping to avoid those in Europe :)

Danu

I have my own machine here, but in this case it's no good :P

Ted Shifrin

Oh :(

Danu

which makes it breaking much worse, of course :P

But okay, my landlord will deal with that I guess.

Mike Miller

Stay here for ten minutes.

@Danu Wait.

Danu

12:02 AM

I'll be around for a while, don't worry.

Akiva Weinberger

...Are you planning on flying to Germany to do his laundry or something

Cows

@Danu One thing led to another and I saw words like higgs and coulomb branch. Can you nudge me in the right direction?

Danu

No, not now.

Cows

ok, will you be able to sometime this week?

or next?

Mike Miller

(1) is trivial (the closure of an abelian group is abelian).
(2) follows because $C(S_w)$ is *the same* as $C(w)$, and it's much much easier to verify that $K_w = C(e^w)$ (where $w$ here is sufficiently small that the exponential map is a diffeo near it). Indeed, suppose $g$ commutes with $w$; then it commutes with $S_w$, and then with the closure. Now if $g$ stabilizes $e^w$, it commutes with $w$, and thus with $S_w$. Conversely if $g$ commutes with $S_w$, then clearly (by definition!) it commutes with $e^w$.

(3) follows immediately from the definition of maximal torus

(4) is just taking the derivative of the definition of the stabilizer

Danu

12:05 AM

(2) was the difficult part

Zophikel

What is a Convex Open St

Mike Miller

The key point I was missing was the explicit identification of the torus involved

Zophikel

What is a Convex Open Set ?

Mike Miller

If it's a random torus it's less obvious how to prove the equivalence between the stabilizer and the centralizer of the torus

Akiva Weinberger

@Zophikel Do you know what a convex set is? Do you know what an open set is?

(Or possibly neither)

Zophikel

12:06 AM

@Akiva yeah

Danu

So why is $C(S_w)=C(w)$? And what do you mean by $C(w)$?

Akiva Weinberger

Yeah as in you know what they both mean?

Danu

did you typo and mean $C(e^w)$?

Zophikel

I know that a Convex Set is a vector space withen $R^{N}$ with line segments joining points withen our vector space

Akiva Weinberger

It's not necessarily a vector space

It would just be a subset of $\Bbb R^n$.

Zophikel

12:08 AM

@Akiva all right

Akiva Weinberger

But, yeah, a set is convex if, for any two points in it, when you draw the line segment between them, every point on that line segment is in the set.

So, like, a donut (in $\Bbb R^3$) is not convex, but a disk and a ball are.

Do you know what an open set is?

Zophikel

@Akiva it's hard to define open set since there's a notion of generality with it so i'll try to give the intuition

Danu

@MikeMiller What's $C(w)$?

Mike Miller

@Danu I got all of those backwards. Being in the centralizer of $w$ is the same notion as stabilizing $w$, more or less. The less obvious claim is that the stabilizer of $w$ is the same as the centralizer of $e^{tw}$ for sufficiently small $t$. (aka for all $t$)

Akiva Weinberger

@Zophikel Specifically in $\Bbb R^n$

Mike Miller

12:09 AM

You could take $C(w)$ to mean "the set of group elements that commute with $e^{tw}$ for all $t$."

Zophikel

@Akiva no i'm not sure :(

Akiva Weinberger

The intuition is, "It does not contain its boundary"

or, rather, "it doesn't contain any point in its boundary"

Zophikel

@Akiva ahhh

Ted Shifrin

Each point is free to wiggle, still staying in the set. :)

Akiva Weinberger

The definition is: A set is open if, for every point $p$ in the set, there's an open $\epsilon$-ball around that point contained in that set.

Zophikel

12:11 AM

so basically a Convex Open Set, is $R^{N}$ with line segments joining points within our vector space that doesn't have any boundary

@Akiva ahh hall right

Akiva Weinberger

An open $\epsilon$-ball around $p$ is just the set of points within $\epsilon$ of $p$.

Mike Miller

Let me rephrase my argument. Consider $S_w$, the stabilizer of $w$. Then it also stabilizes $tw$, and you verify (using injectivity near the identity) that it commutes with $e^{tw}$ for all small $t$, and thus for all $t$. So $S_w \subset C(w)$. Conversely if $C(w)$ commutes with all $e^{tw}$ then it must stabilize sufficiently small $tw$ as well.

So $C(w) = S_w$.

Akiva Weinberger

@Zophikel Yeah, essentially

Zophikel

all right @Akiva sorry I had to look it up

+

Akiva Weinberger

@Zophikel "Joining points within our vector space" You mean "Joining points within our set"

Mike Miller

12:12 AM

Then you observe that (by definition) $C(w)$ means the stabilizer of the line/circle $w$ generates, and then if something commutes with a (not necessarily closed) subgroup, it commutes with the closure of that subgroup.

Zophikel

@Akiva yeah

Akiva Weinberger

the set you want to be convex and open

Mike Miller

Because the closure of $e^{tw}$ is (by definition!) $S_w$, we see that $C(w) = C(S_w)$.

Danu

@MikeMiller For you $S_w$ means stabilizer of $w$? (not the same notation as the picture but that's fine). I can follow your argument I think.

Zophikel

all righ @Akiva got it thanks

Mike Miller

12:13 AM

I was just looking for the definitions from that picture, but then stole the idea from their explanation, @Danu.

@Danu You're right, I just caught what you were referring to

Danu

You are mixing notations; you overload $S_w$

But I can still follow I think

@MikeMiller Here and in the next message you mean $K_w$

Mike Miller

My first message(s) should say "Consider $K_w$" and with $K_w$ replacing $S_w$. In my very last thing

Yeah. I used their notation for the torus and my own notation for the stabilizer

lol

Danu

his notation $S$ for a torus is unfortunate :P

Stupid question: How do I pass to the closure?

Mike Miller

I want you to do that part.

Danu

That should actually be easy

it's like $f(x_n)=x_n$ for all $n$ and $x_n\to x$

Down Christopher

12:19 AM

Just installed a mathjax userscript and thiss alllll looooks good

Ted Shifrin

did you see the link over there >>>>^^^^^, Down?

Akiva Weinberger

@DownChristopher The one given in the room info?

Ted, it's specifically not down, it's up and right

Ted Shifrin

Huh? I said right and up ?

Akiva Weinberger

Ignore me, I was just making a play on the person's name

Physics Guy

Hello there.

Down Christopher

12:21 AM

@AkivaWeinberger github.com/TehFlaminTaco/TacosUserscripts these

Ted Shifrin

oh, duh, I take you literally by mistake so often.

Akiva Weinberger

@DownChristopher There's also a thing over here: math.ucla.edu/~robjohn/math/mathjax.html

(Given in the room description)

I'm curious if your thing is different

Down Christopher

Yeah but the one i linked auto loads for these and gives a preview of the message with it

Akiva Weinberger

(I don't have Javascript installed so I can't check)

Down Christopher

It runs by TamperMonkey on chrome

Physics Guy

12:23 AM

I have a question: Does it seem gay if I say "Hello there"?

Down Christopher

no

Akiva Weinberger

...No

Ted Shifrin

My question in return is: What's wrong if it "seems" gay?

Down Christopher

How do I type mathjax?

Physics Guy

@TedShifrin I just wanted to know to prevent discussions about it in the future.

Ted Shifrin

12:24 AM

Excuse me?

This room is not inhabited by a bunch of teenagers saying "Oh, that's so gay." So please desist.

And, FYI, some of us are gay.

Akiva Weinberger

@DownChristopher How do I run it

I downloaded TamperMonkey but how do I add the script to it

Down Christopher

Go to the link

On github

Akiva Weinberger

OK

I downloaded that

Down Christopher

Click the different files and hit the raw button

Don't download just hit raw

Meow Mix

I use TamperMonkey for my word games.

Ted Shifrin

12:27 AM

no wonder you're lost in game-land so much, Zach!

Akiva Weinberger

Ah, OK

Meow Mix

@Ted At least I'm broadening my vocabulary!

Down Christopher

@MeowMix pls send link

Meow Mix

Send link to what, the script for the game or the word game itself?

Ted Shifrin

are you? :)

Meow Mix

12:27 AM

@Ted I actually am, believe it or not :P

Ted Shifrin

I'll challenge you to WordsWithFriends sometime :)

Meow Mix

Otherwise, I would've not known what "sjambokk" or "roentgenize" meant

Ted Shifrin

When you don't have enough things wasting your time.

Down Christopher

@MeowMix Both?

@AkivaWeinberger You trying it??

Ted Shifrin

well, roentgenize I can guess. No clue on the other.

Akiva Weinberger

12:28 AM

Yeah

Meow Mix

@Down So, the game can be found here: bombparty.sparklinlabs.com

Akiva Weinberger

Is it identical to the other one? In terms of fonts

$\displaystyle\sum_{n=1}^\infty\frac1{n^2}$

Physics Guy

@TedShifrin Okay.....My issue was the following: If it seemed gay and I didn't knew it, then at some point during my lifetime I would have said this, which would probably have lead into a long and boring discussion. So I wanted it to be cleared up now.......Is that an acceptable explanation?

Meow Mix

Here's an explanation of how the game works. Everybody joins one game, and a "bomb" starts. Then you're given a string of two or three letters, and you have to come up with a word that contains those letters in that order.

Akiva Weinberger

Probably

Down Christopher

12:29 AM

@AkivaWeinberger afaik. It just implements the text

Ted Shifrin

No, @PhysicsGuy. I still find it homophobic.

Meow Mix

For example, if the prompt was "ap" you could do "Aptitude", but not "Pastor", because it has to be in the exact order.

Down Christopher

ah

Meow Mix

Now, you can't repeat words, So after I do "aptitude", nobody can do it again

Down Christopher

coolio

that sounded bad

Meow Mix

12:30 AM

The bomb clock goes faster and faster, and when somebody doesn't give a word in time they lose a life.

To get a life back, you have to use all the letters in the alphabet in your words. Like, if I do "aptitude" the letters "a, p, t, i, u, d, e" will be finished and I'll have to do all the other letters before I get another life.

Down Christopher

Userscript?

Akiva Weinberger

@DownChristopher What does the "chat commands" one do

Meow Mix

Let me find it. It just provides an overlay so that you can see how long you've been playing (my record is 58 minutes without dying) and the statistics of how many words and how many lives you've lost and gained and other things.

Down Christopher

@AkivaWeinberger Uhh one sec

Physics Guy

@TedShifrin I'm not homophobic. The problem is that there are many people in the world who consider any sentence which doesn't include how beautiful and great gay people are as homophobic. Why should you think that? Why should you defame me as homophobic when I just wanted to know if a certain sentence or expression might be considered as "gay" in the society?

Ted Shifrin

12:33 AM

Because that suggests that being perceived as "gay" is a negative thing. You get it?

Meow Mix

I don't really think that's considered defamation.

Ted Shifrin

I don't usually go off on this, but seriously ...

ATaco

Hi, I wrote those userscripts.

Semiclassical

comes into chat, sees the conversation, backs away slowly

Down Christopher

@AkivaWeinberger Ask @ATaco

ATaco

12:34 AM

@AkivaWeinberger Yep, it's just an interface.

Akiva Weinberger

@PhysicsGuy It's just strange to consider a sentence to be "gay" or to care that other people might consider it to be "gay"

Ted Shifrin

And I truly resent your characterization that I need to be pandered to, @PhysicsGuy. That's just as bad.

ATaco

@AkivaWeinberger Adds some commands which allow quick writing of ascii emoticons, and LaTeX as images (Which is currently broken)

Akiva Weinberger

Unless the sentence is literally "I am dating someone of the same gender." That sentence I'd consider to be gay.

@ATaco Doesn't seem to work

Typing o_o doesn't do anything but it looks like it's supposed to be one of the commands

Physics Guy

@AkivaWeinberger Ok.

ATaco

12:35 AM

That's probably why I said it's currently broken.

Needs a slash.

Akiva Weinberger

\o_o

/o_o

ATaco

/command

Akiva Weinberger

/shrug

ATaco

Make sure to refresh your page.

Down Christopher

/command

wait what

opps I am dumb

Akiva Weinberger

12:36 AM

ಠ_ಠ

Oh OK it's working now

Thanks

(ノ°Д°）ノ︵ ┻━┻

Down Christopher

Mine did not work

Akiva Weinberger

┬─┬ ノ( ゜-゜ノ)

Down Christopher

@MeowMix bombparty.sparklinlabs.com/play/UUniverselle join

Ted Shifrin

DogAteMy: Do you have a translator for us mortals?

Akiva Weinberger

"(ノ°Д°）ノ︵ ┻━┻" is someone flipping a table

Down Christopher

12:37 AM

@ATaco how do you do the commands? Mine are not working

Akiva Weinberger

and "┬─┬ ノ( ゜-゜ノ)" is someone else putting the table back

ATaco

/command should work, make sure to refresh the page.

logical123

hahaha

Ted Shifrin

Well, that's all clear now, DogAteMy!

Down Christopher

Do I type the emoji?

like O_O

Akiva Weinberger

12:39 AM

The point is that I can now type that by writing "/tableflip" and "/unflip" @TedShifrin

and his code automatically replaces them

Ted Shifrin

wow, pretty cool, except that it's worse than my talking French :D

logical123

@Semiclassical Exactly how i felt, semiclassical lol

Meow Mix

@DownChristopher I can make an English room

ATaco

$user image$

Ted Shifrin

hi logical

ATaco

12:39 AM

Gif Converter now fixed, may take a while to update.

logical123

howdy

Akiva Weinberger

I suppose I could edit the code to put in more shortcuts.

Meow Mix

bombparty.sparklinlabs.com/play/mathse

@Akiva Come play word games with us

logical123

Reveling in a fresh OS install

Down Christopher

@MeowMix I would but I may have to go

ATaco

12:41 AM

I'm usually pingable if you need help with any of my userscripts, if you want to add more commands, feel free to make either an Issue or a Pullrequest, I'm active on the github.

Don't forget to leave a Star!

logical123

I made my first commit on github last night, still kind of baffled at the whole system

Down Christopher

i have to go. Online meeting

LegionMammal978

12:58 AM

By Euler's formula and the definition of exponentiation, $\exp(a+bi)=\exp(a)(\cos(b)+i\sin(b))$, correct? Just making sure

Yeah

yup right1

Yepp

Okay, trying to program this in C using GMP

BAYMAX

GMP?

Meow Mix

1:02 AM

@AkivaWeinberger Oops I clicked on an ad

BAYMAX

ok,here-

GMP is a free library for arbitrary precision arithmetic, operating on signed integers, rational numbers, and floating-point numbers. There is no practical limit to the precision except the ones implied by the available memory in the machine GMP runs on. GMP has a rich set of functions, and the functions have a regular interface.

LegionMammal978

Yeah, using tuples of floats here as complex numbers, so I need to use real operations

Akiva Weinberger

Sorry, have to go @MeowMix

Meow Mix

Me too

Ted Shifrin

Night, young'uns.

Danu

1:10 AM

bubye

I needda go sleep too

Ted Shifrin

Good morning, @Danu.

Danu

My schedule is all messed up

I'm taking a brief "holiday" (still have to study for my upcoming exam) from Sunday onward

LegionMammal978

For the lulz, the actual translation of the formula into C:

mpfr_exp(tmp0, stack[stackTop].re, MPFR_RNDN);
mpfr_sin_cos(stack[stackTop].im, stack[stackTop].re, stack[stackTop].im, MPFR_RNDN);
mpfr_mul(stack[stackTop].re, stack[stackTop].re, tmp0, MPFR_RNDN);
mpfr_mul(stack[stackTop].im, stack[stackTop].im, tmp0, MPFR_RNDN);

10/10 most readable programming language

BAYMAX

hi@TedShifrin

Dodsy

2:30 AM

@TedShifrin Hey, I was wondering if anybody has ever thought of the Riemann Hypothesis in three dimensions? I have a general feeling as to how this would work but haven't done any calculations or anything. Wondering what you think about this, or if you know if anybody has tried to do this. Maybe we can talk tomorrow about it? Also, still working on chemistry just thinking! :)

BAYMAX

"Identifying unknown by knowing the known things "

LegionMammal978

@Dodsy Coming from a relative newbie, how would one extend the zeta function to three dimensions? Some extension of the complex numbers?

Or would it just act on 3-tuples of reals?

Dodsy

@LegionMammal978 I better tell you, I am not even an undergraduate yet! I have a very limited understanding of the Riemann Hyp. But, your idea of extending the complex numbers is exactly where my head is at. I think if the points were plotted it might show some sort of pattern which could be made into an equation of some sort. I think @TedShifrin with his amount of knowledge may be able to have a better understanding.

@LegionMammal978 you probably have a better understanding as: "or would it just act on 3-tuples of reals" means almost nothing to me. I can only infer the meaning.

LegionMammal978

@Dodsy As in, if your extension was based on $1$, $i$, and $j$ (defined through some given relations), then the equivalent 3-tuples would be $a+bi+cj=\begin{pmatrix}a\\b\\c\end{pmatrix}$

We pretty much just separate the 3 parts

Dodsy

@LegionMammal978 I have only considered it using complex numbers, but it sounds like you have a much broader knowledge in how things may work in three dimensions!

LegionMammal978

2:44 AM

Not really, we're just taking 3D coordinates here

Ted would probably know much more than either of us

Dodsy

Ah I see! I don't have "mathjax" installed.

LegionMammal978

Are you on mobile?

Dodsy

No, Chrome.

LegionMammal978

Then try this bookmark while this tab is open

Dodsy

How does this work?

LegionMammal978

2:47 AM

Put it in you bookmark toolbar/list, and click on it while this page is open

BAYMAX

@LegionMammal978 , i am little aware of complex numbers and c,c++,is this sufficient prerequisite to discuss your question ?

LegionMammal978

@BAYMAX Which question?

Dodsy

@BAYMAX We are discussing the reimann hypothesis in three dimensions

BAYMAX

your C question

oh

Dodsy

Yeah....

LegionMammal978

2:48 AM

But yeah, solved that using Euler's formula

BAYMAX

ok , I thought about your previous question you asked Gmp one?

ok

LegionMammal978

2 hours ago, by LegionMammal978

mpfr_exp(tmp0, stack[stackTop].re, MPFR_RNDN);
mpfr_sin_cos(stack[stackTop].im, stack[stackTop].re, stack[stackTop].im, MPFR_RNDN);
mpfr_mul(stack[stackTop].re, stack[stackTop].re, tmp0, MPFR_RNDN);
mpfr_mul(stack[stackTop].im, stack[stackTop].im, tmp0, MPFR_RNDN);

BAYMAX

yup,My bad!

LegionMammal978

@Dodsy But yeah, the Riemann hypothesis concerns the zeros of the zeta function in two dimensions, so a three-dimensional analogue of the function would have to be constructed

BAYMAX

@Dodsy , you can check this video it is nice - [here] (https://www.youtube.com/watch?v=d6c6uIyieoo
)

Dodsy

2:53 AM

@BAYMAX This video is exactly why I'm thinking of this! It shows 2 dimensional cartesian coordinates. Whereas a 3 dimensional may be better

Oh hi @Ted

Hey ted!111

Ted returns!!!

@TedShifrin You can answer my question anytime, I'll come back tomorrow and maybe we can talk about it? I'm heading over to the girlfriend's house and I've got some chemistry to do! I'll talk to you later.

ATaco

3:32 AM

I wrote my own MathJax image service.

Adeek

4:43 AM

@arctictern hi

Secret

[Random] Find a set theory that allow noncommutative inclusions, i.e. $A \subset B \wedge B \subset A =\not\implies A=B$

BAYMAX

7:16 AM

No dream is too big and no dreamer too small !

Tim The Enchanter

7:49 AM

What about the dream in which you dream all dreams that do not include dreaming themselves?

Take that inception

BAYMAX

8:50 AM

oh.. its a snail thing [turbo]! the snail is fast.

@TedShifrin I am talking about the extension problem,now we know that not all continuous functions can be extended from $S$ to $\mathbb{R}$ , in this case what example can I give? next,suppose if $f$ i such a function which can be extended continuously

then is the image bounded

image of $f$

in the sin(1/x) example , it was bounded ,

Martin Sleziak

10:08 AM

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

3

Akiva Weinberger

12:23 PM

Here's an interesting paper.

I guess a motivating question is: consider $\Bbb Z_{100}$ as the set of symbols of the form $[a][b]$ with $a,b\in\Bbb Z_{10}$. (Usually we write, for example, $23$ rather than $[2][3]$.)

The addition formula is $[a_1][b_1]+[a_2][b_2]=[a_1+a_2+z(b_1,b_2)][b_1+b_2]$, where $z$ is $1$ if you need to carry (if $b_1+b_2>9$, essentially) and $0$ otherwise.

What if we change the carrying function $z$? What if we replace it by $2z$? by $3z$? by $5z$? by $0$? by a different function entirely?

And what does this all have to do with cohomology?

Balarka Sen

I am not surprised to see Tim Chow on the acknowledgement :)

Down Christopher

3:29 PM

Hello

robjohn

3:55 PM

Things are certainly slow here today.

Kasmir Khaan

Hi @robjohn

Akiva Weinberger

Yup

robjohn

@KasmirKhaan hello. How are things with you?

Kasmir Khaan

Decent , i got an exam on complex analysis soon and I have troubles understanding laurent series >< how are things with you ?

robjohn

@KasmirKhaan pretty good, at the moment. What troubles you about laurent series?

They are just like power series, with an added bang (calm singularity)

Kasmir Khaan

4:01 PM

Question like find all possible laurent expansions, if they give me something like r<abs(z-z_0) <R i can do them

but how do i determine all possible ones ,and around what points should i do that :/

bang (calm singularity)? =p is that slang?

robjohn

@KasmirKhaan They actually ask for all possible? there are as many laurent expansions as there are points in the plane.

Kasmir Khaan

yes ill give you the example on old exam

robjohn

@KasmirKhaan no, I was just noting that such a singularity can be removed by multiplication by $(z-z_0)^n$

Kasmir Khaan

z^2 / ( 2z+1) ( z+3)

robjohn

@KasmirKhaan maybe they want the ones centered at $z=-\frac12$ and $z=-3$

Those are the ones that have singularities at the point of expansion

Kasmir Khaan

4:04 PM

Yes but from what I understood any annular domain should have such expansion in it

like z between -3 and 1/2

and z between -3 and + infinity

robjohn

@KasmirKhaan Yes, but if you expand about any other points, there is no singular part.

Kasmir Khaan

hmm good point =p

well maybe like you said , only the important ones

like zeros of the bottom of the fraction

robjohn

@KasmirKhaan If I were teaching that class, I would have made that clear about those questions.

A Taylor series is a Laurent series with no singular part.

Kasmir Khaan

Yes =p we get many questions that are very odd

Anyway thanks for the help !:) ill keep working on more examples and see if it gets clearer

robjohn

@KasmirKhaan Ask here if you get stuck

Kasmir Khaan

4:09 PM

I will ! btw are you a proffesor?

robjohn

@KasmirKhaan I used to teach math at UCLA

Kasmir Khaan

@robjohn Nice! you must know Ted then ?

robjohn

@KasmirKhaan I've chatted with him here. Is he now associated with UCLA?

Kasmir Khaan

@robjohn ermm I think he was teaching there but not sure =p forget what I said :D

What do you do now ?

robjohn

@KasmirKhaan I am now developing and maintaining software that is used to teach logic at UCLA and other universities.

Kasmir Khaan

4:15 PM

@robjohn Very nice! I wish you good luck ! now I should go back to do more examples =p thanks for the talk and help again :)

robjohn

@KasmirKhaan Good luck!

mick

4:37 PM

1

Q: Prove that every converging limit $\lim_{n \to \infty} \sum_{k=1}^{a(n)} f(k,n)$ is essentially a riemann sum.

Let $a(n)$ be a strictly increasing function of $n$. Proof that every converging limit $$\lim_{n \to \infty} \sum_{k=1}^{a(n)} f(k,n)$$ is essentially a Riemann sum.

Little Rookie

Hello all, im really confused on the expression of $h(x)$ in the 2 slides, can anyone tell me what is the expression?

arctic tern

4:55 PM

the slide tells you what h(x) is doesn't it?

Akiva Weinberger

@robjohn There are two possible Laurent series for $1/(1-x)$, for example:

logical123

@LittleRookie What do you mean by 'what is the expression', Rookie?

Fawad

Replacing whole RHS as one function to make it simple?

Akiva Weinberger

$1+x+x^2+\dotsb$ and $\dotsb-x^{-3}-x^{-2}-x^{-1}$ @robjohn

arctic tern

there are infinitely many laurent series for 1/(1-x). one for every complex number except 1.

Akiva Weinberger

4:57 PM

They're both around the same point ($x=0$), but the former converges for $|x|<1$ and the latter converges for $|x|>1$.

@arctictern I mean around $0$.

logical123

Could any linux users present tell me how to traverse a directory and save a text file containing the directory hierarchy?

Balarka Sen

@Akiva Isn't the latter around infinity?

Akiva Weinberger

@BalarkaSen Is there a difference for Laurent series?

Besides, I think that if we had two poles, at $1$ and at $2$, we would get three possible series,

for $|x|<1$, for $1<|x|<2$, and for $2<|x|$.

Balarka Sen

I dunno. It seems like you're writing down Laurent series after changing charts from 0 to infinity in $\Bbb R \cup \{\infty\}$ but I am not thinking too hard about it

Transcript for

$Mathematics$ Mathematics