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5:06 PM
@MikeMiller Eh, I interpreted it as a symmetry of the SUSY (+ derived fields) algebra
But yeah, the relevant chapter in Hori et al. is called "Landau-Ginzburg models" ;)
 
Hi everyone!
 
Guys, someon could be give me an axample of a perfectly normal Hausdorff space?
 
@danu I wonder what Landau-Ginzburg means in that context
 
I meant, no only say "all paracompact spaces" or something like that
 
5:15 PM
main thing it means to me is a model of superconductivity
 
I would like to know a precise example, defining the topological space
 
though maybe that's the difference between Ginzurg-Landau and Landau-Ginzburg :p
 
@MatsGranvik Is that a pun on "beet root"?
@user17629 All metric spaces ($\Bbb R^n$ for example) are perfectly normal
Suppose we want to separate the closed sets $E$ and $F$ by a function in a metric space.
 
@AkivaWeinberger I know, you see.. I'm trying to define an example that sastifies: $T_{0} \subset T_{1} \subset T_{2} \subset T_{2 \frac{1}{2}} \subset T_{3} \subset T_{3 \frac{1}{2}} \subset T_{4} \subset T_{5} \subset T_{6}$
 
$x\mapsto\dfrac{d(x,E)}{d(x,E)+d(x,F)}$
 
5:23 PM
that becuase I asked first about $T_6$ and then try to make that sastifies the requirement above
 
Wouldn't anything $T_6$ satisfy all that? Also, I think you want $\supset$, not $\subset$
 
@AkivaWeinberger yes exactly.
beet root
 
Oh, maybe by $T_i$ they mean the list of properties that define $T_i$ spaces, not the class of all $T_i$ spaces
 
@AkivaWeinberger Yep, exactly there are talking about the properties
they*
Sorry for the misunderstanding
@AkivaWeinberger Can you elaborate an example about it, please?
 
5:36 PM
Take $\Bbb R^n$. It's perfectly normal and Hausdorff.
To separate closed sets $E$ and $F$, use the function $x\mapsto\dfrac{d(x,E)}{d(x,E)+d(x,F)}$.
Here, $d(x,E)$ is the distance from a point $x$ to the set $E$, defined as $\inf_{y\in E}\{d(x,y)\}$.
 
@AkivaWeinberger I see, thank you!
 
Note that this is always defined (we never divide by zero):
that's only possible if $d(x,E)$ and $d(x,F)$ are both zero. But, since $E$ and $F$ are closed, that means $x\in E$ and $x\in F$.
But this is impossible since $E$ and $F$ are disjoint.
 
@AkivaWeinberger Got it :)
 
$d(x,E)=0\implies x\in E$ isn't true for all sets, only closed ones, by the way
Take $d(~0,(0,1]~)$ as an example
 
@AkivaWeinberger ok, I gonna try to build it
 
5:42 PM
(And I hope you know how to show Hausforffiness.)
What do you mean by "build it"?
(To separate two points, find the distance, cut it in half, and take the open balls with that as the radius)
 
@AkivaWeinberger I meant the example and try to understand it
 
Ah
It's basically just turning the ratio $d(x,E)~{\large :}~d(x,F)$ into a fraction.
 
Are there any conditions to be satisfied for applying squeeze theorem
 
If $f(x)\le g(x)\le h(x)$ for all $x$ and $\lim f(x)=\lim h(x)=L$ then $\lim g(x)=L$, right?
Don't think there's anything else we need
Well, if we have $\lim_{x\to a}$, then $a$ needs to be a limit point of the domains of everything, but that's just needed for the limits to be defined in the first place
 
Is this inequality true for $x\in[1,10)$ , and the inequality: $2x\leq x^2+1\leq 20x$
 
5:54 PM
~@MikeMiller do you have a hint on this?
 
@ramsay Yes. (I just graphed it)
 
@MikeMiller I tried to recall the proof of stability theorem today.
 
@AkivaWeinberger then when take the limit tends to 0 for this inequality and you will see that squeeze theorem doesn't work for this inequality but why?
 
Hello everyone, I'm new to this chat. Why can't I see what you're writing when it comes to math?
I disabled AdBlock, but that didn't help
 
see the "latex in chat" link in the chat room description
 
6:03 PM
Thanks, everything looks wonderful now
Is this the right chat for a question about Fourier series?
 
sure, though no guarantee anyone will respond
 
Well, it's a basic question that I seemingly can't wrap my head around. Given the Fourier series:
 
@Semiclassical Something supersymmetric
 
@danu well, sure
 
$F(x)=sin(x)+\sum_{n=1}^\infty \frac{1}{5^n} cos(nx)$
How do I determine the Fourier coefficients b1 and b2 from this=
?
 
6:10 PM
well, how are those defined?
the b_n, i mean
 
I know the formulas for $a_n$ and $b_n$, but I'm not sure how to connect the dots
 
@Steve Use \sin and \cos ;)
 
well, i don't know said formulae off the top of my head
 
Oh ok
Give me a min
 
Just look it up on Wikipedia/any textbook
I don't really see what the point is in asking here :P
 
6:13 PM
@ramsay Oh, I see what you mean. It's 'cause $0$ isn't a limit point of $[1,10)$.
 
@Danu
I tried. I understand most of the basic parts of the subject, but this doesn't make sense in my head
 
The inequality isn't true anywhere near $x=0$ @ramsay
 
@Semiclassical $b_n=\frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin{nx} dx$
 
okay. what does that integrate to if $f(x)=\cos mx$?
(for m an integer)
 
@Semiclassical Today I've finalized an $\textbf{EPIC}$ series.
 
6:20 PM
neat
 
I get 0
 
@Semiclassical At least at the level of Ramanujan's series as difficulty. Not sure if I can propose it to some magazine because of the difficulty.
 
@steve right. so the cosine part of the series you quoted before has nothing to do with any of the b_n
that leaves the $\sin(x)$ term.
so all you need to do is compute $b_n$ for $\sin x$
 
@BalarkaSen So you had something to say about it?
 
@user1618033 what ingredients go into it?
 
6:26 PM
@Semiclassical polygamma function (particularly digamma) with different arguments
 
I'm out to jog a bit (like 30-60 min).
 
mmkay, talk to you later
 
@Semiclassical I've had a long day (had an exam earlier) and sorry if I sound thick, but how do I do that for $\sin x$?
 
6:28 PM
BBL
 
uh, compute $\frac{1}{\pi}\int_{-\pi}^\pi \sin x\,\sin nx\,dx$?
 
oh, just a sec
The solution according to my professor is $b_1=1$ and $b_2=0$, but I get $b_2=1$.
 
so you have $\frac{1}{\pi}\int_{-\pi}^\pi \sin x\,\sin 2x\,dx\neq 0$?
 
Ah, now I got it. Thanks. It was just a typo
 
6:35 PM
What CAS program do you use?
 
mathematica
 
7:07 PM
hi
do you know something about von neumann algebras?
i'm not sure if all non abelian von neumann algebras contain $M_2$
 
7:30 PM
@AkivaWeinberger how do you check whether $0$ is a limit point or not?
 
8:03 PM
@MikeMiller You ignoring me? Just tell me to stop bothering you..
 
@AndyMiles I'm not near a computer and find math difficult to read or think about on the phone.
 
@MikeMiller Oh I see (sorry for my impatience)
 
8:26 PM
@MikeMiller Sorry, I left. No, not particularly. Just letting you know that I got work done today (I also worked through some exercises from chapter 1 in no particular order).
I skipped the chapter on Morse theory. Should I read it?
@Huy: Stat is not that bad.
 
8:48 PM
@Balarka Yes.
 
@ramsay It needs to be infinitely close to the set $[1,10)$. The limit points of $[1,10)$ are the points in $[1,10]$.
If you take $\lim_{\Large x=1}\!$, or $\lim_{\Large x=10}\!$, or anything in between, it should work.
 
wait! you mean for $0$ to be a limit point it should be very close to $[1,10)$, right?
 
hi
how do you prove that $\sup (-A) = -\inf A$ if $A$ is a subset of $\mathbb{R}$?
 
First show that -inf A is an upper bound of -A, and then compare it to an arbitrary upper bound.
I can see the second part better if I use contradiction; that might help.
 
@EricStucky Can you show me how you would do it?
 
8:59 PM
@aki A BIG THANKS! (at last i understood after long discussion with many guys)
 
No, but I'll talk with you about it
 
Let $x \in A$. Then if $y = \inf{A}$, we have $y < x$ and thus $-y > -x$.
 
@MikeMiller Done. There isn't much that's proved in that section.
 
mkay
 
Thus, $-y$ is an upper bound for $-A$.
 
9:02 PM
Think you're missing a minus sign
-y>-x, yeah
and that changes the conclusion a little
A little more than that :P
Hmm, actually what is your definition of inf, exactly?
 
@EricStucky It is the greatest lower bound
 
Yeah, I want you to do a little better than that
 
I am not sure how to
 
Is that literally how it was defined in class?
(do you have a textbook? the textbook is usually more helpful for technical things like this.)
 
that's just how I know of it
how do I show that $-\inf{A}$ is the least upper bound for $-A$?
 
9:08 PM
Well, how are you planning on doing that if you can't define $\inf A$?
You need to know what this symbol means.
 
I can define $\inf{A}$. It is the greatest lower bound of the set $A$ with respect to some universal set $U$.
 
Which universal set?
 
It depends upon what type of set we are working with
but for the problem i am doing it is $\mathbb{R}$
 
Okay, that's good.
Now, can you write what 'greatest lower bound' means?
 
Given a set $A$ of real numbers, let the set $B$ be the set of all real numbers that are each less than every element in $A$. The greatest element in $B$ is the $\inf{A}$.
 
9:14 PM
Okay we can work with that
Wait, no
$B$ does not generally have a greatest element, as you've defined it.
 
@Balarka Whence why you should read it. They prove the Morse lemma, and that morse functions are dense, right?
 
Take $A=[0,1]$, then what do you want $B$ to be?
 
@EricStucky Was your proof by contradiction supposing that $-\inf{A} > \sup{-A}$?
 
(It probably ends up that way, but that's not the language I was using.)
(Yeah I think you can argue like this but it feels very stilted to me; it's hard for me to fit the idea into that container.)
I've been having some connectivity problems, maybe this message got lost:
If $A=[0,1]$, then what are you saying $B$ should be?
 
$B = (-\infty,0)$
 
9:27 PM
Okay, so that is consistent with what you wrote, so good on that :)
But it's not good because $B$ doesn't have a greatest element.
 
@BalarkaSen The only extension of the embedding result you got that you should know and isn't in G&P is a function space result. If $\dim N \geq 2\dim M + 1$ (and $M$ is compact), then $\text{Emb}(M,N)$ is dense in $C^\infty(M,N)$. Same with immersions, and you can drop the RHS to $2 \dim M$.
 
u19: I need to get on an airplane, so I may have to leave suddenly. Let me say something about the second step: a very standard-sounding proof starts with "Suppose, for the sake of contradiction, that $-y'$ is an upper bound of $A$ with $-y'<y$. … "
 
@AndyMiles I took a look and I think I would really need to dig deep in the notation to understand what's going on (like, as in looking at the book). I did upvote your question earlier, so hopefully someone else will take a look at it.
 
10:32 PM
can I get constructive comments by you, click this link: math.stackexchange.com/questions/1795982/…
 
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