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6:00 PM
Oh, is that the zero-width space thingy? @TheGreatDuck
 
@SteamyRoot no clue. I was trying to hit backspace when editing, it caused my browser to back a page and when hit the forward button and went back to the page the symbol mysteriously appeared in the edit box.
 
@MikeMiller Hmm...
 
For example $m=5$ (1/5 construction of a cantor set looking self similar set)
$$G(x)=\frac{1}{5^{N_x}}+\frac{1}{5}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{5^n}$$?
 
Zee
I understand, and my question is why is it considered a nessecesry background? I actually like the material but it haven't helped me that much in understanding higher level math
 
@AkivaWeinberger there are several characters like that, but this doesn't seem to render to any width. In the edit box it looks like a square with two rows of digits "00 08".
 
Zee
6:01 PM
Riemann surfaces are a notable exception though
 
Useful in engineering probably
 
hmm @Mike how well do I have to know the stuff in the appendicies
to read these notes properly
 
Anyway my question in english is the following:
 
not at all
 
the ascii code is 8
 
6:02 PM
sobolev spaces well
but not the rest of this crap
it's just juicy
crunchy
 
lol
 
There are plenty of people (I'm not one of them) whose research (e.g., Hardy spaces, Banach algebras) relies heavily on one complex variable analysis.
 
LOL!
 
@TheGreatDuck Alt-0173 is a "soft hyphen" which is usually not rendered in browsers.
 
good sobolev spaces are the things i know best of all this lol
 
6:03 PM
@Danu: You have slightly more than one week of peace left. :D
 
@TedShifrin woop-dee-doooo :D
 
@SteamyRoot it isn't that. It is ascii code 8 which is literally the symbol whose acronym is... "BS"!
 
It has great trolling powers. Like, on many online forum/messageboards, the title you type becomes the link used to access the topic. So if you put the title full of those symbols, the link is invisible and nobody can click it.
 
We knew that the cantor function is constructed iteratively by taking away the middle thirds of each segment. This can be generalised to e.g. taking away the middle 5th s of the segments. What happens to the cantor function as my nth where n grew without bound. Naively I seemed to expect to get the line y=x, but attempt to prove it using the series I wrote above just give me a bunch of zeros?
 
@SteamyRoot it's a special character reserved for programming. It isn't a soft hyphen
guys you're not going to believe it!
apparently the special character is the backspace character
somehow the chat glitched and entered a backspace literally. -_-
 
Zee
6:05 PM
@TedShifrin alright. We're still cool?
 
is there a place to submit glitches to stack exchange?
 
$$\lim_{m\to \infty}\frac{1}{m^{N_x}}+\frac{1}{m}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{m^n}=0+ \lim_{m\to \infty} \frac{1}{m}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{m^n}=?$$
 
is anyone in here a game of thrones fan
 
We were never cool or hot to start with.
Check meta, Duck?
 
Zee
Thanks ted
 
6:08 PM
Meanwhile $x$ in base-m is given by $$x=\sum_{n=1}^{\infty}\frac{a_{nx}}{m^n}$$
 
@EricSilva well, you'll need to know sard-smale eventually, but that's the wonder of being able to learn things
 
So I think I am one term off?
 
oh is that "infinite dim" sard
 
yes
 
Zee
@TedShifrin is Hungerford a good book?
 
6:11 PM
Let me try again if my question is unclear:
 
Which Hungerford?
 
the algebra one
 
We know that the cantor function is given by the following expression:
 
do any of you guys wanna know a secret?
start writing a post and then click in the chat and hit backspace
trust me
 
There are at least two, Demonark.
 
6:12 PM

^^produces a backspace character
lol
 
$$G(x)=\frac{1}{2^{N_x}}+\frac{1}{2}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{2^n}$$ where:
 
oh ok I've seen it before @Mike like a year ago
 
Zee
@TedShifrin "algebra"
 
trying to figure out why the largest free quotient of $\pi_1(\Sigma_g)$ is $F_g$, which I'm pretty sure is true
 
$$x=\sum_{n=1}^{\infty}\frac{a_{nx}}{3^n}, a_{nx}=\{0,1,2\}$$
 
6:13 PM
so i'll just pick it up again when it comes up i guess
 
Zee
I tried reading dummet and foote , it's way too chatty
 
my reason for being kind of advanced for my age is actually kind of dumb
 
So there should exists a family of self similar cantor like functions which is constructed by taking the middle e.g. 1/4, 1/5, 1/6 etc. of each segment away
 
90% of it is that the local university was closer to me than my high school, so i just took all their math classes instead of going to my school
 
and in general they should be given as:
 
6:14 PM
@Ted I was kidding
 
I like Hungerford less than I like Dummit and Foote, but my taste is not what rules the world.
 
in the les enfants terribles project, two mathematical brothers were made as clones from the genes of a super mathematician
but one of them, Eric, got all of the analysis genes, and one of them, Balarka, got all of the topology genes
 
$$G_m(x)=\frac{1}{m^{N_x}}+\frac{1}{m}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{m^n}$$
 
Neves likes it more than D&F
I disliked both of them :D
lmao
 
<--- didn't even get blue genes
 
6:16 PM
I like Hungie more and hate D&F
But I dislike both
 
likewise, $x$ is expressed in base $(m+1)$
 
I don't really think there are many good textbooks
 
Zee
@TedShifrin that's quite impressive actually
 
You take what you get
 
I really dislike Hungerford, just as I dislike Conway's Complex Variables text. They're more watered-down rewrites ...
 
6:16 PM
$$x=\sum_{n=1}^{\infty}\frac{a_{nx}}{(m+1)^n}, a_{nx}=\{0,1,2,...,m\}$$
 
You can say that about CVI, but not II
 
I don't really think i've liked an algebra book other than Jacobson
 
CVII is a really nice reference to have
 
Now, if I take the limit as $m\to \infty$
 
Zee
Rewrite of which book?
 
6:18 PM
$$\lim_{m\to \infty}G_m(x)=\lim_{m\to \infty}(\frac{1}{m^{N_x}}+\frac{1}{m}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{m^n})=0+ \lim_{m\to \infty} \frac{1}{m}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{m^n}=\lim_{m\to \infty}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{(m+1)^n}$$
 
oh actually I also had some book I think it was by some dude named pinter or painter or something that i read in high school that I thought was cute
 
Zee
@EricSilva is Jacobson chatty?
 
O wait a sec... I got $\lim_{m\to \infty}G_m(x)=x$
@zee Do we expect as the constant intervals in each iteration of the construction of a cantor like function become smaller, the function will tend towards the function y=x?
 
how difficult to read these math codes?
Why can't they appear in what are intended to represent?
 
Sorry typo:
$$\lim_{m\to \infty}\sum_{n=1}^{N_x-1}\frac{a_{nx}}{m^{n+1}}$$
but then it is not $x$ cause my denominator is one unit off
 
6:22 PM
@CaptainBohemian see "latex in chat" in the room description, upper right corner
 
Heya tern!
 
@arctictern thoughts on why the largest free quotient of $\pi_1 \Sigma_g$ is $F_g$?
 
heya
 
do I need to set up some software to see them?
 
Zee
@Secret I can't read latex here. It will get closer to x=y but that's it
 
6:23 PM
@Captain @Zee Just read the link.
 
No, @CaptainBohemian. Read the stuff at that link.
 
@MikeMiller none
 
@Zee ok I see
 
I guess the kernel has to be free of infinite rank
 
Coudl someone explain to me why here: https://math.stackexchange.com/questions/9580/linearity-of-convergence-in-distribution-of-random-variables
$X \to X_n$ in distribution. It seems really wrong to me =(
 
6:25 PM
Such a boring day.
 
@MikeMiller The kernel corresponds to $\pi_1$ of a cover of $\Sigma_g$, which is a surface of infinite genus because it's got infinite index.
And surface of infinite genus have free $\pi_1$ right?
 
Noncompact?
 
There are not many compact surfaces of infinite genus :)
 
yeah
but that's not much help lol
@BalarkaSen that's nonsense
 
Oh oops
 
6:26 PM
you can pick an infinite index F_2 in F_2
but kernel is free
 
@ZirconCode lmfao that's a post from 2010, and the guy you replied to hasn't been here since 2011.
 
$X$ and $-X$ have the same distribution in that answer
 
I read but don't understand what it means.
 
Actually, I take my oops back. Which bit of it are you establishing as nonsense?
 
@Dodsy one can dream... well that's why I'm here I guess
 
6:29 PM
surface is infinite genus
idk why that should be true
it's noncompact though
 
Ahh. I suppose you are right.
So @Ted's right for pointing that out to me. Sorry.
 
Anyway I don't know where this argument is going and I need to get back to work
2
Q: maps of (real) surfaces

Nick LindsayLet $S$ be a closed surface of genus $g > 0$ and $[S,S] = Hom(\pi_{1}(S),\pi_{1}(S))$ be the monoid of (homotopy classes of) continuous maps from $S$ to itself. Consider the sub-monoid of elements that induce the $0$ map on $H_{2}(S,\mathbb{Z})$. Question Is there a "nice" set of generators for...

Someone should answer this question, it shouldn't be too hard
 
Can a surface group really have a free group of finite index as a subgroup? I don't think so.
 
bro
oh finite index? no
but this isn't finite index
There's a theorem of Kneser that says maps of closed oriented surfaces of degree 0 are homotopic to non-surjective maps, so this is equivalent to saying your map factors through the 1-skeleton, aka you're trying to get a nice set of generators Sigma_g -> Sigma_g that factor as maps pi_1(Sigma_g) -> F_{2g}
I think that should be a feasible classification if you can understand a little bit better what the image can be; it's necessarily of infinite index, say
In general I think you should be able to prove that up to a diffeomorphism of the surface the map factors through the handlebody
 
That seems believable
 
6:36 PM
In other words, his maps are determined by a change of basis and then just $g$ elements of the fundamental group to send the loops to
Since the point-preserving mapping class group is finitely generated and so is the fundamental group, maybe that's enough
I gotta go, you and @AkivaWeinberger should work this out :)
 
Work what out? Huh?
just woke up
 
@Akiva We want to prove the largest free quotient of $\pi_1(\Sigma_g)$ is $F_g$
 
By the way, on the $1+\dotsb+x^{p-1}$ thing, I actually remember seeing that done before now
You sub in $x\mapsto x+1$ and use some sort of criterion
 
@BalarkaSen Well, I mean you should answer that MO question
 
Oh ok
 
6:39 PM
Eisenstein I think (IIRC what that is)
 
But I think that's the first step
 
Yup, DogAteMy, re algebra.
 
Remind me what Eisenstein is?
 
He was a person.
 
@Dodsy Lel
 
6:40 PM
@BalarkaSen Free quotient…?
 
Gotthold Eisenstein
 
Meaning, a group that is a quotient of that and is free?
 
Battleship Potemkin was a great movie
@Akiva Yeah
 
Something about $p|a_i$ for all $i$ except the leading term, DogAteMy, and $p^2\nmid a_0$.
Think about the proof reducing mod $p$. Best way to remember it. See Artin.
 
And $\pi_1(\Sigma_g)$ is $\langle a_1,b_1,\dots,a_n,b_n\mid [a_1b_1]\dotsb[a_nb_n]=1\rangle$, right?
@BalarkaSen
 
6:42 PM
Yes.
 
@TedShifrin Who is "DogAteMy"
I feel unhip around you Ted.
 
Akiva
 
@Dodsy Me.
 
Ninja'd
 
Oh.
 
6:44 PM
Loose hips sink ships, Nate.
 
Hey @Alessandro
Kek @Ted
 
imgur.com/a/4OMjL @TedShifrin @AkivaWeinberger
 
How's it going?
 
@Dodsy …Thanks
 
6:46 PM
No problem
I'm always here to help
 
@Akiva ples put as your picture :P
 
Funfact, that's my hand
 
I thought G was different
 
Turns out, G is open
 
6:48 PM
I actually kinda do want to change my picture at some point, but I'm not sure what to
 
@AkivaWeinberger it's not very formal, he's being handwavy
 
(If you know about Borel sets now is the time to cringe at my comment)
@Alessandro LOL
 
@AlessandroCodenotti strongly reacts
 
@Daminark ugh
 
As in on queue
 
6:49 PM
@Daminark I'm confused
 
G is the set of *open sets or something
 
Open, but yeah
 
I thought we were talking about hand signs for some reason.
 
When you do the Borel hierarchy
 
@Dodsy We are
But we're doing math puns on them
 
6:50 PM
It was a sort of psuedo-pun
@Akiva you're sniping me so much on this chat
 
@Daminark …Because G is the first letter of the word gopen
 
Ah, $G_\delta$
 
You're being dangled.
 
so much for that suggested problem
 
(As opposed to $F$ which is flosed.)
 
6:51 PM
heya @Alessandro
 
@MikeMiller Yeah I have no idea how I'd start that
 
Rip @Mike
 
\begin{align}(x-1)^4&=x^4-4x^3+6x^2-4x+1\pmod5\\ &\equiv x^4+\phantom4x^3+ \phantom6x^2+\phantom4 x+1\pmod5\end{align}
 
looks like not going to my galois theory class is about to bite me in the butt
all these problems are number theory
gg rip me
 
6:59 PM
#rekd
 
goodbye happy life
 
In fact, $(x-1)^{p-1}$ should be $x^{p-1}+\dotsb+1$ mod $p$
(Simply through rewriting the latter as $(x^p-1)/(x-1)$ and using $x^p-1\equiv(x-1)^p$, the Frobenius thingy)
 
@EricSilva looks like i've got too much fuckin writing to do anyway
 
(This is me trying to use the proof of the Eisenstein criterion directly without doing the substitution thingy first)
 
7:02 PM
In any case, that means that if $x^{p-1}+\dotsb+1=fg$, then, $f$ and $g$ have to be powers of $x-1$ when reduced mod $p$.
 
I'll have to write a short thing about Frobenius theorem for an uni project this semester, looks rather interesting (not the one Akiva mentioned though)
 
So, $f(1)$ and $g(1)$ have to be multiples of $p$ I guess?
Which means that $f(1)g(1)$ is a multiple of $p^2$, which contradicts the fact that the value of $x^{p-1}+\dotsb+1$ at $1$ is $p$.
That's cool. It actually looks cleaner if you do do the substitution first, though, doesn't it
 
does anyone know a good book on algebraic number theory lol
 
cassels and frohlich
 
omg nevermind this homework is not due on friday
what a relief
 
7:08 PM
big fan
 
thanks @Mike
 
(it is not what you're looking for lol)
 
@EricSilva Aww
 
10 mins ago, by Akiva Weinberger
\begin{align}(x-1)^4&=x^4-4x^3+6x^2-4x+1\pmod5\\ &\equiv x^4+\phantom4x^3+ \phantom6x^2+\phantom4 x+1\pmod5\end{align}
I can look at individual terms to get an identity in terms of binomial thingies, right?
So $\binom{p-1}k\equiv(-1)^k\pmod p$ I think
 
7:10 PM
Well, in that case, I know the ANT class uses Ireland/Rosen
Not sure how helpful it is but hey
 
i should learn number theory at some point in my life
i know 0
 
A few of us are throwing around the idea to do a bit of Weil over the summer if you want to hop in
 
@Mike this book spooks me
depends when @Daminark, at the beginning of summer i will have my hands full with other things
 
you're not much of a weil guy I don't think...
 
why do you think so?
 
7:17 PM
Hm, the puzzle set introduces the Eisenstein criterion two problems later, and makes us show that the $(x^p-1)/(x-1)$ thing is irreducible using it
 
@Daminark do you mean "Basic number theory"
 
which means that, the first time, I'm supposed to do it without Eisenstein.
 
Number theory for beginners
 
Which, presumably, would end up being how Gauss did it or something.
 
I didn't originally know there was a difference
And was rather surprised
 
7:18 PM
I remember hearing about a number theory book, possibly "Number Theory for Beginners," that immediately talks about insanely advanced topics
 
ok lol thank christ
 
Basic number theory is class field theory, number theory for beginners is the basic stuff
 
Ah, so it's the other one.
 
the table of contents dont tell u anything
 
Also, sniped you I guess
 
7:20 PM
In NTFB? Yeah it doesn't, but it's basically a book that starts from scratch but goes very quickly
 
honestly i feel so bad about my lack of number theory knowledge
it all looks like gibberish to me
 
do you like number theory?
 
Hey, small question here :
what are the main fields in Linear Algebra, which are important for Computer Science, especially machine learning and neural nets?
 
me or Daminark? @Mike
 
Who's that addressed to?
Good lord the sniping
 
7:23 PM
u slow
 
you
 
Merp
@Balarka true
 
Anyway, uh, I'm not sure how many levels this will go before the intended recipient is revealed, so in order to not find that out, it's seems fun from a very brief taste, but it's early to say
 
@ak
@AkivaWeinberger thanks :)
 
7:27 PM
I don't know enough to say whether or not i like it
i literally know nothing beyond some very elementary stuff
 
lmfao.
 
lol
 
I am fucking pissed
I am livid.
I am frothing at the mouth.
 
why
 
Whoa, what happened?
 
7:29 PM
Someone parked their car over two spaces
 
The guy from Queens called me
basically said tough luck
 
What guy from Queens?
 
The school I applied to
 
Oh that's jacked up
 
I spent 320 bucks to apply to a school
and they didnn't review my marks
 
7:30 PM
…Oh, Queens college, not Queens, New York City
 
Queens University.
 
@Dodsy overseeing the stuff that you had said in chat about this that strikes me as pretty inappropriate
 
Yes it was.
 
@Akiva the first time Queens came up I was confused as well
 
He wants me to oay another 160 bucks to apply for part time studies.
 
7:31 PM
@Daminark comment from Keerthi on the new hw "
However, I recommend getting an early start." rip me
 
Think if some new yorker called me up and pissed me off.
 
wtf
 
Gg @EricSilva
 
And @Dodsy really? That's just bad
 
7:33 PM
@Dodsy Why did he reject you without reviewing your marks
 
Because of an administration issue. Basically my new marks never showed up on my application
and then when they received them they didn't review them
but when they received my hs it was on my application so they reviewed that.
So naturally I posted on redit and ranted about it
it already has an upvote.
 
that's pretty bad
sounds like they're money grubbing you tbh
 
Yeah man, I'm done. I'll go to UWO next year in London.
 
Sorry to hear, @Dodsy. You do have other universities you applied to as alternative options, right?
 
Yeah UWO
which is considered a better school.
But I like Kingston more.
 
7:36 PM
Nice.
 
and have too much family in London.
2 of my cousins go to UWO...
anyways I'm off
better go watch a comedy
Thanks for the support guys
you guys da reel mvp
 
Enjoy
 
Well, my heart goes out to you, and hope it all goes well. Have fun Nate!
 
So apparently if I have a polynomial whose roots are a bunch of distinct integers, and I subtract 1, I get an irreducible thingy
($\prod(x-a_n)-1$, $~a_n\in\Bbb Z$ distinct)
 
How do you prove this?
 
7:40 PM
I don't know yet
That's the next thing on the puzzle sheet.
It says, as a hint, to use the Gauss Lemma
(That's the one that says factoring over $\Bbb Q[x]$ is the same as factoring over $\Bbb Z[x]$)
 
oh hahaha
 
I wonder if taking derivatives might be useful
Clearly, I start with saying "Assume it's equal to $fg$ with $f,g\in\Bbb Z[x]$."
 
@BalarkaSen On a fiber bundle, the vertical subbundle is always integrable simply because the fibers are submanifolds, right?
 
@AkivaWeinberger hint: multiply a bunch of numbers together and subtract 1. Is the result not coprime to all the original factors? Dare I not conjecture, that the same is true for polynomials?
 
7:44 PM
yeah
 
Yeah I think so. It should just be the foliation given by the fibers
 
@TheGreatDuck So?
 
Aigh't
 
So it's not a multiple of any of the $(x-a_n)$. That doesn't mean it can't be factored without them
 
@AkivaWeinberger irreducability is the same as not being able to factor into integer roots, right?
 
7:45 PM
@TheGreatDuck Not quite
 
oh
nvmd then
also fyi
 
$x^4+2x^2+1=(x^2+1)(x^2+1)$ is not irreducible, but it has no integer (or even real) roots.
 
is the universe of quack chat starts acting funky dont freak out
 
Irreducible just means you can't factor it into things of smaller degree.
 
7:46 PM
I'm using the browser page inspector to track down the cause of that bug from earlier
i think the issue is with the type of box they used. I'm gonna isolate that code and see if it still does it.
 
@TheGreatDuck Say again?
Use the sandbox to test the chat.
 
@AkivaWeinberger open chat in a new page. Click in the chatbox then click out. Hit backspace. congrats, you discovered a bug.
 
7:57 PM
@TheGreatDuck Here's a question: Is $x^4+4$ reducible?
Actually, there's a fairly large chance you know this one already
 
uum
I think?
 

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