@MarianoSuárez-Alvarez Hola.

Spivak tiene unos ejercicios sobre integrales que me van a mantener entretenido por un buen rato =D

@PeterTamaroff No hay que llorar.

@PeterTamaroff Are you doing a math bachelor?

@GustavoBandeira I'm not whining... ?¿?¿?¿

@GustavoBandeira I'm not sure what it is.

Licenciatura.

@PeterTamaroff Oh, in Brazil, licenciatura is a course for teaching high school.

@GustavoBandeira Mmm not that.

@PeterTamaroff Can you send me your course's webpage?

uba.ar/download/academicos/carreras/lic-csmatematicas.pdf

exactas.uba.ar/academico/…

@PeterTamaroff 6 años!

@GustavoBandeira Yes...

Gotta go eat!

@PeterTamaroff k, cya.

@GustavoBandeira Maybe Mariano can tell you

@PeterTamaroff What?

@GustavoBandeira About the math career in the UBA

@PeterTamaroff Yep.

@robjohn

@PeterTamaroff I guess I need to say something.

@robjohn Are you there?

@JasperLoy It' s OK. I'll go to now.

It'd be better to ping us with a message we can read later than to ping us over and over again until you happen to be on chat the same time we are.

Hey, quick question here about smooth structures on manifolds.

If the manifold itself has a single homeomorphism onto an open subset of R^N, does that imply that the atlas given by that single map gives rise to a smooth structure?

Intuitively it feels wrong, but I can't see what part of the definition of a smooth structure is violated.

Any help would appreciated =)

do homeomorphisms preserve smoothness? (disclaimer: not having studied diffeo geo, I have no idea what I'm talking about)

What do you mean by smoothness?

I'm talking about an abstract manifold so it might not make sense to talk about a manifold being inherently smooth.

Anyhow, Spivak defines smoothness as a condition on the transition between charts, so if there is only one chart, it must be smooth.

But then any jagged open curve segment in R^2 full of cusps is given a smooth structure by any single homeomorphism from an open subset of the line, which feels like contrary to the definition of a smooth structure. I mean, how do you talk about a tangent space with an arbitrary parametrization?

*not defintion, I mean "intuition"

I am aware of that, Jasper.

To give a specific example of a question, consider the open interval (-1,1) and the image M of this interval under the function f(x) = |x|. Obviously, this function is not differentiable at the origin, and yet it would appear that f^-1 gives M a smooth structure as a manifold.

I guess smooth structure means something different than my naive impression.

Dedekind's cut is so complicated!

This is going to be a long shot, but does anyone know anything about Dvoretsky's theorem?

Eh, what is going to be a long shot?

I have trouble in English.

I'm reading Rudin's Principles of Mathematical analysis.

Theorem Closed subsets of compact sets are compact.

Proof Suppose $F\subset K\subset X$, $F$ is closed (relative to $X$), and $K$ is compact.

@FrankScience Okay, what do you not understand about it?

@JayeshBadwaik I'm writing the context.

@FrankScience okay. continue. sorry to interrupt.

Let $\{V_\alpha\}$ be an open cover of $F$.

If $F^c$ is adjoined to $\{V_\alpha\}$, we obtain an ...

What does be adjoined to mean?

hmm, I am not sure.

let me see what it can mean,

2.35 Theorem

You can trace back to that page and get the full context.

yup, I am guessing it is just a union, but I am not sure.

It should be just a union from what it appears.

$F^c\subset \bigcup_\alpha V_\alpha$?

No.

It is like this

@PeterTamaroff I am now.

@FrankScience We have an open cover $\{ V_{a}\}$

@JayeshBadwaik What's the relationship between $F^c$ and that cover?

now adjoining is just an open cover $\{ V_{1} , V_{2} ,.... ,F^{c} \}$

$F^{c}$ is open in $X$ as $F$ is closed in $X$.

Ah, got it. Thanks. It's really obscure.

Hehe, its rudin after all.

@FrankScience You might want to see this

especially if you are self-studying.

or did I already tell you about it?

If so, I am sorry.

Yeah, you did.

Hmm, okay, I will remember now that I told you. :-)

These theorems are not so easy, thus I failed to prove any of these.

Bonjour, my friends!

I have a question about covering spaces. I have a problem with proof of fact, that two liftings of path with a common start point coincide or don't intersect

@JonasTeuwen Good morning. How are you today?

Hello. How does the community feel about questions like "I don't understand the following (one line) argument from a textbook"?

Hi

@JasperLoy Never reply to a joke literally unless you want to people to mock you.

Hi all

@OldJohn Hi.

@JayeshBadwaik A quiet day in mathland, I think

@OldJohn Its still not bad. Day before yesterday, there was a 10 hour stretch with just a single message, that too from me I think. :-)

@JayeshBadwaik aha - so you prevented it from being a record-breaking day :)))

@OldJohn :-) :-)

I read somewhere that in Finland there is no such thing as an "uncomfortable silence" - if people have nothing useful to say, they just remain silent :)

:-)

@OldJohn This is so true for me. My thesis title matches exactly this description. :P

@JayeshBadwaik Mine too, I think

My thesis title is "dune-devicelab : A framework for hybrid simulation of nanoelectronic devices"

"Boundary behaviour of continuous and fine continuous functions in Euclidean space" - I missed out on the witty bit :(

@OldJohn :-)

@OldJohn : there is a function $f : \mathbb{R} \to \mathbb{R}^2$ and as $x \to \infty$ $f(x)$ moves along a straignt line, I dont know whether to call it asymptotic as $f(x)$ can lie on both sides of the line. If we can still call it asymptotic (I am not sure of def of asymptotic), How do we characterize this property mathematically in equations?

BTW $f$ is smooth

It can have infinite intersections with the line

My research was all about "asymptotic paths" which concern paths $\gamma$ which either go off to infinity or to the boundary of of a region where a function $f$ is defined and such that as we go along the path $\gamma$, the function $f$ approaches a limit "the asymptotic value of $f$ along the path $\gamma$"

functions I dealt with were mostly things like $f:\mathbb{R}^n \rightarrow \mathbb{R}$ where $f$ was continuous (or fine continuous)

your situation could probably be turned into the same sort of thing

@OldJohn I am making a picture for it. But I wonder how to draw free hand with mouse on Geogebra?

@RajeshD No idea, I'm afraid :(

Isn't your situation something like a curve approaching an asymptote?

@OldJohn yes the curve almost becomes a straight line, but it can have infinite number of intersections with the line, the deviation of the curve from line becomes smaller and smaller as we go to infinity

@RajeshD Is it a curve from some control system?

@OldJohn what is fine continuous?

@RajeshD That is the standard definition of an asymptote - mostly the curve does not cross the asymptote, but it is perfectly OK for it to do so

I don't have any physical system in mind, I just want to characterize this property mathematicaly in symbols

Fine continuous is a bit hard to explain, unless you know about discontinuous subharmonic functions :(

@OldJohn Ok got it, an asymptote be intersected with the curve, its allowed I guess

Yep - perfectly acceptable (to all authors I know of)

@OldJohn Was this message for me? I don't get it

@RajeshD no, it was for me.

ok

e.g. $\sin x /x $ would have the x-axis as an asymptote - and crosses it infinitly often

Sorry! - sent to wrong person

@OldJohn Okay. I know that subharmonic functions are something like positive deviation from the laplace. (Is that correct?)

@OldJohn fine, understood. this example sums up all I wanted

@RajeshD Great!

And hence, they may not go to zero at infinities.

@JayeshBadwaik ??

@RajeshD Again for @OldJohn. Sorry.

@OldJohn Thanks Oldjohn

@JayeshBadwaik oops

@JayeshBadwaik That is fine for continuous ones - for the general case the wikipedia article has the appropriate definition

@OldJohn Okay, I think I get it.

@OldJohn I guess, you can provide me a definition of fine continuous now.

I can atleast try.

@JasperLoy Because no one is using chrome. :P :P

@JayeshBadwaik OK - the fine topology is the coarsest topology on $\mathbb{R}^n$ which makes all the subharmonic functions continuous

@OldJohn Okay.

@JasperLoy Firefox heere. :P :P

@JasperLoy Munkres cautions the readers against it repeatedly in the beginning.

good definition here

I have lost count of how many times I have read the pebbles and truck lines.

@OldJohn I was searching for fine continuous functions and was not getting it. thanks.

@JasperLoy Indeed - but potential theorists all agree on the same definition, luckily :)

@JayeshBadwaik so, a function is fine continuous if it is continuous with respect to the fine topology

@OldJohn Okay. :-) Nice.

so fine continuous functions include all the discontinuous subharmonic functions and superharmonic functions and lots of others

There are 5 day old quotes still on the starboard. unacceptable

When I was a teacher, there was a move to use "negative 3" - so I tried to get kids to use it - then gave up and went back to "minus 3"

@JasperLoy Yup, and I was actually referring to the dearth in quality.

this statement is not starred

@JasperLoy People star anything in other rooms.

going off for a few minutes, while I do my monthly pc reboot.

Time for a break - back later

@JasperLoy Back up now. No, I send my computer to sleep.

I upgrade weekly and hold back kernel upgrades to once a month.

Then, when I upgrade the kernel, I reboot the PC.

Why do you think I should shut down by the way?

@JasperLoy I remembered what I wanted to tell you yesterday. I wanted to say that, in the beginning, MSE probably wanted to ensure that they reach a critical amount of information to get started the popularity of MSE, and hence it encouraged self-answering by giving points. Now that MSE is quiet successful, it has probably stopped doing so to prevent abuse. This is something similar to how Reddit popularized its platform.

@JasperLoy Sleep mode saves electricity. It is not completely on. It is sleep state consuming 0.2% of the normal consumption. Mine is a laptop, for desktop I agree though.

@JasperLoy Great. Now take a pic fast. :-)

@JasperLoy Which one have you got?

@JasperLoy Deadline?

@JasperLoy Hence, you should always be logged in.

@JasperLoy nice. It is very similar to what I have. HTC Wildfire.

@JasperLoy Try to empty browser cache and see if it changes something. You cannot disable it, but probably clearing it might help.

Yeah.

but then he will say, "There is this setting which I will implement, but not now!!"

:P :P

@JasperLoy What??

No, I do not think so.

Okay, do you mean a flag in the menu?

@JasperLoy Or just use an "alias"

I used that for chrome when it did not interface well enough with KDE proxy settings.

@JasperLoy Hmm. The thing is the session "expires" as you closed the window.

and all the states are deleted.

But you have got a point, it can be easily added, no one thought of it, I guess.

All browsers do so, so seems like a design issue for all of them.

But then, rekonq has multiple windows, with different setting for each of them.

So, by default, it is off, but if you want to enable it for a specific window, you can enable it.

@JasperLoy Flash works alright on firefox, even on linux.

@JasperLoy Linux people know. I do not (did not) know. Jonas uses chrome for that. I use VLC, the added advantage is any video I watch is automatically downloaded and converted into an ogv format and stored. But as you said, some video cannot be linked, so my reference was only for youtube, dailymotion and similar sites.

@Charlie Good morning. Good morning.

Or is it afternoon already?

@JayeshBadwaik Thanks!(it's afternoon :P)

@JayeshBadwaik Afternoon.How are you?how was your sunday?

@Charlie Lazy. :-/ :-)

@JayeshBadwaik good.it's a beuatiful day here!

@Charlie good. cloudy and cool?

or sunny beautiful?

@JayeshBadwaik sunny beautiful!

@Charlie good. you are not going out then?

@JayeshBadwaik no no...i took a walk.but i'm fine.you?

@Charlie There has been some heavy rains today here so I went out for a walk, it was nice.

@JayeshBadwaik delicious!

@Charlie :-)

@JayeshBadwaik a bird told me you like celine dion and metallica?is that rue?

@Charlie Yes I like them.

@JayeshBadwaik :)))))

@Charlie So, I guess you like them too? :-)

exit light,enter night take my hand...rest in neverneverland

@JayeshBadwaik YES

My favorites are the Unforgiven

@Charlie Hmm, mine are master of puppets, enter sandman, nothing else matters.

@JayeshBadwaik of course.great ones!!good taste you have!

@Charlie you too. which ones do you like by celine?

@JayeshBadwaik many.i like i surrender

@Charlie I like the complete album of "All the way, a decade of song." and then "all by myself". Its all coming back to me now (the seven minute version) taking chances among others.

@JayeshBadwaik :)

@Charlie checkout the photographs, sketches etc here I like them.

Actually this is a better link to show all his stuff.

@JayeshBadwaik Beautiful!

@Charlie Aren't they? There is one he has done very similar to "The Starry Night" by Van gogh. Its really beautiful.

@JayeshBadwaik starry nught is my fav pic

@Charlie !!!

@JayeshBadwaik :0

:)))

This is the one

@JayeshBadwaik sweet!!!

@Charlie yeah.

@JayeshBadwaik DO YOU LIKE VAN GOGH?

@Charlie Yes! Why the caps?

@JayeshBadwaik i forgot it...

@Charlie Hmm. I like almost all of van gogh paintings. I haven't seen all, but his self-potrait is really good. Something similar as that of da vinci.

@JayeshBadwaik oh!I love the sunflowers!

@Jay?

@Charlie Sorry, be right back. mom's call.

oh,sorry!

back.

ok

Hi!!

again!

Heya

@N3buchadnezzar hi

@JasperLoy Great!!

Hmmm

Any tips on solving the equation x(x+1)(x+2) = 60 ?

I know the solution is easy to find by guessing, but I am looking for some clever substitutions =)

well I do not want to guess a solution then use polynomial division...

...

This equation is trivial to solve

I was thinking of using a similar trick to sovle mine.

@N3buchadnezzar Trivial is in the eye of the beholder.

I want to include clever tricks and techniques!

@JasperLoy It is too messy. I usually apply ferrari's method to solve a quartic.

@N3buchadnezzar Have you seen ferrari's method?

@JayeshBadwaik x = y - b/3a ?

@N3buchadnezzar yup.

Well

My equation is on that form!

@JasperLoy :)

@N3buchadnezzar Yup.

Now what?

nothing what?

Now, you can see why such method would fail. Because, you have nothing to group in this sense.

@Charlie it was for n3.

@JayeshBadwaik sorry :|

I know, I think I will have to use trig:/

Especially when you are learning.

@JasperLoy Yeah.

@JasperLoy good!

Hello.

Hi Jasper.

@JonasTeuwen Hi.

@JonasTeuwen HI!!

@JasperLoy forgetting is bad like that.

i hate to forget...

O.o

@Charlie That's easy for you to say, you are not Polar! ROTFL

@JayeshBadwaik yoiu read!!!

@JasperLoy Hurt.I.Am. Will.I.Am.Not.

@JasperLoy Jayesh liked...

@JayeshBadwaik I alwys thoought about that :) hehehheh

@Charlie :P :P

@JasperLoy Good!

@JasperLoy we all do

@JasperLoy sparks?