of course there's this 20 comments in the last x days, too.

yeah, that one too

hopefully no user keeps count :)

@tb That's a heap'o'comments :-)

@tb is it other than 3 days?

I only saw two types of those: either some users engage in a discussion with a cr...eative mind, or there's a discussion on math going on. I never saw an actual fight there (I'm sure they happen, but I guess that those get flagged as offensive or whatever anyway).

there are fights

most of what I've seen have been chat-worthy rather than fights.

However, as I've noted, my experience may be colored by the questions I've chosen to answer.

For a long time I didn't want to join chat (for many reasons), and that option to transfer a comments discussion to chat wasn't there from the beginning. It was added some time during the summer last year, if I remember correctly.

@tb I joined last summer, so I didn't realize it wasn't always there.

@tb I hate that thing. Nowadays at that point I usually expand my answer.

@BrianMScott There is not usually enough to move to chat. Both parties just have a small amount to say at each point.

I was catapulted to a chat room a few times without actually wanting to go there (I probably clicked on the link instead of "add new comment") and I left immediately. There still are some users who do not want to use chat, and I think that's everyone's own call and we should respect that. As long as the discussion remains civil, I see no problem with 50 comments, except that, as Brian says, it might at some point be better to expand on the answer (or write an answer, for that matter)

@robjohn And often (when I run into) the problem actually can be cleared up by giving a longer explanation in the answer.

In chat, it would be a comment or two and then it would end.

@BrianMScott Sometimes, the comments do indicate where an improvement to the answer can be made.

Some one please help if you have access to this paper or the same paper on a different site here

@robjohn Yes, and then I just fix the answer and thank and notify the commenter.

@RajeshD Can you access this?

@BrianMScott Indeed.

@robjohn : I have accessed it there but the pdf is not scanned properly, not in a useful condition there

@RajeshD what do you need from your PDF?

i want to read it

see any new formulas and results on Fourier series which are not known to me earlier and also see what they are doing in the paper

@robjohn : never mind if its not possible to download

@RajeshD I assume you don't have access to SpringerLink from your institution.

no not all areas

@RajeshD I can read the copy from NASA. What's wrong with it?

wait

@robjohn : can you see page 2 or 3 and subsequent pages from there of? strangely i can't its all jumbled up!

@RajeshD It looks fine to me.

Try downloading again?

@RajeshD That looks like a mistransmission.

@RajeshD Not to sound creepy, but why are you on Eitan Tadmor's page?

it is very annoying that an oriented atlas of a manifold is not contained in a unique maximal oriented atlas...

as opposed to what happens with atlases

Oh? What's a good example?

dunno

@robjohn : problem rectified....earlier i was viewing in google chrome...that was the problem when i saved the page as pdf and reopen it in adobe its fine

if it worked, the union of compactible oriented atlases would be an oriented atlas, and I don't think that's true

@RajeshD I always download and read.

hm, maybe it is

I'm a bit surprised.

How could it not be true? :-|

But I haven't thought about atlases in a long time.

The definition of compatible I have in mind is "union of compatible oriented atlases is oriented"...

That's what I would try, too. And what could possibly go wrong?

@tb we have a nice one from National Geographic.

@Alex Youcis : I happened to google for some papers i came across a pdf but then i trucated the url to see if there are any more papers on this "Spectral recovery and detection of edges in spectral data"

OMG : Did anyone see this

I am scared

@RajeshD see what?

Watz goin on.....some one please update me

/.me thinks

@robjohn : This How is this possible ?

....you posted a screenshot, guy.

And, Eitan Tadmor is from the school I attend.

Although, on second thoughts, fear my hacking powers.

Hahha hmm.....Honestly you really scared the hell out of me ...I thought my comp was infiltrated by you.............wasn't that a very subtle hint in the screen shot, you could have played around with me more thank god

@Alex

I don't know man, I could have made you post that screen shot.

OMG

off for lunch

@RajeshD have a good lunch

@ZhenLin Didn't you have a question related to the symplectic form on the cotangent bundle a while ago? It was somehow swallowed by the general soap opera happening at the same time. Would you mind asking it again?

@tb soap opera? here? I'm amazed!

I did, but reminding me of $d(p \, dq)$ was enough.

But there was some follow-up, no?

Anyway, I don't insist :)

Well, it's not coordinate-free...

You can easily make it coordinate-free.

Oh?

First you need to think about the canonical $1$-form on the cotangent bundle.

Which comes from the bundle projection $\pi: T^\ast M \to M$.

Which gives a map $d\pi :TT^\ast M \to TM$ by differentiating.

Let us write $d\pi (q, \varphi): T_{(q,\varphi)} T^\ast M \to T_{q}M$.

Yes, I think I have a vague idea of how to do it now.

And we want to define a $1$-form $\lambda$, and it suffices to specify it pointwise.

At the point $(q,\varphi)$ set $\lambda(q,\varphi) = \varphi \circ d\pi(q,\varphi): T_{q,\varphi}T^\ast M \to \mathbb{R}$.

Remembering that $\varphi: T_{q}M \to \mathbb{R}$.

This defines a $1$-form and the (canonical) symplectic form is the exterior derivative $d\lambda$.

That's it.

Thanks!

good morning, Zhen and Theo

and a good one to you, Ilya

good morning!

sleepy

I have to give an inspection for the exam again today, 3.5 hours are just lost

That's a very funny and productive thing, yes. Can I have half a point more here? No. But I need one point for passing. So? If I don't, then ... Yes, that's pretty bad, but I can't do anything about it. Why not? ad nauseam.

Was a summary of the THC ever posted? I looked on meta, but I don't see one.

@robjohn a few hours ago :)

hey guys

@tb Moshe got the binomial theorem thingy yesterday.

I saw, thanks.

yeah so he got how to calculate the coefficient, which things to multiply etc

it was just that he made some errors in the arithmetic before that was causing some problem

but other than that it turned out ok for him!

Yeah, that's what I saw in his room, too. He got something like $(3x)^3 = 6 x$

yeah that was not right

Is BV a subset of HBV? (BV bounded variation and HBV harmonic bounded variation)

I got it . The answer is yes. somewhat trivial

@robjohn How do you post images here?

@AméricoTavares to the left of the text box where you're typing, there is a button 'upload...'

there is an upload button next to the send button ------------------------->>

Hey @Ilya you beat me

Wooo......who is at the hottest place game : its 41C here!

@RajeshD game?

I did not know this all this while, did not come across in standard books but i am surprised to see this today, its so important! Any idea where i can find a proof : quick links are appreciated

@Ilya : not a lengthy game, just have to tell the temperaure at your city

@RajeshD quick links: [8] and [12]

@RajeshD 14C here

huh in that same paper?

as it is suggested there

@Ilya I'll google for them for sure, just wondering anyone had any idea already...anyway its a good observation

@RajeshD another good observation would be to tell what is $S^\prime_n(g;\theta)$

Heya

derivative of the nth partial sum of the Fourier series of the function $g$.

@RajeshD and $\theta$

@n3bu: Hey Ya

$\theta$ is the independent variable of the function $g$

Could anyone explain to me exactly what a line integral is? =)

It's an integral along a line. Nothing special.

You're missing a line element $\mathrm{d}s$.

Hello everybody!

First of all: a curve is a (at least) once-continuously-differentiable function $\gamma : [a, b] \to M$.

So you need to pick a parameterisation of $C$.

I am able to calculate it, but what it means I am having trouble with. Can it be be seen as a line, with some "density" beneath it given by 4x^3 ?

@Ilya Thanks!

Is there some easy way to test if the continuous bijective mapping between two compact spaces is open?

@Nimza: If we're talking Hausdorff spaces then I think that's already a homeomorphism.

@N3buchadnezzar Yes, precisely.

@ZhenLin wow! Why?

I don't really know, but it can be proven!

Just a test.

@N3buchadnezzar To evaluate an integral over a curve, you basically use substitution: $$\int_\gamma f \, \mathrm{d} s = \int_a^b f (\gamma (t)) \| \dot{\gamma} (t) \| \, \mathrm{d}t$$

Yeah, I got it. But I would like to understand what I am doing more than just, pluging into formulas.Although my book is not that great at explainging these concepts.

@Nimza: OK, the precise statement is this: a continuous bijection from a (quasi)compact topological space to a Hausdorff topological space is a homeomorphism.

@ZhenLin yeah, thanks :) I found a proof

@N3buchadnezzar Well, first of all, do you understand the trivial case $f = 1$ and why it calculates the arclength?

I am quite sure it comes from the pythagorean theorem

eg $\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$

Well, given that it's almost a definition I don't think you can really say it comes from some theorem or other.

The question is, do you think it's a reasonable definition?

Then you sum up an infinite amount of these, so yeah, I think it gives meaning.

Although $$ \int_a^b \| \dot{\gamma}(t) \| $$ is a tad hard to grasp, since I tend to view $ \| \dot{\gamma}(t) \| $ as the size of the accelereation, but I guess I have to drop that definition.

Right. So an integral over a curve is just the same thing, except with a density.

@N3buchadnezzar: $\dot{\gamma}$ is a velocity, not acceleration.

$\ddot{\gamma}$ is the acceleration.

@ZhenLin I meant velocity sorry about that =) btw what haveyou studied, or what branch of mathematics are you focused on ?

Category theory and topos theory.

... $(t_{k}, $u_{k}, v_{k})\in C$.

which is proper grammar: (1) 6 divide by 3 equals 2

(2) 6 divided by 3 equals 2

(2) is proper grammar.

I prefer "6 divided by 3 is equal to 2."

Is there a way to combine \mapsto with \xrightarrow{}?

maybe I'll try \stackrel

I think I saw something for that in symbols-a4.

You could use \mapsfromchar in stmaryrd.

I'm planning to use this on MSE :)

I think extpfeil might have that

Then you'll probably have to hack up something using \mid and \rlap...

mathtools has \xmapsto, convenient.

Meh. Anyway, can someone help me understand jmc's comments to me here?

Concerning modulo residues and elements of the profinite integers $\widehat{\Bbb Z}$.

An element of the profinite completion of $\mathbb{Z}$ is indeed determined by its residues.

@ZhenLin thanks.

@anon That is my personal choice, but do you know why? Is it because (1) has not subject of the sentence?

@MaoYiyi "divide" is a verb, so must be conjugated correctly.

@ZhenLin Yes, but it's not obvious why one must use a past participle here.

Sure it is. 6 doesn't divide 3, 3 divides 6!

Yes, and when 3 divides 6, it equals 2. Doesn't seem very obvious to me.

Actually, I don't think you'd say it's 3 that's doing the dividing. The subject of the verb is unspecified. To write it in full, you'd say "When one divides six by three, it becomes two".

Hey @anon

Which you can then re-word as "When six is divided by three, it equals two" without specifying the subject. Then you can remove the "is" and the "it".

If one wants to say it in full, one should replace "by" with another preposition.

"When one divides <cake> by three" doesn't sound right...

Sounds right to me. If I have cake and I want to share it with two friends, I would divide it by three.

hey Raj.

I was idling wondering to myself what are necessary and sufficient conditions on a set $A\subseteq \Bbb N$ of moduli for an $x\in\widehat{\Bbb Z}$ to be uniquely determined by its residues modulo the elements of $A$. Intuitively I think the image of $A$ under all $p$-adic valuations must be infinite.

I would divide it into three parts.

@anon: That's easy. $A$ has to be cofinal in $\{ 2, 3, \ldots \}$ considered as a poset ordered by divisibility.

I don't know what cofinal means. :P Does it mean every number has a multiple in $A$?

(In terms of the divisibility construction)

If $X$ is a poset and $A$ is a subset, then $A$ is cofinal in $X$ just if for every $x$ in $X$ there is $a$ in $A$ such that $x \le a$.

The notion of cofinality is probably the biggest stumbling block when trying to understand subnets

Cofinality isn't very hard. It's basically just denseness.

That makes sense. That's definitely sufficient, but I have a feeling it may not be necessary.

It is necessary. It has to do with the universal property of $\widehat{\mathbb Z}$ as an inverse limit.

Let $B$ be the set of all numbers not divisible by either 2 or 3. Then let $C,D$ respectively be the set of all powers of $2$ and all powers of $3$. Define $A=B\cup C\cup D$. Now the residue of $x$ modulo $2^b3^c n$, with $(n,6)=1$, can be determined by its residues modulo $2^b$ and $3^c$ and $n$ via CRT. However, the number $6$ has no multiple inside $A$.

Or does CRT not apply to $x\in\widehat{\Bbb Z}$? Hmm.

render

Hmmm... actually, yes, it probably does depend on the objects in the inverse system.

No, it should apply, because it only needs to apply to the cyclic groups in the inverse system. Hrrrmmm.

Sorry, the problem is that $\{ 2, 3, 4, \ldots \}$ is not an inverse system.

render

@ZhenLin how did you do that?

@MaoYiyi phpSyntaxTree

@ZhenLin is that the websites name or a php program?

Both.

Agh, forgot that we take the opposite order when we actually form the inverse system...

very confusing.

The presence of composite numbers is confusing. If we only look at prime powers then it's definitely the case that we need a cofinal set of prime powers.

But if we allow composite numbers then I guess we need a more sophisticated kind of "dense" subset.

If A contains only prime powers it needs to be cofinal - I can see that.

Hmmm... I feel like introducing a Grothendieck topology. :p

Basically, the Chinese remainder theorem is telling us that $a b$ doesn't need to be in the system if $a$ and $b$ separately are in the system.

So let's say that $a_1, \ldots, a_n$ $J$-cover $x$ when $x$ divides $a_1 \cdots a_n$.

render

We modify the cofinality condition to be "for all $x$ in $X$ there exist a $J$-cover of $x$ by elements of $A$".

Yes that looks like it's what we need.

does tex not work in chat? Just asking because I see the latex code.

@MaoYiyi: math.ucla.edu/~robjohn/math/mathjax.html

We use that to render latex inside the chatroom.

And if I'm not mistaken, that's equivalent to my intuitive guess that the image of $A$ under $\operatorname{ord}_p$ must be infinite for every prime $p$!

@anon I clicked on the site and it worked on the website. Hmm.

You "clicked on the site"? I think you clicked the blue text on the site. You need to drag that blue text to you bookmarks bar on your internet browser, and it will create a button you need to click to render latex on whatever tab you're on.

@anon Yes, looks like it.

@anon I tried that but it just opened a google search.

Another way of putting it: $A$ is "dense" if $\{ a_1 \cdots a_n : a_i \in A \}$ is cofinal.

@MaoYiyi what browser are you on?

@jm cHROME

@ZhenLin Nice.

I hate idiots

Self-hatred is never healthy.

@anon did you just call me an idiot?

@Ilya why they don't compete for your job.

:P

@anon is it 'yes'?

@MaoYiyi There should be a bar in there storing bookmarks...

@MaoYiyi some of them already has a job

@Ilya Sounds like grounds for a duel at twelve paces...

@Ilya As they say on 4chan, thatsthejoke.png. (But no, I do not think you are an idiot.)

then I don't see any connection between my message and your answer

@jm but twelve paces in which space?

@JM its working now!

"What kind of idiot do you think I am?" "The usual kind, of course..."

@Mao How dare you write 'JM' in small letters!...(I am just kidding!)

@RajeshD don't read with a microscope.

@Ilya non sequitur, good friend... ;)

@Mao : Its better to use one's head while typing!

@RajeshD I thought fingers work better?

using your head can give you a headache.

@MaoYiyi Right, and you also get unsightly marks on the forehead...

funny

@anon: I think this is the general statement for inverse systems: if $A$ is a subset of an inverse system $X$ with the property that for all $x$ in $X$ there is a subset $B$ of $A$ such that the maps $\{ x \to b : b \in B \}$ are jointly monic (injective), then $A$ is "dense" in $X$.

@JM 呵呵

@ZhenLin Jointly as in we consider it like an injective map from $x$ to $\prod_B b$? I was trying to figure out the general statement too.

Yeah.

Yes, that's the correct "density" statement.

88, I gota sleep.

@ZhenLin: No, wait. Perhaps it should be that there exists a $y$ higher up on the poset than $x$ such that $y\to\prod_B$ is injective, not necessarily $x$ itself.

I'm going to get confused if we talk about inverse systems as posets, so let's stick to arrows.

If $y$ is "higher" than $x$, then that means there's an arrow $y \to x$... but one strange property of the inverse system used to construct $\widehat{\mathbb{Z}}$ is that all the arrows appearing in it are epic (surjective).

I think you're right though, since if $y \to x$ isn't surjective, then the part that isn't in the image doesn't make it into the inverse limit anyway.

In the case of $\widehat{\Bbb Z}$, our moduli set $A$ need not generate every integer under products. Rather, for every integer $x$, there is a multiple of $x$ that can be expressed as a product of elements of $A$. Take my previous example and modify the sets of powers of $2$ and $3$ to instead be sets of powers of $2$ and $3$ greater than $100$.

Well, $\mathbb{Z}/a \times \mathbb{Z}/b$ may not be isomorphic to $\mathbb{Z}/(ab)$, as you know...

Oh, right.

but it is guaranteed to contain enough information to reconstruct the residue modulo $\textrm{lcm}(a, b)$.

@Mao : "Its better to use one's head while typing!", "Its better to use one's head for typing!"......there is a subtle difference between these two sentences.

I usually type with my fingers...

@N3bu : did you understand the difference between the two sentences ?

@RajeshD Yeah... Immature though :p

sometimes fingers are not enough

The first sentence does not tell you to use your head instead of fingers for typing

@anon Hm, that means my original definition of $J$-covering was flawed. But I think you can figure out how to fix it.

@anon : that is funny but off topic though

@RajeshD There's a topic?

@N3bu : What do you mean by 'immature' ?

@anon Fingers and lube are usually enough for me though, I rarely have to use my head. Of course I am talking about when I am doing this.

@JM no there isn't

@N3bu : Its getting worse...I guess i stop it here for heaven's sake!

My de-stressing toy.

@ZhenLin destressing? Looks more like a stressing toy! 8-).

yes

Admittedly, yes, it is stressful when I can't remember the necessary algorithms.

Distressing.

just by the sight of it!

I liked the Tower of Hanoi. Mindlessly following the solution method is somewhat de-stressing.

@anon : mathematical induction ! whooo

If I didn't want stress from the Tower of Hanoi, I'd limit myself to two discs...

@anon: If I wanted mindless I'd do the ordinary Rubik's cube. Although lack of practice means I screw up the PLL step more often than I'd like...

Yeah, when you know all the algorithms it is really detressfull

Then you can as well have a nice beer...

@JonasTeuwen Now that is de-stressing...

I find a large amount of beer stressfull, there are many beverages that are far more pleasant to consume.

Fire water!

Hmm, I wonder how long before this page is entirely composed of 20k+ people...

@JonasTeuwen what is the difference between a beer and a nice beer ?

@RajeshD A nice beer is actually nice.

well said

nice bear

Of course, we have to define "nice"...

I define which beer is nice or not.

(8-))

@JonasTeuwen 'This requires tests!

Hi.

@Jonas: hi. Your definitions worked for me

Man, Greens theorem confuses me. Can we or can we not use greens theorem over a region that has holes in it. Or does it deppend on what kind of holes we have?

@N3buchadnezzar holes? you should use the integral over the whole boundary - so if holes have boundaries that's ok

@N3buchadnezzar: if I'm not mistaken, that's how it works

@Ilya Yeah, but It seems as we can just ignore the hole.

@N3buchadnezzar what do you mean?

Reading about it here, having some problems understanding what I am reading.

Oh I see, greens theorem works even though we have holes in the surface. All we need is C to be a positively oriented, piecewise smooth, simple, closed curve and have continuous second derivatives.

@Ilya :-).

I'll take a walk

@AméricoTavares sorry for taking so long (I took a nap). you just give a link to the image in a comment by itself.

@robjohn: the guy mentioned that he is fighting with TA over this problem

should I perhaps reduce the full answer to hints only?

@Ilya what was their fight about?

@robjohn I have no clue :)

They weren't disputing the result, surely

so you think my answer is ok in the sense it is full? it's not a homework, so I thought there is no need to play in hints

@Ilya did you mean $X(\omega)\in A$ instead of $X(\omega)>A$?

:-) I see it's fixed

@robjohn yes, thanks - I've fixed it

@Ilya Yes, your answer looks to be complete.

@robjohn thank you, robjohn.

@Ilya He's not going to like your answer, though :-)

@robjohn too many formulas?

@BenjaminLim I mean that you are a very dedicated student.

Or that you are very devoted to the study of Atiyah Macdonald.

@FortuonPaendrag no not really soon I will be switching to miles reid :D

Hahahaha! Come over to the dark side

@Ben Do you mean reid's commutative algebra or algebraic geometry?

@RajeshD Hello Rajesh! How go your mathematical activities?

@FortuonPaendrag I meant the commutative algebra book. I would like to see some noetherian rings.

Yes,yes. Of course you would.

@Fortuon : Got some peculiar and new ideas which at best seem just plausible...just goin with them whenever time is there...

@FortuonPaendrag man I got to say atiyah macdonald has been a baptism of fire for me

@Ben Well, from now life will be easy-peasy for you :)

hahahahaha

What about you?

@FortuonPaendrag But I learned a lot from atiyah macdonald

@Fortuon : IIRC you are from Hyderabad right?

@RajeshD Plodding through Rudin's Real and Complex Analysis at a Snail's pace. Avunu :)

@Ben Yes, my point precisely. You seem to have spent hours of painstaking study on it.

man its really hot in here

And its not even Rohini Karte yet!

@FortuonPaendrag What is it? name of a month?

Its a legend. On the day of Rohini Karte, usually the 21st of may , the sun is most furious.

Oh ! got to take care not to get out on that day

Haha, yes. Im coming to Hyderabad in the first week of June. Hopefully it is cooler by then.

@FortuonPaendrag Well that's the way no?

only if it rains

pray to rain god

@BenjaminLim Yes, indeed. :) One has to do similar things to Rudin's Real and Complex Analysis to learn anything. His proofs can be a bit obscure at times!

@FortuonPaendrag that's what you are using now?

@Ben Yes. My class deals with some parts of it and I self study some other parts of it.

Stuck : (