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10:09 PM
Hello everyone! I have never used this chat; I do not know how it works here. So I just post a little questionchen and somebody will answer or do I need something special?
 
LOL @Vishal: It's really that small a question? A little questionchen? :P
Heya DogAteMy :)
 
haha :) got me there
btw, never knew its existence, glad to have found out. MSE, MO are among the best collective efforts out there :)
 
DogAteMy: I just saw your Möbius $\times\Bbb R$ question. For starters, that won't be compact.
What's your question, @Vishal? In general, people usually just ask, although you may get more response if you put it on the main site.
 
I don't have a question. Got here from the meta site asking about chat rooms. Just exploring right now :).
 
Ohhhhh ...
 
10:23 PM
Well thanks anyway for the explanation, see you around I guess :)
 
Okey dokey.
 
hi Ted
 
Hi @Kasmir
 
I have a small question about analytic functions
I still did not go to deep into the material but
 
small, huh?
 
10:25 PM
:D
 
im BACK
(and im zach!)
 
okay lets take f(z) = 1/ (z-1)
 
tuve que dormir un poco
 
where is it analytic?
 
Rehi @Zach
 
10:26 PM
i was thinking about projective geometry during my nap
duality is like
 
Everywhere except at $z=1$, where it has a pole.
 
we can represent it by pwer serie when abs(x) < 1
yes but my question is
 
Get used to $z$, not $x$.
 
$\Bbb P$ lines correspond to $\Bbb P*$ points, and vice versa?
or is it the other way
 
why this definition , every serie that can be represented by power serie is analytic
yes sorry z =p
 
10:27 PM
The point of duality is that it goes both ways, @Zach. (This is special to the plane, by the way.)
 
i meant
nevermind
:P
 
@Kasmir: You have to ask where you can expand it in a series. The series you wrote down (centered at $0$) converges only on $|z|<1$. But you can expand it (and you should know how to do this) in a series centered at $z=2$ or $z=i$ or ... any $z=z_0$ with $z_0\ne 1$.
 
I should start a math club at my school. there's a lot of kids i know interested in learning math
 
Lots of high schools have math clubs, @Zach, so, sure. Get a teacher to be officially an adviser and do it.
 
what about middle school?
do you think they'd let me?
 
10:29 PM
LOL, oh yeah, I keep forgetting.
Isn't this like your last year? Or is there one more?
 
last year
 
Well I can expand it everywhere
 
Probably too much hassle for just a month or two. But why not just gather kids you know are interested and do some math after school? You might talk to a teacher about it.
@Kasmir: Everywhere? And do you know how to do it, say at $z=2$?
 
@Ted heh, so i sent an email to the mathcamp people a few days ago, when i was considering applying about financial aid. they said financial aid is flexible and that we wouldnt have to pay the whole fee in the case we can't afford it
so i guess that's good to know for next year
 
hmm 1 / (1-z ) = sum _ 0-> inf z^n
right ?
 
10:33 PM
That's only good on $|z|<1$, @Kasmir. I want a power series centered at $z=2$. (That means it's a series with powers of $(z-2)$.
 
Hmm I really dont get this stuff yet
We had many defintions of analytic
So I lost the connection
analyic : integral are path independent
 
No, it's not the definition. You need to know how to find the power series centered somewhere else.
 
analytic means the power series converges at that point...?
idk, i only did miniscule complex analysis
 
In a neighborhood of the point, @Zach.
 
Its same formula Ted
 
10:35 PM
NOOOOO, @Kasmir. It is not.
We already know that your series converges only for $|z|<1$. I've said this 3 times already.
 
Yes yes but we can find nighborhood of radius 1
around any point
 
You're not making any sense.
 
abs (z-a) < 1
Iam very comfused indeed
 
No, if you take the point $z_0=1.1$, the power series centered there will have radius of convergence $.1$.
 
Okay
 
10:37 PM
@Ted sorry, i'm not a topologolologist!
 
You need to do the high school algebra of converting $z-1$ into $(z-2)+?$. Then what?
 
z-2 +1
 
Irrelephant, @Zach.
 
neighborhood is that thing that contains an open set
let me look up definition
 
So now how do you do a geometric series for $\dfrac 1{1+(z-2)}$, @Kasmir?
 
10:38 PM
because i is not 2 smart
 
Nah, @Zach, it's just a word for an open set.
 
oh :P
so "Neighborhood of P" <=> "Open set containing P"?
 
well we write it in this form right ? 1 / ( 1- something )
geometric series
 
Right.
Yup, @Zach.
 
Hmm okay =p but that is not really my issue
I want to understand the anatomy of it
 
10:40 PM
Well, do it, @Kasmir. And do it for $z_0=i$ and for a few others.
 
Welcome to the topologists' office; there are many open-- i mean clopen doors of opportunity.
 
I mean more like , what are analytic functions really
 
Find the radius of convergence of those series.
Not true, @Zach. Very few clopen ones.
 
I know how to find radius of convergence using radio test
 
I'm telling you to actually sit down and compute a few of these series so you understand what's going on.
You need this skill to do Laurent series later, for sure.
 
10:41 PM
Oh yeah :)
Okay I ll do that :D
Thanks Ted !
 
@ZachHauk some people do define nbhds of a point $x$ as sets containing an open set that contains $x$
 
are the only clopen sets in the standard topology (on $\Bbb R$) $\emptyset$ and $\Bbb R$ themselves?
 
Doing concrete examples usually increases understanding more than you think.
 
@AlessandroCodenotti ah, so a closed non-open set could really be a neighborhood too by that definition?
 
@Alessandro: I don't know any mathematician who thinks a neighborhood needn't be open.
Wow ... tooooo many negatives in that sentence.
 
10:43 PM
My topology professor used this definition and then specified "open nbhd" pretty often
 
NBHD sounds like some deadly chemical
 
@Alessandro: Maybe it's like the silly French with $0$ being positive and negative.
 
"A space is Hausdorff if every pair of points has a pair of disjoint open nbhd blablabla"
 
> A group is a groupoid with 1 element
 
So, Zach, did you have anything substantive to say/ask about projective geometry? :)
 
10:45 PM
Are you saying my nonsense wasn't substantive?
/s
 
Well, it might have been, but you withdrew the question.
 
no, i just am procrastinating, to be honest
which is a habit
because
i have to do an art project
 
Habits are for nuns, not for math geeks.
Ah, that explains it.
 
hmm
this is a perspective drawing
 
Hi ! I'm been proving some theorems on linear algebra and I stumble upon one that I dont understand. It says if $A$ is a matrix m x n with $m < n$, then $AX = 0$ has at least one non trivial solution.
I get that if $m < n$, and if we name $r$ to the non null rows then $ r \le m$ then $ r < n$, and here it says as $ r < n $ we know it has at least one non trivial solution, why ? Is it because I have a variable dependent of the others ?
 
10:48 PM
In English, @Maks, we say there have to be free variables because there are too many columns. That means you'll have parameters in your answer, so infinitely many answers.
Note that because you're looking at $AX=0$, rather than $AX=b$, the system is consistent — i.e., has solution(s).
With a $b$ there it might be inconsistent — i.e., have no solution.
 
@TedShifrin In the case if have for example $ x_1 + x_4 = 2/3 $ and $ x_2 + x_4 = 5$ then $x_4$ would be a free variable ? That's the concept ?
 
Yes, and $x_3$ is free, as well.
 
Yep
Now, can you help me think this one ?
 
wtf
i was listening to a playlist of songs
 
smacks @Zach for impudence
 
10:50 PM
and this one has like a bunch of math in it
 
Oh, you should listen to Tom Lehrer's songs :)
 
the song's name is aleph-0
not sure you'd enjoy the actual song, but i'll link it anyways
 
I have to prove that if I have two matrix $A_{r x n}$ and $B_{n x m}$ prove that if $ m > n $ then $ABX = 0$ has non-trivial solutions
 
@TedShifrin You know me now xD
 
I was thinking that $(AB)_{r x m}$
 
10:51 PM
@Ted there was even an affine transformation in there!
 
Oh, math in the visual part, not the words. Listen to Tom Lehrer, anyhow, for the words, @Zach :P
 
My phone died. It's probably better to just define nbhds as being open and it's surely pointless to discuss about this :P
 
But I cant deduce anything from $n$ with that logic, neither find a relation between m and r
 
Yeah, and all sorts of fractal pictures, @Zach.
 
@Ted a little bit ago i listened to "Simple Finite Group of Order 2"
but that means they must be inverses :/
 
10:52 PM
@Maks: If $B$ is $m\times n$ and $m>n$, what do you know about solutions of $BX=0$?
 
the 2 elements of a group of order 2 are not each other inverses
 
actually...
forget that
i realized an element can be it's own inverse
 
Well, @Vishal, I guess that means we don't need to talk math :P
 
If $B$ is $m \times n$ and $m > n$ I know that $BX = 0$ has at least one non trivial solution
 
Right, @Maks. Now you're talking about $(AB)X=0$. How can you rethink that?
 
10:54 PM
But here I have the inverse $B$ is $n \times m$ and $m > n$
 
Only a few minor fixes were needed on my Baire spaces thing, I'm quite satisfied
 
Oh sorry, I reversed the letters, @Maks. But the concept is identical.
Yeah, I said it wrong, dammit.
 
@TedShifrin I have more columns than rows, not more rows than columns
 
@TedShifrin Munkres for example, off top of my head. btw, I really admire your patience in answering all the questions here. I could not do that
 
If $B$ is $n\times m$ and $m>n$, then ...
 
10:55 PM
@TedShifrin I dont know what happens then hahaha
 
@ZachHauk You can even construct infinite groups as big as you wish where every element is its own inverse
 
@Vishal: Seriously. In all my research life and pedagogical life, I've never seen the other use.
 
My desk right now has: 2 phones, a raspberry pi computer, my math notebook, a bunch of supplies like compass and ruler, fortune cookie wrapper, 22 guage wire, colored pencils, the art project, and of course my computer
 
I have more variables than equations
Do I have infinite solutions?
 
Right, so with more columns than rows you're guaranteed a nontrivial solution to $BX=0$. Now go back to my question with $(AB)X=0$.
 
10:56 PM
Well ... (AB) is $r \times m$
 
oh and also a Jacob's ladder :D
 
But I dont know if $r > m$ or $ r = m$ or $ r < m $
 
Don't go there, @Maks. I led you a specific way for a reason.
 
@TedShifrin Munkres for example, off top of my head. btw, I really admire your patience in answering all the questions here. I could not do that
 
You could say his work is very Tedious
 
10:57 PM
No, @Vishal, Munkres explicitly says "neighborhood of $x$" is another way of saying "open set containing $x$"!!
 
ba dum tsss
 
sends Zach through the projective duality machine
 
help me
all of my points are now lines
 
@TedShifrin Mmmm
 
@TedShifrin really? i dont have one lying around tonight. will check tomorrow. I always thought that is where I picked this notion up when I first learned topology
 
10:58 PM
This is a super-important point, @Maks. How can you rewrite or rethink $(AB)X$?
I'll bet you 20 whatever unit of money you use. :P @Vishal
 
20 bitcoins
(22591.80 USD)
 

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