Mathematics - 2012-08-21 (page 1 of 3)

Peter Tamaroff

00:10

@JonasTeuwen Edison is disappoint.

Jonas Teuwen

Edison can kiss my shiny metal ass.

Peter Tamaroff

@JonasTeuwen So now you're a robot?

Edison uses thunderstorm - it is very effective

Jonas Teuwen

Pretty much.

mixedmath

00:39

y u downvote come at me bro — kxx 3 hours ago

sharing the wealth

does this make you feel his pain, or want to downvote? hmm...

Henry T. Horton

lol all his answers are "WolframAlpha says..."

Peter Tamaroff

@mixedmath Let the downvote begin?

@mixedmath I flagged some stuff

Henry T. Horton

American flags I hope.

mixedmath

perhaps that's what all his schoolwork looks like. W|A syas such and such

it's likely true enough of my calculus students

Peter Tamaroff

@mixedmath You need to bring down the hammer, stranger professor.

mixedmath

00:50

ah, professor is a reserved word here. As a grad student teaching undergrads, I'm called a "teaching fellow" - which means that neither I nor my students really know how they should refer to me

so they call me either David or professor, anyway

but to cope with homework, the powers that be (I don't make this decision) make homework nearly meaningless to the grade for calc I and II

for better or for worse

Peter Tamaroff

@mixedmath hmmm, that's a good way out... in the end HW serves them, not you!

@mixedmath I'm about to try and state a theorem on step functions. They are so messy to work with XD!

mixedmath

on step functions? Are you learning how to integrate or something?

lebesgue or something?

Jonas Teuwen

The Apostol way to Riemann integration I suppose.

Peter Tamaroff

@mixedmath Its Apostol's approach to integration. I'm not learning per se, but re learning, settlings things in

Jonas Teuwen

Sigh, have to get up soon. Good night guys!

Peter Tamaroff

01:02

@JonasTeuwen Night. I'm starting uni tomorrow, so 6.a.m it will be!

Jonas Teuwen

Good luck!

For me also 6AM, but is already 3AM.

Babai 8-).

Peter Tamaroff

@JonasTeuwen REDBULL AND COFFEE?!

Night

skullpatrol

later

Argon

01:47

@PeterTamaroff I want to sit in at some lectures at a university. How do you suggest I choose a class to attend?

Peter Tamaroff

@Argon Hold on. How old are you, again?

Argon

@PeterTamaroff 16

Peter Tamaroff

@Argon OK. And what are you studying in maths highschool now?

Argon

@PeterTamaroff This year I am taking a course on functions and advanced functions.

Peter Tamaroff

@Argon You have "courses" in highschool?

Argon

01:51

@PeterTamaroff A class.

Peter Tamaroff

@Argon And how are you doing?

Argon

Two classes, "functions" and "advanced functions."

@PeterTamaroff In what sense?

Peter Tamaroff

@Argon Academically.

Argon

@PeterTamaroff Quite well. Above average.

Peter Tamaroff

@Argon What would that be?

Argon

01:52

@PeterTamaroff 90%

Average is about 82%

or so

Peter Tamaroff

@Argon And you find the material "not challenging"?

Argon

@PeterTamaroff Not at all.

Peter Tamaroff

@MarianoSuárez-Alvarez Hola.

Argon

@PeterTamaroff I'm constantly bored in math class and my teachers do not do a good job keeping me occupied.

Peter Tamaroff

@Argon OK. Well, I can personally give you this advice: it is not something hard to learn to use maths: integrate, differetiate, use series, calculate sums, solve ODEs, solve linear equations, finds roots, whatever. In fact, I can see you even know how to evaluate improper integrals with residue theory. But it is important, if you want to study, learn, and eventually work or investigate that you follow a certain "learning path". With

this I mean that there is a much more interesting and important theory and material to grasp before movign on to complex analysis, or real analysis, fourier series, multivariate calculus, and

that means understanding what you're doing, why it can be done, what theorem avails it... understanding the thoery, following a proof, producing "little" proofs and stuff. There is a process of maturing that is important. It is great you're interested and know all that, but I think it will do you much better if you could hold your impulse to absorb "data" and worried (if you're not doing it already) about understanding rather than knowing. IT will be much more rewarding.

And that is just my opinion. I might be wrong.

Maybe other more experienced people can guide you better,

Zhen Lin

02:01

Did someone just come in and bump a whole bunch of PDE questions?

Peter Tamaroff

@ZhenLin YES

@ZhenLin I'm on the edge.

Zhen Lin

I see. Hmmm.

user19161

02:43

@PeterTamaroff Are they giving good answers?

Peter Tamaroff

@JasperLoy What do you mean?

I'm leaving BTW.

user19161

@PeterTamaroff OK bye!

Peter Tamaroff

@JasperLoy But what do you mean?

user19161

@PeterTamaroff I thought the person bumping the PDE is answering them.

Peter Tamaroff

@JasperLoy No, no, he's just retagging.

Byes.

BenjaLim

03:01

@JonasTeuwen Hey

J. M.

03:43

@HenryT.Horton Mr. Clean?

leo

please close this

J. M.

@robjohn I call it the "onion helix".

@leo *plays Taps on the bugle*

leo

@J.M. ?

btw the Cannabis curve already come from Math World

and please someone with power see this consecuence of the previous one

have good night fellows!

cya

Jayesh Badwaik

05:29

@BenjaLim Did you get an estimate on this or are you not working on it right now?

Jonas Teuwen

06:30

I woke up to Joshua Bell's play. 8-).

user19161

@JonasTeuwen I just woke up to Jonas Teuwen's words.

Jonas Teuwen

8-).

Not nearly as beautiful as the thing I woke up to.

user19161

Jonas Teuwen is more beautiful than Joshua Bell. QED.

Jonas Teuwen

I love ze violin.

user19161

I love the cello.

Jonas Teuwen

06:33

But Jonas would rape the violin instead of play it 8-(.

user19161

But I can't read music.

user19161

I know what CDEFGAB is though.

user19161

I am still marvelling at the latexmk solution, really wonderful.

Jonas Teuwen

What? It looks like a terrible hack to me!

Jayesh Badwaik

@JasperLoy It is better to use Kile (KDE)

user19161

06:36

In fact, the option is hidden in the latexmk manual but of course I needed someone to tell me the solution still.

Jayesh Badwaik

The real name of KDE is Konig Desktop Enviornment (Bow down to your king)

user19161

@JayeshBadwaik I tried it. It is bloated. All I need is the simplicity of TeXworks.

user19161

In fact now I only need one click on one toolchain to typeset everything. I don't need all those symbols or auto-completion. I prefer typing everything myself.

Jayesh Badwaik

Well TeXworks does look nice.

user19161

I hope it does not become bloated in future.

user19161

06:39

And texmaker is still better than kile anytime.

user19161

In fact, I think the author of texmaker is the original author of kile.

user19161

So you will see many similarities between the two.

Jonas Teuwen

@JayeshBadwaik That's a horrible advice.

@JasperLoy Use emacs.

user19161

@JonasTeuwen Only for those who also do other programming...

Jayesh Badwaik

@JonasTeuwen I was messing around with Jasper. You missed my konig part.

Jonas Teuwen

06:41

Yeah, wow, that was some quality trolling.

@JasperLoy ... what?

@JayeshBadwaik Click... emacs? Trolling again holy cow.

user19161

@JayeshBadwaik No man on earth can mess with me. They can kill me or imprison me but I am not afraid of anyone.

Jayesh Badwaik

@JonasTeuwen I meant, one keyboard shortcut

Jonas Teuwen

Puurfect.

C-c C-c RET.

And there is your .pdf.

user19161

Wow, I keep cleaning up my own TeXworks configuration until now I have the shortest most beautiful configuration code.

user19161

One click to rule them all.

Jonas Teuwen

06:43

@BenjaLim Hi.

@PeterTamaroff Redbull? That's for metrosexuals.

@JasperLoy AucTeX also has one...

Matt

Quick good morning to everyone.

Jonas Teuwen

Can also detect if LaTeX needs to be run again.

@Matt Sup bro.

user19161

@JonasTeuwen Ah, I think I tried all the major editors. I came to the choice between texworks and texmaker.

Matt

@JonasTeuwen I haven't contacted the CA lecturer. I haven't managed to bring myself to writing that email. I don't know what to write or how to phrase it.

Jayesh Badwaik

I hated TexMaker

user19161

06:45

I introduced kannappan to texmaker and now he uses it fulltime.

user19161

I don't know why people like texniccenter.

user19161

I think texniccenter is very unpolished.

Matt

@JonasTeuwen Otherwise: not much up. Starting to write thesis today. What about you, bro?

Jonas Teuwen

@Matt Still having a holiday.

Jayesh Badwaik

@JasperLoy Well, I use Vi for short scripting work, Emacs for programming and Kile for LaTeX stuff.
Advantages of Kile over Emacs that I want are mainly related to a nice window pane showing current files and the ability to click on symbols to get them when you forget the sequences.

Jonas Teuwen

06:46

@Matt Oh why not...? Do it.

I've told you what to write basically.

@JayeshBadwaik What the fuq, why would you need that when you have this: i.sstatic.net/aTRG8.png

@Matt Good luck with the thesis!

Matt

@JonasTeuwen Yes.

@JonasTeuwen Thanks.

Jonas Teuwen

Click on symbols? That's really sick. Don't you have a memory?

user19161

I can just use ctrl T in texworks instead of click.

Jonas Teuwen

Otherwise we have detexify (spawn a browser within emacs!) 8).

Jayesh Badwaik

@JonasTeuwen I almost always type from memory, but I hate having to search when I need that one symbol.

Jonas Teuwen

06:48

And you mean that's not searching...?

Jayesh Badwaik

@JonasTeuwen Dude, my internet connection is not as great as yours.

at least currently.

Jonas Teuwen

Oh?

user19161

@JayeshBadwaik Don't you have a symbol list with you?

Jonas Teuwen

Just basic fibre glass here.

user19161

You should have a symbol list printed in paper!

Jonas Teuwen

06:49

No you should not.

user19161

Paper books are for real men.

user19161

E-books are not.

Jayesh Badwaik

@JonasTeuwen I have moved to another city recently, my broadband is not enabled yet.

@JasperLoy Ebooks are the future, then I guess you would say the future is not for real men. Yes, go die, you neanderthel :P

Jonas Teuwen

A discussion between metrosexuals... I'm off (to the physiotherapist).

3

Jayesh Badwaik

@JonasTeuwen Babai.

user19161

06:52

@JayeshBadwaik E-books are for fiction. One needs paper math books. Math is for real men. Fiction no.

5

Jayesh Badwaik

It looks like one metrosexual scared another and he ran off. :P

Ilya

Metrosexuals are metrosexy

Jayesh Badwaik

07:08

Ring is a 3-tuple $(R,\cdot,+)$ where R is a non-empty set and $+,\cdot$ are binary operations on $R$ such that $(R,+)$ is an abelian group and $(R,\cdot)$ is a semigroup with multiplication($\cdot$) distributive over addition ($+$).

Is this correct?

yunone

I usually see a ring $R$ written as a $5$-tuple $(R,+,\cdot,0,1)$ where $(R,+,0)$ is an abelian group, and $(R,\cdot,1)$ is a monoid, not just a semigroup, if you want to assume unity exists.

Jayesh Badwaik

@yunone Artin says" A semigroup S is a a set with an associative law of composition and with an identity."

Then I would say artin is in error here?

yunone

@JayeshBadwaik Well, it may just be a matter of preference. But the wiki article on semigroups seems to disagree with Artin in that semigroups need not have an identity, but monoids do.

Jayesh Badwaik

@yunone okay.

Zhen Lin

07:34

I think it's because "semigroup" used to mean monoid.

Ilya

07:49

@ZhenLin I usually read about Markov semigroups (with identity) and the often comment is sort of "formally, these are monoids"

@JonasTeuwen isn't he on vacations? ;)

Jayesh Badwaik

@ZhenLin Ohh. Till when?

Zhen Lin

Probably Bourbaki...

Jayesh Badwaik

Okay. so in modern context, monoids are the one with identity and semigroups may not have one?

Zhen Lin

Yes.

David Wheeler

08:05

@JayeshBadwaik the distinction isn't as strong as you might think. you can always adjoin an identity to a semigroup, in much the same way as you can add 0 to the positive natural numbers without changing anything: you simply adjoin an element e, and define: a*e = e*a = a.

Jayesh Badwaik

@ZhenLin Okay. Thanks.
On a different topic: In http://math.stackexchange.com/questions/183458/approximation-of-elements-in-arithmetic-progressions-by-logarithms-of-integers question, I think given any $ak+b$ a sequence $\alpha_{n} = e^{b}*e{kn}$ should match the line exactly, and I am not understanding what the fuss is all about. I guess I am being just as stupid.

Zhen Lin

Ick, hard analysis.

Jayesh Badwaik

@DavidWheeler Thanks. The actual terminology is of course superficial, just wanted to know the current prevalent usage.

@ZhenLin ohh! No problem. I intended to ask to the complete chat anyway. :-)

David Wheeler

because of the pervasiveness of category theory, i think that monoid is becoming more common than "semi-group-with-identity".

Jayesh Badwaik

@DavidWheeler Okay

David Wheeler

08:12

one caveat about monoid homomorphisms: you have to add the identity-preservation rule separately, the multiplicative property isn't enough (unlike with groups)

Jayesh Badwaik

@DavidWheeler Ohh, right, That is a good point. Almost missed that.

Artin sends you off like a hitchhiker in a galaxy without a guide.

pretty interesting when you have people to discuss around with, but painful otherwise

David Wheeler

however, if a semi-group homomorphism between monoids is surjective, then it does preserve identity.

Jayesh Badwaik

Okay, nice result. just proved that too.

David Wheeler

given any set X, you get a generic monoid, called the monoid of transformations of X, or End(X). i like to think of End(X) as: things you can do (in X). this has a sub-monoid of bijections, Sym(X) or Aut(X), which i think of as: reversible things you can do (in X).

sort of the difference between: one-way processes, and two-way processes.

you can make this more formal by considering M-sets (and their "reversible" cousin, G-sets)

a lot of things can be M-sets...for example computer programs, where you have a concatenation of instructions that operate on a data structure.

Jayesh Badwaik

08:35

@DavidWheeler That is what I was thinking of :-)
Irreversible programs belong to End(X) but not to Sym(X)
and hence, it is possible to rollback those instructions if there is an error and such.

And one can then formally define which instructions are reversible and stuff, just by looking at the final output.

David Wheeler

exactly. so it is preferable to work with Sym(X), if you can.

Jayesh Badwaik

No, that point was different. Sheesh, I mixed up two different things. Well, it is getting confusing for me now, I should go back and practise a little myself. Will come back later for more stuff.
Btw, G-sets are just group actions right?

David Wheeler

well, obviously sometimes you might want to discard data. that is a one-way process, but it can improve efficiency.

yes, given a G-set, you can explicitly come up with a homomorphism G-->Sym(X), and vice-versa.

equivalently (Cayley's theorem): every group is a permutation group.

Jayesh Badwaik

@DavidWheeler every group is a subgroup of permutation group

David Wheeler

the trouble is (for finite sets), that Sym(X) has order |X|!, which is often much bigger than |G|, so it's not an "efficient representation".

yes, i should have said "subgroup" as "permutation group" often means: a full symmetric group.

Fortuon Paendrag

08:46

@DavidWheeler Can I ask you a number theory question?

Jayesh Badwaik

@DavidWheeler hmm, I know only lang and artin's terminology so may not understand some other normal conventions.

David Wheeler

@FortuonPaendrag you can ask, but i may not know the answer.

Jayesh Badwaik

@DavidWheeler Thanks for the stuff, will read more into it and ask here if i come up with some difficulty.

David Wheeler

@JayeshBadwaik lol, that's ok, those are both "classics" so i'm sure you can "get by" only knowing what's in them :)

Fortuon Paendrag

Ok. Here it goes. Can every $p$-adic number $x$ such that $|x|_p \leq 1$ satisfy a monic integer polynomial?

Jayesh Badwaik

08:48

@DavidWheeler lol, see ya, bee back later.

Fortuon Paendrag

Absurd. Im spouting nonsense. I just realized. @DavidWheeler

Sorry, don't worry about the question.

Alexander Amenta

09:10

'evening everyone

Gustavo Bandeira

@HenryT.Horton Oh, this is you, Henry?

@AlexanderAmenta Morning here.

Alexander Amenta

@GustavoBandeira 'morning

Gustavo Bandeira

@AlexanderAmenta I thought you were from Greece.

Alexander Amenta

why greece?

Jayesh Badwaik

@AlexanderAmenta The great one was from macedonia, and hence probably ;-)

Gustavo Bandeira

09:21

@AlexanderAmenta Alexander.

Alexander Amenta

i've never really associated the name Alexander with Greece. understandably i guess

Gustavo Bandeira

@AlexanderAmenta Haha

Alexander Amenta

are you (both of you) from greece? just asking

Jayesh Badwaik

@AlexanderAmenta hahah. No, I am from India.

Gustavo Bandeira

We at MSE should unite someday to record a song and videoclip like this.

2

@AlexanderAmenta Brazil.

Ilya

09:24

@AlexanderAmenta see here

@GustavoBandeira oh no, gangnam style

op-op

Alexander Amenta

india and brazil, sounds good. what's up

Sabyasachi

hmm.

@Ilya, are you Ilya Vinagradov?

Ilya

@Sabyasachi vinOgradov?

Alexander Amenta

@Ilya what's to say i'm not the late prince alexander of belgium?

Sabyasachi

Sorry,

Alexander Amenta

09:26

@Ilya what's to say i'm not the late prince alexander of belgium?

Gustavo Bandeira

@Ilya We could make it in Brazil, a blend of carnival and electrohouse music!!!

Sabyasachi

I mean $Vinogradov$

Ilya

@Sabyasachi no, I am not

Fortuon Paendrag

@AlexanderAmenta: Too obscure, no?

Jayesh Badwaik

@AlexanderAmenta Well, for one, you have to use zsh for that. Do you use zsh?

And every day it gets harder to fight the urge to su to the user and freak people out.

Gustavo Bandeira

09:28

@AlexanderAmenta Not much, just doing some homework and studying piano.

Jayesh Badwaik

@AlexanderAmenta I am currently struggling with monoids and semigroups.

Alexander Amenta

i have four messages to reply to =O i won't get anything done at this rate

@FortuonPaendrag: in person i'm probably more obscure than that. i don't have a wikipedia page
@JayeshBadwaik: i don't even know what it is but i like the ghost in the shell pun
@GustavoBandeira: cool
@JayeshBadwaik: also cool

Gustavo Bandeira

@AlexanderAmenta Congratulations, your earned some social skills points!!!

m. k.

pokey the penguin is cool

Gustavo Bandeira

@AlexanderAmenta Achievement unlocked: Talking to 4 persons at the same time!!

Alexander Amenta

09:32

@m.k. finally some pokey appreciation!! i quoted him in my honours thesis
@GustavoBandeira i don't know why i thought going on chat would be a good idea while trying to do integrals

m. k.

that's awesome

Jayesh Badwaik

@GustavoBandeira There is one like that? Jeez, I should do that sometime, just for the sake of it.
@AlexanderAmenta and now that you have taken your revenge by pining me twice. I will go back to my monoid so that I can work and you can too. bye , see ya later. :P

m. k.

i quote pokey the penguin every time i say "yes"

Gustavo Bandeira

@JasperLoy OMG! I'll post this on facebook!!!

user19161

@GustavoBandeira Um OK!

Alexander Amenta

09:34

@m.k. only if you can audibly put a line through it

user19161

Sorry I bumped my old questions. I am just trying to retag them the way I deem best.

Alexander Amenta

@JayeshBadwaik cheers =p

Gustavo Bandeira

09:48

$n^{-\infty}$ equals zero?!

Alexander Amenta

@GustavoBandeira makes sense. $n^{-\infty}$ is the number of functions from a set with $-\infty$ elements to a set with $n$ elements. and there are none of those

Gustavo Bandeira

@AlexanderAmenta I've learned exponentiation pretty mechanically, I can't see it that way.

@AlexanderAmenta I've tried to - don't know If it's the right procedure - algebrize it(I guess this is the name for it) like: $a^b=?$ but I can't perform this

Alexander Amenta

@GustavoBandeira it's a joke - i have no idea what n^{-\infty} is supposed to be!

Gustavo Bandeira

@AlexanderAmenta Consider $n$ as any number.

integer I guess

Alexander Amenta

but if you look at n^{x} and let x go to -\infty then it makes sense. unless n=0

where is this coming from?

Gustavo Bandeira

09:55

@AlexanderAmenta I guess you need the $.

@AlexanderAmenta E.J. Barbeau - Polynomials.

@AlexanderAmenta It's an exercise where he asks a polynomial of degree $-\infty$

Alexander Amenta

haven't read that one, but i've seen it around. looks fun

what's the definition of 'degree'?

Gustavo Bandeira

@AlexanderAmenta Supose you have a polynomial: $a_nt^n+a_{n-1}t^{n-1}+\cdots+a_1t+a_0$. It's a polynomial of $n$ degree. If $a\neq0$

Alexander Amenta

what if a_n = 0?

Gustavo Bandeira

Yep, forgot that.

Alexander Amenta

ok good, that's an important point

did you find a degree -\infty polynomial?

Gustavo Bandeira

10:01

@AlexanderAmenta I tried to do $77x^{-\infty}+1$

Alexander Amenta

x^{-\infty} isn't well-defined though, and polynomials aren't allowed to have negative exponents anyway

and no $-\infty$ terms appeared in your definition of the degree. is there another way to define the 'degree' which could allow for negative infinite degree?

Gustavo Bandeira

@AlexanderAmenta Then I've answered as I told and went to check for the answers, the answer was 0... I was like: "WTF?"

Alexander Amenta

indeed.. what's the degree of the zero polynomial?

evidently $-\infty$, but what do you think it should be?

Gustavo Bandeira

@AlexanderAmenta I was thinking that degree mean the exponent of the leading coefficient, like $x^{this.is.its.degree}$

Alexander Amenta

you're right in all cases except for the zero polynomial =)

what's the leading coefficient of the zero polynomial?

Gustavo Bandeira

10:06

@AlexanderAmenta It does not exist?

Alexander Amenta

exactly! but you've defined your 'degree' in terms of leading coefficients. so by your definition, the zero polynomial doesn't have a well-defined degree

Gustavo Bandeira

@AlexanderAmenta I have no idea. I wanted to solve it like this: $2^5=2\cdot2\cdot2\cdot2\cdot2$ then I would try to algebrize it, like: $a^b=a\cdot a\cdot a\cdot a\cdot a...$ but this gave me no insight.

Alexander Amenta

there's a more precise definition, which does include the zero polynomial - recheck your book and see if it's there (it should be, if this question is in there)

Gustavo Bandeira

@AlexanderAmenta Yep.

A zero of a polynomial p(t) is any number r for which p(r) takes the value 0. When p(r) = 0, we say that r is a root or a solution of the equation p(t) = 0. There are many situations in which we need to have information about the zeros of a polynomial, and considerable amount of attention is devoted to methods of solving equations p(t) = 0 either exactly or approximately. In particular, knowing the zeros of polynomials is often helpful in graphing a wide variety of functions and obtaining inequalities.

Alexander Amenta

it doesn't mention degree, but it does give one possible interpretation of the 'degree' of a polynomial: the number of (complex) roots (counting multiplicities)

in which case constant nonzero polynomials, having no roots, would have degree 0 - but the zero polynomial has infinitely many roots!

this doesn't answer your question though

the definition of degree i'm thinking of is 'the largest integer n such that the coefficient a_n is nonzero'

Gustavo Bandeira

10:14

@AlexanderAmenta What's the name of the process I called "algebrize"?

@AlexanderAmenta Yes, I'm trying to understand what you wrote.

Alexander Amenta

i think 'algebrize' is the right word, but i don't know if it's a legitimate english word

Gustavo Bandeira

@AlexanderAmenta This is intractible. Haha

BenjaLim

@AlexanderAmenta Hey

For tomorrow I would like to discuss based on:

1

A: Function is pointwise limit of integrals

Here is a hint on how to do the problem. Because $\phi_n(x) = 0$ for $|x| \geq 1/n$, you can say $$\int_{-1}^1 \phi_n(x) dx = \int_{-\infty}^{\infty} \phi_n(x) dx.$$ Now choose $x \in\Bbb{R}$. Then $$\begin{eqnarray*} |f_n(x) - f(x) | &=& \left|\lim_{N \to \infty} \int_{-N}^N \phi_n(x...

0

A: estimation for Dirichlet kernel.

For (1), I am not sure what kind of estimate you're after. Perhaps this may be helpful: Recall that $$(f \ast D_N)(x) = S_N(f)(x)$$ where $S_N$ is the $N$ - th partial sum of the fourier series of $f$. Now by direct calculation I find that the fourier coefficients of $f$ are $1/2$ if $n =0$ and...

Alexander Amenta

@BenjaLim sup

BenjaLim

Both of those are related to Fourier analysis

Alexander Amenta

10:24

yep sure. i don't know how useful i'll be since this is the first time i've really looked at the dirichlet kernel, but i'll try

BenjaLim

hahahahahahahahaha

@AlexanderAmenta I'm blue in the face trying to construct some damn covering space of $S^1 \vee S^1$

Alexander Amenta

how about $R \vee R$

BenjaLim

@AlexanderAmenta I have to construct one as follows. If $a,b$ are the generators of $\pi_1(S^1 \vee S^1)$, I have to construct a covering space corresponding to the normal subgroup generated by $\langle a^2, b^2 ,(ab)^4 \rangle$

@AlexanderAmenta And to ensure normality

I have to throw in the conjugates of each of the above

so for example I would need $ba^2b^{-1}$

$ab^2a^{-1}$

$b(ab)^4b^{-1}$

But now I don't need to conjugate by $a$ for the last one

because I get it for free from the rest

You have no idea how many pictures I have drawn

Alexander Amenta

i can't remember how to do these

BenjaLim

Considered close to 20 different figures....

Alexander Amenta

10:29

i can't remember how to do these

BenjaLim

All not working out

How was the meeting today

Alexander Amenta

i'm thinking wrap the first S^1 around itself twice to get a^2, do the same for the second to get b^2

BenjaLim

yes

Alexander Amenta

then wrap S^1 \vee S^1 (viewed as one loop) around itself 4 times to get the rest

and connect them all up

BenjaLim

But the problem now is the rest of them

but you get too many relations

that is the problem

That was the obvious one to go for

Alexander Amenta

10:30

yeah like i said i've forgotten how to do these

i do analysis now

BenjaLim

yeah

Alexander Amenta

just integrals

BenjaLim

exactly

@AlexanderAmenta Yesterday night I was trying to show the dirichlet kernel is unbounded in the $L^1$ norm

Guess what I did when I got stuck

Alexander Amenta

=p

i am doing an estimate now and i have split the integral into two bits, just like a real analyst

what did you do

BenjaLim

INTEGRATED BY PARTS.

Alexander Amenta

10:32

how many times?

Gustavo Bandeira

@AlexanderAmenta I'm "almost" undertanding it.

BenjaLim

once

and the estimate fell out

Alexander Amenta

@GustavoBandeira good =) it's a subtle thing, i think. but the most reasonable 'degree' for the zero polynomial is $-\infty$

Gustavo Bandeira

@AlexanderAmenta This: $a^{-\infty}=\overbrace{a\cdot a\cdot a\cdot ...}^{{-\infty}\text{ times}}$ kinda helped me to understand.

Alexander Amenta

really? that's more confusing to me.. how do you multiply something by itself -infinitely many times?

Gustavo Bandeira

10:35

@AlexanderAmenta Exactly! There are no $a$'s to multiply, I guess.

Alexander Amenta

usually an empty product is interpreted as the number 1 though

swair

@Leon , (real analysis pen-friend)..reply!

Alexander Amenta

the point with my definition is that for all integers m, the coefficients a_m, a_{m+1}, ... are all zero, and so the degree has to be less than m

so the degree has to be less than every integer. -infinity is then the only candidate (despite it not being an integer)

most definitions i've seen explicitly define the degree of the zero polynomial to be -infinity

swair

@Leon ping

Leon

@Swair, finally found you :-). If you interested we could set up something like weekly meetings or something like that, even skype!

swair

10:40

@Leon ok sure. Though I have some time constraints (newly started job!) so once a week is good. drop me a mail at swairshah_at_gmail

Leon

Congratulation on the job! I am in the same position, just got a job offer so will be quite busy! I will drop you an email!

Alexander Amenta

@GustavoBandeira here's another thing to think about: if you multiply a degree m polynomial with a degree n polynomial, what's the degree of the result?

swair

ok sure then. talk to you soon!

Gustavo Bandeira

@AlexanderAmenta Processing...

swair

@Leon also give me your mail address in case you forget mine or something :P

Gustavo Bandeira

10:53

@AlexanderAmenta I've made a question and while I made the question, I thought about other hypotheses, look

@AlexanderAmenta Still processing your question...

Matt

Hi @Ilya.

BenjaLim

@AlexanderAmenta Hey

I have tried and tried till I'm blue in the face

totally out of ideas

Matt

bbl

Ilya

@Matt: hi, don't leave

which questions to ask and which to answer in order to get 10+ votes

Matt

@Ilya I'm here.

Ilya

10:56

@Matt I'm not :-p

user19161

@BenjaLim Well, how long have you tried?

BenjaLim

A long time

Like maybe 4 hours

No joke

I started at about 4pm

user19161

@BenjaLim If it is less than a day, it is not long.

user19161

I am just speaking in general.

BenjaLim

@JasperLoy I have to hand in the assignment on friday.

Transcript for

$Mathematics$ Mathematics