Mathematics - 2012-05-09 (page 1 of 5)

Mariano Suárez-Alvarez

12:00 AM

that is just "calculus"

/me wishes xypic \ar's defaulted to ^- and _- to place labels :/

Peter Tamaroff

1:23 AM

@ClarkKent This is ghost town!

Rajesh D

1:38 AM

Looks like the election season is on

guys...Looks like people are asleep!

I wonder canvasing for the mod election allowed in here?

user19161

@PeterTamaroff Well, this is not a 24-hour come-here-if-you-are-lonely chat you know!

Rajesh D

Hey @Clark

user19161

@RajeshD Hi!

Rajesh D

you seem to be new in here, Welcome!

user19161

@RajeshD I am not. I changed my username. I know you!

Dylan Moreland

1:48 AM

Well, that's hardly fair.

user19161

@RajeshD I cast my votes already.

user19161

@PeterTamaroff Maybe it is Nightmare on Elm Street. Those movies are all the same. It's always Freddy, Freddy and Freddy.

Peter Tamaroff

@ClarkKent What are you saying now? I really didn't like your last remark.

user19161

@PeterTamaroff Which part? I am just saying all the Freddy movies are similar.

Peter Tamaroff

@ClarkKent That

user19161

2:01 AM

@PeterTamaroff Oh dear. Are you upset?

Peter Tamaroff

@ClarkKent Not really.

Rajesh D

@Clark : I think I know who you are, it was trivial indeed

user19161

@PeterTamaroff OK good.

user19161

@RajeshD OK good.

Peter Tamaroff

@ClarkKent So you realize the comment wasn't really friendly right?

user19161

2:03 AM

@PeterTamaroff No, I don't. I am just stating my opinion of some movies.

Peter Tamaroff

@ClarkKent I mean this comment.

Rajesh D

@Clark : Sorry I didn't, I thnk i am wrong, but you had something like what in a name kind of a thing

user19161

@PeterTamaroff No, I don't think it is unfriendly. Maybe you misread it.

Peter Tamaroff

@ClarkKent Most probably, then. Could you explain it?

user19161

@PeterTamaroff It just means what I said. Take it with a sense of humour!

Peter Tamaroff

2:07 AM

@ClarkKent Ha well! I couldn't at that time! =) I was having a hard time doing a mock midterm and something wasn't working out. My bad.

Rajesh D

@Peter : Could you please explain it to me

Peter Tamaroff

@RajeshD Explain what?

Rajesh D

about the comment from @Clark . Whats the joke in it ?

user19161

@RajeshD There's nothing to explain anymore. It is nothing really!

Rajesh D

ok thanks

user19161

2:10 AM

Oh dear, I think we are all on different wavelengths...

Rajesh D

@Clark ; I have put a bait to know your identity, see my previous comment !

user19161

@RajeshD I think I get it now. Well, I will fall for the trick then by telling you that there should not be a space before the question mark or the exclamation mark.

Rajesh D

hahhaha gotcha!

Later...(gone for breakfast)

Peter Tamaroff

@RajeshD Breakfast? What time is it there?

user19161

@PeterTamaroff Well, good luck for your actual midterm then!

Peter Tamaroff

2:18 AM

@ClarkKent I think I'll do well. How are you involved in mathematics?

Rajesh D

@PeterTamaroff 7:45 AM

Hi @JM

Good morning

J. M.

@RajeshD Oh, hi!

Rajesh D

2:36 AM

not in contention this time ?

@JM The gravatar is cool...(not due to any math reasons, i don't know topology) but its kind of aesthetic

J. M.

@RajeshD What contention?

Rajesh D

mod elections!

J. M.

@RajeshD Aesthetics was kind of the point... :)

@RajeshD I was never interested in modding this place. :)

Rajesh D

ok

I am pleasantly surprised to see robjohn contesting.

Zeeshan Mahmud

3:00 AM

Hello

Can I ask a question?

David K.

Yes

There's no guarantee that you'll get an answer though :p

Zeeshan Mahmud

Any questions similar to transcendental numbers raised to itself infinite times such as pi 'staircase': pi^pi^pi....?

J. M.

@ZeeshanMahmud Yes, that exists.

David K.

You can ask any question you like. But whether or not you get an answer depends on who is here, and what we collectively know.

Zeeshan Mahmud

@JM Mind sharing the link? :)

Mariano Suárez-Alvarez

3:04 AM

it is impossible to know if we can answer your question before you actually ask it...

J. M.

@ZeeshanMahmud Have you seen this?

mixedmath

good evening

Zeeshan Mahmud

@J.M. I remember seeing it, for some reason I didn't bookmark it. But it would be a start.

J. M.

Hi mix!

Zeeshan Mahmud

Anyone knows what happened to Srivatsan?

Dylan Moreland

3:16 AM

He said he wouldn't be around as much. There was an announcement.

J. M.

@ZeeshanMahmud He's on vacation IIRC.

Dylan Moreland

And it turned out to be pretty accurate.

Zeeshan Mahmud

Sorry to hear that. Per Erdos he died, then.

J. M.

@ZeeshanMahmud Note that $\pi$ is outside the range $[e^{-e},e^\frac1{e}]$, so your power tower doesn't converge.

Zeeshan Mahmud

@J.M. Tex not appearing ok, no I need to install anything?

Ok I pasted it in Lyx.

J. M.

3:25 AM

@ZeeshanMahmud Well, you'll need this.

Zeeshan Mahmud

By the way, as an aside my question was inspired after Vi Hart's youtube video. :)

@J.M. Thanks.

ok leaving.. Thanks for the help everyone

Eric Gregor

would someone explain to me why $\hat{f}(0)=0$ here: math.ucdavis.edu/~jlirion/course_notes/Prelim_Solutions.pdf (ctrl+f for "Riemann")

t.b.

4:00 AM

good! not too much drama in the past few hours. people seem to have found more worthwhile ways of wasting their time...

Rajesh D

Hi @tb

t.b.

hi rajesh

how do you do?

Rajesh D

fine what about you

t.b.

tired to the bones, honestly.

Rajesh D

then a good sleep is the best remedy

Eric Gregor

4:04 AM

hi @tb, would you be able to help me with that question i asked? it would probably take you literally a glance, i am just missing something obvious

i don't want to bother you if you're tired, though

t.b.

@RajeshD yeah, that's what I thought 5 hours ago, but after three hours of sleep Mr Sandman left the building, and after 2 hours of lying awake, I've had enough.

Eric Gregor

my fourier analysis question will put you to sleep :)

t.b.

@EricGregor I glanced at it, but there's too much notation I don't understand, so I quickly gave up. I'll look again in a few minutes.

Eric Gregor

i'll translate. there is some weird notation

t.b.

By the way: there's this story that Littlewood supposedly described the proof of Riesz-Thorin as "the most impudent idea in analysis" :)

Eric Gregor

4:08 AM

lol, what?

t.b.

No joke, it appears in Bennett-Sharpley, but they don't give a source, and I wasn't able to locate it. --- "It" being the original quote of Littlewood

Eric Gregor

$H$ is an operator on the space of Schwarz functions to $L^2$, $Hf=i\text{ sgn}(\xi)\hat{f}(\xi)$. the goal is to show that for $f$ in the Schwarz space and $Hf\in L^1(\mathbb{R})$, that $\int_\mathbb{R}f(x)\ dx=0$

what a strange comment!

J. M.

I wonder what prompted the harsh adjectives... (and good morning!)

t.b.

Hi, J.M.

Eric Gregor

$P(\mathbb{R})$ in the file denotes the space of Schwarz functions

t.b.

4:11 AM

Ah, clicking the mathjax thingie, might help.

Eric Gregor

he claims that by Riemann-Lebesgue one can see that $\hat{f}(0)=0$ in order to prove the claim

t.b.

@EricGregor why? because this symbol $P(\mathbb{R})$ is so conveniently used in so many completely unrelated ways by so many mathematicians and it has so manifestly nothing to do with Laurent Schwartz?

Okay, lemme look at that question, now.

Eric Gregor

i agree it's stupid notation. Folland uses $\mathcal{S}$

t.b.

which is what everyone else uses.

But it is always a good idea to be creative, especially when it comes to not using completely standard notation. Lest one could understand the argument too easily.

</rant>

Eric Gregor

i was just scanning old qual solutions to prepare for an exam and stumbled on this, and i didn't understand it...i get the use of R-L to show that $\hat{f}(\xi)$ tends to 0 as $\xi$ tends to $\infty$

like using $\Sigma$ as a variable

$f(\Sigma^2)=\Omega^2 \Sigma$

J. M.

4:16 AM

@tb Tsk, tsk... :) you've had your bellyful of crap notation, I take it.

t.b.

Eric, sorry, what's $C$? Does that perchance denote the class of continuous functions?

Eric Gregor

yes, i think so

that is the R-L theorem

but yeah, again strange notation

i hope this is making you sleepy, tb

t.b.

I agree, but they might have chosen to call continuous functions, say, $Y(\mathbb{Q})$

Eric Gregor

lol

t.b.

No, I'm just not focusing. Now trying for real

Eric Gregor

4:21 AM

maybe this will backfire. maybe it will make you sleepy and have nightmares about notation

Rajesh D

@Eric : Pondering over something and find the notation isn't clear or evident! Argh...

Eric Gregor

this is how experts collude to prevent further entry into crowded fields

Sb Sangpi

which site is the best to upload maths graph to reference in mathemactics question?

Eric Gregor

@Sb a gif?

Sb Sangpi

yea, gif,png,jpg..etc

t.b.

4:24 AM

@Eric: Isn't this simply the fact that $\widehat{Hf}$ is skew-symmetric and continuous in zero?

or whatever this property is called when an i is involved?

Eric Gregor

@Sb imgur?

Rajesh D

I find [this title] of this paper strange! Could someone tell me whats it all about in few lines?

t.b.

@SbSangpi: for asking a question here?

Sb Sangpi

no..i mean, I want to reference the graph in asking question from maths Q & A site.

Rajesh D

@Sb Sangpi : There is an upload facility on the question entry page of math.se

J. M.

4:26 AM

@SbSangpi There is a button for uploading pictures in the panel where you type your questions...

Rajesh D

@JM : I beat you

J. M.

@RajeshD That's fine. :)

Sb Sangpi

awaw,...yeayea.. my bad.! thx :D

Eric Gregor

@tb you mean that $Hf(0)=0$ because it's skew symmetric? is this obvious?

Rajesh D

I find this title of this paper strange! Could someone tell me whats it all about in few lines?

Eric Gregor

4:31 AM

@RajeshD what do you find strange about it?

J. M.

@RajeshD What doesn't sound straightforward from the abstract?

t.b.

@EricGregor Well, you know that $\widehat{Hf}$ is 1) skew-symmetric by definition 2) continuous because it's by definition a Fourier transform of something

Eric Gregor

i get 2), not 1), @tb

oh

i'm a fool

i see it now. thanks @tb

Rajesh D

What is meant by singularities of Fourier and Taylor series. Does he mean the singularities of the function that they are converging to ? If so the taylor series does not exist for functions with singularities no? then i am confused what he means?

t.b.

@EricGregor No, sorry, I messed up a little bit, but the idea is simple

If you have a function: $i\hat{f}$, which is continuous, then you change the sign of it on the negative real axis and leave the sign on the positive real axis alone and still end up with a continuous function $\widehat{Hf}$, the only way this can happen is that $\widehat{Hf}(0) = 0$, and hence $\hat{f}(0) = 0$.

J. M.

4:35 AM

@RajeshD What about $\log(1+x)$?

Rajesh D

and " The method can also be applied to Fourier series of analytic periodic functions " Does analytic periodic functions have singularities?

t.b.

there's no way of being smoother than analytic

Rajesh D

at x = -1 ?

Eric Gregor

thanks @tb, i appreciate it. i will now sleep in your stead. goodnight!

t.b.

good night, Eric

Eric Gregor

4:37 AM

try valerian root and less light before bed :)

J. M.

Valerian is a bit smelly, no?

t.b.

Luckily I don't have a cat

So this would be an option.

Valerian, not a cat.

J. M.

I was going to suggest warm milk, but I think you already know that...

Rajesh D

@JM I didn't know that functions with singularities can have a taylor series converging to them. Any way I'll get back to it after some reading. But what on earth is this " The method can also be applied to Fourier series of analytic periodic functions "?

Agh I hate milk but i like coffee

J. M.

@RajeshD Unfortunately I can't access the paper, so I don't know why they'd want to Borel-transform a Fourier series...

t.b.

4:40 AM

@JM Warm milk with honey and nutmeg, yeah, makes sleepy but doesn't help sleeping.

Anyway, thanks for the suggestions, but I have to get up in half an hour anyway, so I could as well stay up.

The town hall chat digest is being posted right now.

Rajesh D

@JM : tell me something, what is the thing about taylor series for functions with singularities. How can they exist?

J. M.

@RajeshD The singularities would be at the boundary of the disk of convergence, of course...

Rajesh D

@tb : by any chance I was wondering you have access to this

t.b.

what happens if a bounty question is closed as NARQ?

Rajesh D

@JM ok

J. M.

4:44 AM

@tb I think the bounty is still awardable. All closing does is to prevent new entries...

Rajesh D

never mind about my last comment

t.b.

@JM I'm sorely tempted to "take care" of one, but it'd just be mean.

leo

hi there

mixedmath

hiya

t.b.

hey leo, and mixedmath

J. M.

4:50 AM

Hello and hi. (Pick which one you want.)

mixedmath

hmm... hi, I think

It's stunning to think that MSE has over 50k questions

leo

Do you think that my comment to kuku in here is understable?

I mean, is it some broken english?

J. M.

@mixedmath ...and they just keep coming. :o

Brian M. Scott

@mixedmath And just over 86% of them have at least one upvoted answer.

mixedmath

which correspondingly means that thousands have no upvoted answer - another crazy thought

t.b.

4:54 AM

@leo It is/should be understandable I believe.

mixedmath

@leo: I follow it just fine too

Brian M. Scott

@mixedmath A little over 7000. But I'm pretty sure that a lot of those were actually settled in the comments.

leo

Thanks :-)

t.b.

you might follow up on the hint, though.

Brian M. Scott

@leo No, it's fine.

leo

4:56 AM

It seems that kuku don't follow it

t.b.

that's why I said that you should explain yourself once more :)

leo

but my the english is not the reason

ok

t.b.

no it's not the English at all, it's completely clear what you're saying

leo

5:10 AM

@tb, just for a quick check, this is my explanation:
you have replied to tb that the (exactly) reason because $g=h$ (almost everywhere) is $\int f(g-h)=0$, so it seems that from $\int f(g-h)=0$ your are deducing that $(g-h)=0$. What I'm trying to say you is that this not always hold. I mean, it is **not** true in general that $$\int f(g-h)=0\implies g-h=0.$$ However the above implication holds for some particular $f$. For instance, if you can find $f\in L^2[0,1]$ such that $f(g-h)=|g-h|$ then the above implication holds.

t.b.

looking.

leo

:-)

t.b.

@leo The comment is fine, but...

I wouldn't insist on explaining your previous comment too much, but let me look at the exchange in the comments again.

@leo Okay, here's what I'm trying to say. I think you can simply say something along the lines

Mariano Suárez-Alvarez

The obvious answer to math.stackexchange.com/questions/129254/… is simply «Concrete Mathematics»...

t.b.

@leo Yes, it is true that $\int |g-h| = 0$ implies that $|g-h| = 0$ almost everywhere (what happens if it's not?).

J. M.

5:20 AM

@MarianoSuárezAlvarez I have a hard time coming up with another example that has the requisite irreverent graffiti in the margins...

t.b.

@leo "The point is that the statement "because $\int f (g-h) = 0$" is not precise enough: do you mean for some $f$ or for all $f$?"

Sorry that I'm being so slow :)

This is exactly the kind of question that one should just leave alone, in my opinion. What a lazy OP... :/

mixedmath

yes t.b., I completely agree

I've been thinking more about that issue, i.e. about a homework policy of some sort

leo

@tb My point is that it seems that the OP believes in $$\forall f\in L^2[0,1](\int f(g-h)=0\implies g=h)\tag{1}$$

Brian M. Scott

@tb Lazy and unreasonable, but at least it's clearly not homework!

leo

That's no true

t.b.

5:28 AM

@BrianMScott well, if there's no work involved how could it be homework? :)

Brian M. Scott

Ha. You do have a point.

leo

:-) yes I have

SauravTomar

a simiple question is $\int sin(1\t) = -cos(1/t)*-t^2$

ah got the mistake

t.b.

@leo well, it would be true if you knew that $(g-h)$ is locally integrable. That's the fundamental lemma of calculus of variations. But proving that $g-h$ is locally integrable is as hard as to prove that $\int |g-h| = 0$ which directly gives the desired conclusion.

Or are you making an a.e. versus everywhere point?

@mixedmath do you care to elaborate?

Unfortunately the question already got an answer...

mixedmath

and a reasonable answer too...

when I was considering homework questions at first, all I was really hoping for was that if someone wrote a hint-style answer, that in general (but not necessarily always) others try not to undermine it

but in thinking more about it, I don't even know if even that's actually possible

which, in a sense, is very sad.

leo

5:37 AM

@tb Yes, but now I have realized that my early comment predicate something about $g$ and $h$ (there exist $g$, $h$, $f$ such that...) So in (1) $g$ and $h$ must be not free in order to say that (1) is not true.

it must be:
$$\forall f,g,h\in L^2[0,1](\int f(g-h)=0\implies g=h)\tag{1}$$
is not true

mixedmath

It reminds me of the question of where we put the blame. It's not our directive to actively avoid helping others with homework. It is instead one's own prerogative to maintain one's own honor code, or rules, or what not. At least, that's one way of rationalization. But at the end of the day, I still won't give complete solutions, others likely will, and I think that will be that.

J. M.

"I think that will be that." - pretty much, yes.

t.b.

@leo well, that doesn't make much sense, does it? It's actually a true statement because the premise is false...

mixedmath

@JM Really, this comes as no surprise though, right? A year ago or so, Qiaochu really took to the idea of a homework policy, and he's very dedicated, has rep to flaunt, and is a mod. When he couldn't, it was a big sign.

t.b.

@mixedmath I'm really torn. I like to write (what I consider) good answers and I'm actually restraining myself not to give the easy one-shot answer that many others give. I think it's utterly unhelpful to insist that some premise is missing and not elaborating on how to massage the statement into a correct one.

J. M.

5:44 AM

@mixedmath Not at all. People view homework differently, period. JDH's words (which Bill quoted in meta) come to mind...

mixedmath

starting next year, I will be teaching classes to students. Yet only in the last year have I learned about the incredible ease of getting answers to most basic questions online, quickly. One thing I'm going to have to start thinking about is how to accurately judge my own students skills.

Gigili

Morning.

t.b.

Just an example of what I have in mind: if someone asks a somewhat elaborate question "why is it true that every division ring is a field?" and fails to mention finiteness in the statement but clearly implies it in the rest of the question, it's simply not a helpful thing to do to write "what about $\mathbb{Z}$?" into the answer box, and I cannot understand how such answers often garner more than a few upvotes.

J. M.

I distinctly recall one of the members grumbling that now he won't be able to use in his classes an example that got asked, and was given a relatively generous answer, on math.SE...

@Gigili Good morning...

t.b.

'ello Gigili!

Mariano Suárez-Alvarez

5:46 AM

@tb, that makes a sensible comment...

t.b.

Yup, that's basically what I'm saying...

Mariano Suárez-Alvarez

it can be a useful comment, too

forcing the OP to realise he is missing something

that might be all that is needed for him to answer himself

generally, tho, it only results in the hypotesis being added to the question

but if you enjoy doing things yourself (and are a bit generous) then you will probably want the OP to maybe have that little joy

leo

@tb I'm confused. What it is true is that in the scenario of the question we have that for all $f\in L^2[0,1]$ $\int f(g-h)=0$. What I'm trying to point is that if we take $f=\operatorname{sign}(g-h)$ that says that $\int |g-h|=0$ and so $g=h$ a.e., as we wanted.

t.b.

@leo I agree completely :)

@MarianoSuárezAlvarez yes, I think we agree completely there. :)

It is more than a bit unfortunate in my view that many people think it's more important to follow some SE-policies like, no discussion in the comments, and it is important to have as many questions as possible answered, etc than to indulge oneself in the fun of doing math and let others have a bit of that fun, too.

6

Gigili

Tee-Bee completely agrees with everyone, today.

t.b.

5:56 AM

I agree completely with you, too, Gigili :)

Isaac Solomon

Hey guys. Whats a good way of thinking about F_8, the finite field with 8 elements?

Like, a concrete way of thinking about its arithmetic.

Brian M. Scott

@tb Because Aha, gotcha frequently overrides common sense.

t.b.

Probably.

leo

6:16 AM

@tb now this is my comment:
_You have that for all $f\in L^2[0,1]$, $\int f(g-h)=0$. It is true that $\int |g-h|=0$ implies $g=h$ almost everywhere. So, it is enough to find $f\in L^2[0,1]$ so that $f(g-h)=|g-h|$ in order to conclude that $g=h$ almost everywhere._

t.b.

@leo very good!

Brian M. Scott

@leo That's certainly very clear.

leo

@tb thanks for review my comments :-)

t.b.

@leo By the way, I see no mistakes in your English in both comments, don't worry about the mistakes too much.

leo

that's a relief

I have never taken an english course other than the given on high school

t.b.

6:25 AM

I distinctly remember that discussion we once had after one of your questions (was it the Haar basis thing?), and you were really unsure of your English skills. Don't be. Just try to do your best. The person you talk to will most likely be willing to listen carefully and will ask for clarifications.

David Wallace

@leo The best way to improve your English is practice. Posting questions and/or answers on StackExchange is a good way to do that.

And don't worry about making mistakes - there are people here who will edit your question or answer for you if it has English errors in it.

Brian M. Scott

And considering some of the English that we get, you have nothing to worry about; you may occasionally sound foreign, but you're very rarely at all hard to understand.

leo

however, my spoken english must be pretty bad. Perhaps like this

t.b.

@leo oh, a classic :)

leo

:-)

t.b.

6:29 AM

but I only knew it in written form

leo

To improve my english is one of the reasons to join to this site

and because I like math

Brian M. Scott

@leo In that case I'll mention that the idiomatic expression is simply reasons to join this site.

leo

@BrianMScott I see, thanks :-)

t.b.

@leo You are not alone. Practicing, improving and feeling more comfortable with math exposition is one of several main motivations for me to participate.

leo

Other reasons are:

answer a question is a good excuse to TeX an exercise that you have solved.

and finally I think this way is the best to talk about math. Have you tried to talk about math with other person by phone? for example. There is a lot of ambiguity

t.b.

6:44 AM

Talking about math on the phone is in fact a matter of practice and it seems to me that it helps if you know each other quite well (but of course a video call with chat is easier). But I had many very long and fruitful discussions with friends about math before the internet became an usable resource in this respect; also pretty technical stuff, that is to say: actual computations.

Mariano Suárez-Alvarez

@mixedmath, most of your flags regarding oldish questions would have benefitted from becoming comments to the questions :)
I ended up writing those! :P

leo

I say that because of things like: "consider x over x plus 1 plus x over 1 plus n..."

mixedmath

@Mariano: Ok - thanks

leo

you can interpret many things

Mariano Suárez-Alvarez

flags commenting on closure rarely end up in our closing the question

because when we get the flag it rarely has more than one or twovotes

t.b.

6:49 AM

@leo nobody prohibits the use of parentheses and more words :)

Mariano Suárez-Alvarez

(and if we start closing unilaterally then chaos ensues!!!)

apart from obvious things (spam and so on) posts ought to be closed by the actual users, no?

mixedmath

this is true

did you keep the one with 11 questions in the question?

that one seemed interesting to me

Mariano Suárez-Alvarez

likewise, flags on low quality in the immense majority of cases should be just downvotes

I personally see many, many posts which I deem to be of low quality

leo

@tb you are right, but sometimes it happens to me with some class mates. Things like that just pop up and then we have to find some board or something to write

Mariano Suárez-Alvarez

and everyone would be shouting at me if I deleted those (and very correctly!)

yet flags seems to indicate that people want me to delete them for them? :D

If I am going to be shouted at, at least it should be because of my own judgement :)

t.b.

6:55 AM

I'm actually a bit confused by the flags I see in the mod tools. I never know whether they are mostly auto-generated or other user's flags.

mixedmath

at Phil.SE, the mods often practice a sort of flag-passing - so that all the mods might voice their opinion on some flags sometimes

does that happen at all here?

Mariano Suárez-Alvarez

not really