Mathematics - 2012-09-18 (page 1 of 2)

then you can make the isomorphism one of (in the first case, quotient) monoids (under multiplication) instead of just as sets (which just uses bare cardinality for an isomorphism between classes of complex numbers and reals)

If you take 0 out of C you get an abelian group isomorphism $\Bbb C^\times/\sim\,\,\cong (\Bbb R,+)$ via $[z]\mapsto \log|z|$

Peter Tamaroff

Yes, true [0,inf) seems better

anon

also you can put topologies on errything and retain the isomorphisms

Peter Tamaroff

What is a monoid?

anon

set with associative multiplication; no inverses necessarily but an identity element

Anybody know a name for the set $E_a=\{0,1,\cdots,a\}$ with operations $x\oplus y:=\min\{x,y\}$, $x\otimes y:=\min\{x+y,a\}$? I believe it is a commutative ring (so associativity and distributivity hold).

Peter Tamaroff

Like what you add to a group to get a ring

Mariano Suárez-Alvarez

1:06 AM

a semiring, maybe, anon

because you are not going to have additive inverses

there is a famous semiring, called the tropical algebra, or the max-plus semiring

which is somewhat similar

Peter Tamaroff

@mariano Hello

Mariano Suárez-Alvarez

Hello :-)

Peter Tamaroff

Reading soc y estado here. Oh god why

Mariano Suárez-Alvarez

jajaja

Peter Tamaroff

So a ring with no inverses is called a semi ring?

Mariano Suárez-Alvarez

1:15 AM

with no additive inverses

$\mathbb N_0$ with its usual addition and multiplication, for example

Peter Tamaroff

Okey

MaoYiyi

1:55 AM

to show e^x for latex one types $e^x$ however, on the title of my question it showing as exp(x). why?

anon

2:07 AM

Because you never typed e^x in the title*, you typed `$exp(x)$` which Byron changed to $\exp(x)$.

*Unless you originally had e^x in the title and then deliberately changed it to exp(x) within five minutes (edits in the first five minutes are not publicly documented).

MaoYiyi

2:17 AM

@anon wow, I'm a moron.

Zhen Lin

3:00 AM

I have a proof to write and the expressions are so absurdly long that one term fills up a whole line...

leo

3:25 AM

hello there!

leo

3:37 AM

Given a compact set $K$ of $\Bbb R^n$ and an open nhbhd $U$ of $K$, I'm trying to prove that there is a $C^\infty$ function $f:\Bbb R^n\to \Bbb R$ so that $f_{|K}\equiv 1$ and the support of $f$ is contained in $U$.

I've proved that given an open bounded set $A$, there exist a function $f:\Bbb R^n\to \Bbb R$ such that $f$ is $C^\infty$ and is positive over $A$. So I've tried take $g:\Bbb R^n\to \Bbb R$ of class $C^\infty$, positive over $U\setminus K$ which is open (and bounded without loose) and then consider $f=\chi_K+g$. This function is identicaly $1$ over $K$, but it can fail to be $C^\infty$

so, I'm stuck

any idea?

anon

3:53 AM

seems like a job for some form of mollification. not something I have experience with.

Mariano Suárez-Alvarez

this is done in Lee's book on smooth manifolds, which you can find online

leo

what I've typed above it's almost the finish. All start with bump functions

Mariano Suárez-Alvarez

you need to be very lucky to to this with a mollifier

because the inf of the distances from points of K to the complement of U might be zero

so using a mollifier (which has a fixed support) is going to be impossible

leo

4:09 AM

I've done some things with mollifiers before, but no idea how to approach this in that way

@PeterTamaroff heyo!

Gustavo Bandeira

4:32 AM

Is there a way to avoid this:

anon

how did you even get that

\frac{ac-b^2}{a-2b+c}=\frac{bd-c^2}{b-2c+c^2} should work fine, no?

Gustavo Bandeira

@anon Lemme try it.

Nope. Same thing.

anon

works for me on main. whatever is rendering it is the problem for you. is it on mse? if so what browser etc?

Gustavo Bandeira

Chrome.

Lemme try another browser.

It works fine on firefox.

user19161

6:41 AM

@MarianoSuárez-Alvarez Hehe, that's only a pre-first edition draft though. Anyway the second edition is out, since you mentioned it. It's one and a half times as long as the first edition.

user19161

@GustavoBandeira Yes, Chrome has problems with chat and math for me. I guess it is because it is not so responsive.

Jayesh Badwaik

@WillHunting My TeX says command \mathscr not found. Strange! \mathcal, \mathfrak work though.

Zhen Lin

You need a package like rsfs

Jayesh Badwaik

@ZhenLin Okay. According to this link , amsfonts should have been sufficient, but it is probably wrong then. My distribution does not have rsfs by default it seems, will have to get it separately I guess.

Zhen Lin

\mathscr is sometimes used as an alias for \mathcal, but I guess you want them to be different fonts.

Jayesh Badwaik

6:53 AM

@ZhenLin Yes.

Gustavo Bandeira

11:31 AM

@robjohn I made this question:

0

Q: A list of books for discovering mathematics using computer software

I'm searching for books that allow one to discover/experiment with mathematics by using computer environments such as Mathematica/Magma/Magma/Pari-GP, etc. Until now, I discovered these: Discovering Mathematics: A Problem-Solving Approach to Mathematical Analysis with MATHEMATICA® and Maple(T...

Someone suggested that it would be better to make it CW. What do you think?

Parth Kohli

@Gustavo you have interesting questions!

Gustavo Bandeira

11:47 AM

@ParthKohli Yep. I'm trying to build a body of mathematical culture.

2

By knowing names, ideas, etc.

Parth Kohli

Very interesting, @Gustavo.

Parth Kohli

12:24 PM

Hey, quickie question.

What's the proof? "The powerset of an arbitrary set $\mathbf{A}$ with $n$ elements contains $2^n$ elements."

BenjaLim

@ZhenLin I posted an answer to one of your questions.

The torsion part is isomorphic to $\Bbb{Z}/n\Bbb{Z}$

Parth Kohli

12:52 PM

Darn, you Math people!

robjohn

@JonasTeuwen: I see you have returned :-)

Michael Greinecker

1:22 PM

@Path By induction. Let $S$ be a set with $n-1$ elements, assume that $|P(S)|=2^{n-1}$ and let $x\notin S$. Then you can show that $|P(S\cup\{x\}|=2^n$ by showing that to each subset containing $x$, there corresponds exactly one subset not comtaining $x$.

robjohn

1:33 PM

@MichaelGreinecker I don't think that will ping @Parth

Off to the park. BBL

Michael Greinecker

BBL?

robjohn

2:09 PM

@MichaelGreinecker Be Back Later.

Michael Greinecker

@robjohn You are back from the park?

robjohn

@MichaelGreinecker Indeed. It is later :-)

Michael Greinecker

@robjohn I need a clock that shows later.

Parth Kohli

2:40 PM

@Michael that was easy. Thanks!

I know that this question is too ambiguous, but here: How did you become a Math guy?

Michael Greinecker

@ParthKohli Actually, I'm an economist. But I like thinking a lot.

Parth Kohli

@Michael: Great, thinking is to mind what exercising is to the body :-)

By the way, the above question was for all of you!

Jayesh Badwaik

@ParthKohli I don't have much time and have to go for dinner but my personal advice is probably start doing math and always try to do challenging stuff. In mathematics ,the quote "The work will teach you how to do it." cannot be overstated enough.

Parth Kohli

Of course, I do a lot of Math!

advice taken

Jayesh Badwaik

Then you are good! And patience is second quality I guess which is very important.
Anyway, really have to go now, see you later.

Parth Kohli

2:52 PM

Thank you, @Jayesh.

Jaydon Zhao

3:51 PM

if $n = \frac{a}{b}$ where n, a, b are integers, is $n$ a rational number, or does $gcd(a, b)$ have to equal $1$?

anon

@JaydonZhao a/b is a rational number. it can also be a natural number. It is not necessary for gcd(a,b)=1 for a/b to not be a natural number: it is only necessary that b not evenly divide into a (which is not equivalent to their being relatively prime; consider nonreduced fractions, like 10/6).

Parth Kohli

We consider $\gcd(a,b) = 1$ in certain proofs by contradiction. For example, prove that square root of 2 is irrational.

Jaydon Zhao

okay, thanks :)

Gustavo Bandeira

I THROW [STACK OVERFLOW]( math.stackexchange.com/users/8671/graphth) ON THE GROUND!!!!!!

Michael Greinecker

4:12 PM

@GustavoBandeira ?!?!??

Jayesh Badwaik

@GustavoBandeira ?????

Gustavo Bandeira

I THROW STACK OVERFLOW ON THE GROUND!!!!!

Now it's corrected. =)

Michael Greinecker

@GustavoBandeira ?!?!?!!

Gustavo Bandeira

Just look the links again.

I was in a tablet, then I couldn't type precisely.

Michael Greinecker

@GustavoBandeira I can read. So ??!?!??!?

Gustavo Bandeira

4:25 PM

Nothing.

I'm out.

Peace.

Parth Kohli

4:58 PM

Hello there.

Noah

Hello

Sosi Kun

hello guys!

Parth Kohli

Hey, @Noah and @SosiKun.

Sosi Kun

could someone tell me of some very good resources to study algebra and calculus?

Noah

@SosiKun try the Khan Academy

Sosi Kun

5:05 PM

i'm sorry if this is covered in a question somewhere, but I can't seem to find a reference URL and I wonder if this was covered somewhere

Noah

@SosiKun khanacademy.org

Sosi Kun

@Noah, thanks, I'll check it out!

do you know of any other? :)

Noah

I think this is one of the best

it's vid based

and is well known

Sosi Kun

mhm :)

thanks so much!

i'm checking it out now!

Sosi Kun

5:19 PM

wow man, that's a really cool website!

Parth Kohli

It actually is, yeah.

Sosi Kun

thanks guys, I'm off! Thanks for the input

Jonas Teuwen

6:05 PM

@robjohn I have!

@robjohn A couple of people e-mailed me which made me feel somewhat actually liked here 8-).

5

And since that is for most people quite a rare feeling I decided to embrace it. Kinda.

4

robjohn

@JonasTeuwen Of course you are liked here. I could not find the reason you were suspended. Was it just chat or the main site?

Jonas Teuwen

@robjohn I said the following, I believe: "We need less of these Sovjet Russia jokes and more gay themed ones!" (a joke with reference to the statement we need less of the last ones...)

4

I am unable to see something offending in that...

Of course, cultures are different, but that might quite well be one of the least offending jokes you could make in this country 8-).

robjohn

@JonasTeuwen touching any sensitive subject (gays for instance) can set some people off

Jonas Teuwen

It never occurred to me as a sensitive subject.

But maybe people might think this country is really crazy allowing euthanasia, abortion and even gay marriage.

robjohn

@JonasTeuwen for some people it is, and so I usually refrain from commenting on it

Jonas Teuwen

6:10 PM

Yes, but I still don't see it as a reason to flag. Who here would do that? They make much harsher statements!

And that is clearly a joke. Not even containing a deadly disease (which I would understand).

robjohn

@JonasTeuwen definitely there are a lot of people in the US that would find those awful.

Jonas Teuwen

@robjohn Yes, I know.

(here too, but way less, I suppose as there is almost no political party against it)

robjohn

@JonasTeuwen there are a lot of visitors here. As I said, I could not find any information (not that I could tell what I found if I did)

Jonas Teuwen

Can also not signed-in users flag?

And who the heck would approve that.

If you get suspended for something like that!

robjohn

@JonasTeuwen you mean unregistered users? I don't know. I wouldn't think so.

Jonas Teuwen

6:12 PM

That's ridiculous. Sure, it might be sensitive for some people, but I never targeted anybody directly so they should just mind their own business (I think).

At least! I would not feel very comfortable around somebody like that!

But oh well, happened. Fine!

I'm okay. Back to maths.

robjohn

@JonasTeuwen That is the best strategy :-)

Jonas Teuwen

But I do wonder what the stereotype Dutch looks like to Americans :-)).

Americans are fat, ugly and stupid. Also very religious. (Don't suspend me!)

John Junior

@JonasTeuwen I'm glad you're back :)

Jonas Teuwen

@JohnJunior Hello. John.

robjohn

@JonasTeuwen I don't know of any stereotype for the Dutch. I generally think of windmills and sabots when I think of Holland.

Ed Gorcenski

6:14 PM

I think of the scene from Austin Powers.

Jonas Teuwen

@robjohn Those are actually here!

But a bit rare, though 8-(.

But there are some really classical Dutch villages where you would see those 8-).

Ed Gorcenski

"There's only two kinds of people I can't stand: those intolerant of other cultures, and the Dutch"

Jonas Teuwen

Does that mean that the Dutch are not intolerant of other cultures?

I beg to differ!

They are tolerant as far as other cultures does not influence theirs in their own country.

Ed Gorcenski

I think it was mostly juxtapositional irony :)

Jonas Teuwen

For example immigrants demanding restriction of freedom of speech, or more strict drugs rules and so on.

They turn into like war machines if that might happen. Not so tolerant 8-).

And for the rest, other cultures are interesting in the way looking at monkeys is. It is fun to see, but you would not want to be in their cage.

Monkey.

I wonder how people look at the "USA for Africa" song nowadays.

To me it looks very sincere, but also a bit like they seem to think there are only clay huts in Africa 8-). It is like the biggest continent on the world with people. And there are like big cities and such. And hospitals and universities. Crazy eh...?

This is a weird country actually: the political leader goes on a bike to work 8-).

user19161

6:31 PM

Hello bros!

John Junior

hi

user19161

@JonasTeuwen This sounds pretty offensive to others! But I am OK.

Jonas Teuwen

@WillHunting Hey. Sexy thing.

@WillHunting That's the stereotype!

All Singaporians... uh like to hang foreigners for throwing a can on the floor!

user19161

To me, most humans are rather stupid and evil, I am sick and tired of this world.

Jonas Teuwen

@WillHunting But you believe in reincarnation, right?

So, if you die.

You will end up in the same shit again.

So do not die.

user19161

6:33 PM

I am especially sick of the people around me. And I have been abused by many people in this life.

user19161

One thing most people lack is...empathy.

Jonas Teuwen

@WillHunting They are not worthy, so don't give them the attention they do not deserve.

Gustavo Bandeira

@WillHunting Learn to ignore. If you don't, you're gonna kill yourself.

John Junior

@WillHunting In other words "grow-up."

Jonas Teuwen

@JohnJunior Practice what you preach! 8-).

user19161

6:39 PM

I am not sure what this question is anymore! math.stackexchange.com/questions/198641/…

user19161

It is a disaster even after edits.

user19161

There are at least three different intentions in there.

John Junior

@JonasTeuwen Preach what you practice! 8-)

Jonas Teuwen

@JohnJunior So I should tell people to talk shit...?

Why?!

user19161

Oh, I realized that after I reached 2k, the system no longer prompts me to comment when I downvote!

John Junior

6:41 PM

@JonasTeuwen I didn't mean it personally :)

I was just playing with words.

Jonas Teuwen

@JohnJunior Neither do I.

user19161

Have any of you read any LaTeX book? Not tutorial, a real book.

user19161

I am looking at the LaTeX books on Amazon.

user19161

I guess the two best choices seem to be Kopka and Gratzer.

user19161

But I think LaTeX 3 will be out in a couple of years, so maybe I will postpone getting any book now even though it will still be mostly relevant.

Peter Tamaroff

6:47 PM

@WillHunting I answered that one!

Jonas Teuwen

@WillHunting Don't get a book for bloody software, just start typing!

user19161

@PeterTamaroff Yeah. I don't understand why people are doing analysis when they don't even know induction.

Jonas Teuwen

If you wonder "could this be done easier" start running Google.

user19161

@JonasTeuwen Yeah I know you would say that!

user19161

It is quite weird that I see questions asking very basic things when the asker seems to be doing a way more advanced course.

user19161

6:49 PM

They are trying to fly before being able to walk.

Jonas Teuwen

@WillHunting Ask the easy questions.

Peter Tamaroff

@WillHunting Bitches be crazy.

Jonas Teuwen

Perhaps they only figure out they don't know this by trying to answer harder questions: good learning process.

@WillHunting YES! DO IT! TYPE SOMETHING.

Something you know already. Or would like to know, make your own notes - in LaTeX.

And google for answers.

Okay, I need a beer, see you guys!

user19161

@JonasTeuwen OK, have a BIG one!

Peter Tamaroff

@JonasTeuwen I'm beginning to think you have a drinking problem.

Or maybe I have one.

I drink too little!

user19161

6:51 PM

@PeterTamaroff beginning

Jonas Teuwen

@PeterTamaroff I have a drinking problem now.

As there is no beer, that is why I am going. Get it? Good.

user19161

@PeterTamaroff I drink almost none. I have tea.

Peter Tamaroff

@WillHunting Well, you're a Singaporean. Only the Devil knows what you drink.

Has anyone used Google's Goggles?

user19161

@PeterTamaroff Nope.

user19161

Even my swimming goggles don't work too well.

Peter Tamaroff

6:55 PM

@WillHunting You take a pic of something and Google searches for matches of it, like brands and similar stuff.

user19161

@PeterTamaroff Oh, google search by image, I know about it.

user19161

@spareoom I see you are in this room too!

Spare Oom

It's a total accident. I wondered where the ping came from.

John Junior

@WillHunting Go to the google image search page and click on the "camera" beside the search space.

Spare Oom

@WillHunting As I clicked the EL&U when I signed in, someone here talked just as I clicked, so I ended up here and apparently forgot to leave.

user19161

7:00 PM

@SpareOom Oh, OK. Back to the other room then.

user19161

@JohnJunior Technology is so advanced. I wonder what mathematics is involved in this.

user19161

Definitely heavy analysis.

John Junior

@WillHunting Face/pattern recognition?

Peter Tamaroff

@JohnJunior I use the camera of my phone!

user19161

@JohnJunior Yes.

Peter Tamaroff

7:04 PM

@WillHunting Yes, vectorized images and stuff

user19161

@PeterTamaroff My phone has no camera!

Peter Tamaroff

Database searches.

John Junior

They even have tune recognition now...

Peter Tamaroff

@JohnJunior Sony Ericsson had that way back

user19161

@JohnJunior have

John Junior

7:07 PM

^too late, I beat you to it :)

user19161

Next time, Junior!

John Junior

Keep trying :-D

user19161

Now we sound like Indiana Jones!

John Junior

In the Temple of Doom?

user19161

Well, the pop is always calling the son Junior.

John Junior

7:09 PM

Later pop ;-)

leo

some help on:

15 hours ago, by leo

I've proved that given an open bounded set $A$, there exist a function $f:\Bbb R^n\to \Bbb R$ such that $f$ is $C^\infty$ and is positive over $A$. So I've tried take $g:\Bbb R^n\to \Bbb R$ of class $C^\infty$, positive over $U\setminus K$ which is open (and bounded without loose) and then consider $f=\chi_K+g$. This function is identicaly $1$ over $K$, but it can fail to be $C^\infty$

I'm stuck

Peter Tamaroff

@leo I saw this today. It's beyond my ken!

leo

@PeterTamaroff thanks man!

no problem

I guess @robjohn have some experience on it

Peter Tamaroff

@leo But, what is the original problem, what are you trying to accomplish?

leo

16 hours ago, by leo

Given a compact set $K$ of $\Bbb R^n$ and an open nhbhd $U$ of $K$, I'm trying to prove that there is a $C^\infty$ function $f:\Bbb R^n\to \Bbb R$ so that $f_{|K}\equiv 1$ and the support of $f$ is contained in $U$.

Peter Tamaroff

7:20 PM

@leo =P OK.....!

@leo The support of $f$ is $\{x:f(x)\neq 0\}$. OK

leo

@PeterTamaroff yes

Peter Tamaroff

@leo And you not the restriction of $f$ to $K$ by $f_{\mid K}$?

I usually use $f\mid K$

leo

some authors says that the support is the closure of the above set

@PeterTamaroff right

Peter Tamaroff

@leo Did I tell you I refubrished my version of Calculus of Tom Apostol?

leo

@PeterTamaroff como?

Peter Tamaroff

7:25 PM

@leo refubrished = refaccionar

(O algo asi)

leo

@PeterTamaroff ahhhh, no

@PeterTamaroff how it looks?

Peter Tamaroff

@leo just a sec =)

@leo Hmmm things are slow...

leo

@PeterTamaroff it looks nice!

Peter Tamaroff

@leo Yes, I had the title and stuff printed in gold (I didn't know it was actual gold till I got it). I really wanted it not to wear away.

that one is nicer.

leo

yes indeed

Peter Tamaroff

7:32 PM

@leo I need to get a better hold of what uniform contiunity is

John Junior

@PeterTamaroff Would you recommend Apostol in preperation for Spivak

Peter Tamaroff

@JohnJunior Both are "advanced" calculus texts, I guess. I like them both.,

When I say "advanced" I just mean they are very rigorous, and particularily, Spivak has very challenging exercises. Spivak is a little more "user friendly", IMO.

And he made me laugh once or twice!

leo

@PeterTamaroff there is a book of Spivak where his hint to one of the exercises is Sugerencia: hay una forma fácil de hacerlo.

Peter Tamaroff

@leo LOL Really, Spivak?

John Junior

translation please

leo

7:42 PM

@PeterTamaroff yep. I don't remember which book. Calculus or Calculus on Manifolds

Peter Tamaroff

@leo I remember one when he says something like: "Si se esta tardando demasiado en resolverlo, algo esta mal." o "Dése una demostracion de una línea".

@JohnJunior "Hint There is an easy way to solve it."

John Junior

@PeterTamaroff Thanks :)

leo

@PeterTamaroff yes, things like that. "only if absolutely necessary, use exercise XX"

Peter Tamaroff

@leo This one is priceless:

John Junior

What language was Spivak originally written in?

Peter Tamaroff

7:47 PM

@JohnJunior English.

leo

@PeterTamaroff ja ja yes

Peter Tamaroff

"En adelante ya no habra mas mentiras."

@leo conoces la identidad de Boya?

Ayman Hourieh

@leo Urysohn's lemma allows you to construct a continuous function with your requirements under a locally-compact Hausdorff space. I wonder if its proof can be amended to produce a $C^\infty$ function in $\mathbb{R}^n$.

Peter Tamaroff

$$2\pi = {e^\gamma }\prod\limits_{n = 2}^\infty {\exp \left( {\frac{{\zeta \left( n \right)}}{n+1 \choose 2}} \right)} $$

leo

@PeterTamaroff algo con factoriales

Peter Tamaroff

7:53 PM

@leo Es esa de ahi!

@AymanHourieh Hey. I wonder if you can help me grokke uniform continuity.

Ayman Hourieh

@PeterTamaroff Hello! Nice book you have there.

Sure.

leo

@AymanHourieh I think the objective of the exercise is not to appeal to Urysohn's Lemma. I almost finishing it. It is a quite long exercise. All start with bump functions

Peter Tamaroff

@leo You're finishing? Good!

@AymanHourieh Yes. A nice lady refubrished it.

leo

@PeterTamaroff only that part is left

Ayman Hourieh

@PeterTamaroff So what about uniform continuity?

leo

7:59 PM

@PeterTamaroff Profesor Boya was in here teaching some courses

past year he teach Álgebra Lineal I

creo

Peter Tamaroff

@AymanHourieh Well. I know the definition allright, and can apply it. But I want to understand in less technical terms what it means.

The idea is that there is a threshold $\delta$

And whenever $x$ and $y$ are withing that threshold, then $f(x)$ and $f(y)$ will be $\epsilon$ close.

And $\delta$ works for any pair $x,y$ of the interval $A$ where $f$ is UC, right?

(for a given $\epsilon$)

Ayman Hourieh

Have you studied sequences and series of functions? The power of uniform convergence is evident there as it ensures that properties like continuity and integrability are preserved by the limits of sequences and series of functions.

Peter Tamaroff

@AymanHourieh Yes, I've studied that. A while ago though. But I'm talking about uniform continuity, though.

Ayman Hourieh

Ah, sorry. Should have read that more carefully. Yes, your understanding is correct.

Peter Tamaroff

@AymanHourieh So the idea is that if $y$ is in a $\delta$ nbhd of $x$ (or conversely) $f(y)$ will be in a $\epsilon$ nbhd of $f(x)$.

hhh

8:07 PM

Good LaTex command to make an Alert! -notice?

Peter Tamaroff

@AymanHourieh Now, I want to prove $x\sin x$ is not UC on $\mathbb R^+$

@hhh ??

hhh

@PeterTamaroff I want to just stress some writing out of the other writing somehow...

Peter Tamaroff

@hhh Oh, you're asking a question!

"What is a good LaTeX command to make an Alert! notice?"

hhh

@PeterTamaroff Yes :)

(for writing math so hence here)

Peter Tamaroff

@leo Alguna vez oiste de "casi cotas superiores"? Supongo que Spivak esta tratando de $\limsup$ y $\liminf$ pero con palabras extrañas. =P

Ayman Hourieh

8:15 PM

@PeterTamaroff Intuitively speaking, as the value of $x$ increases, the graph of $x \sin(x)$ becomes more spread out vertically. So no matter what $\delta$ you choose, you can shift the interval on the $x$ axis unit you find values $x$, $y$ within $\delta$, but their images aren't within a fixed $\epsilon$.

Does it make sense?

Peter Tamaroff

@AymanHourieh Oh, yes.

Ayman Hourieh

@PeterTamaroff A rigorous proof isn't very different from what I said above. Just try adding $2\pi k$ to $x$, $y$ and see what happens to $|f(x) - f(y)|$ while $|x - y|$ stays the same.

Peter Tamaroff

@JonasTeuwen

@AymanHourieh So basically no $\delta$ will work for any given $\epsilon$ (it seems)

Ayman Hourieh

@PeterTamaroff Indeed. You can always add enough $2 \pi k$ to $x$, $y$ preserving $|x - y|$ but making $|f(x) - f(y)|$ as large as you want.

Peter Tamaroff

@AymanHourieh Sweet.

I'll finish the proof, then!

Ayman Hourieh

8:26 PM

@PeterTamaroff Cool!

MeAndMath

8:36 PM

Hi everyone!

Peter Tamaroff

@AymanHourieh I think it it'd be wise to choose $x=0$ and $y\in (0,\pi/2)$ and exploit the fact that $x+2\pi n$ will always be a root and $y+2\pi n$ will grow large.

MeAndMath

@JonasTeuwen
http://www.williamsportmd.gov/images/fireworks4.jpg

Peter Tamaroff

Then I can keep $|x-y|<\pi /2$ but make $f(x)_f(y)$ really large.

MeAndMath

@PeterTamaroff Hi Pedro!I was eating something and remered you

Peter Tamaroff

@MeAndMath That sounds creepy! Do I look edible?

MeAndMath

8:39 PM

@PeterTamaroff It was dulce de leche

Peter Tamaroff

@MeAndMath Hehehe with what did you have it?

MeAndMath

@PeterTamaroff cheese

Peter Tamaroff

MeAndMath

@PeterTamaroff white cheese with no salt

Peter Tamaroff

@MeAndMath Oh, that's better.

Ayman Hourieh

8:41 PM

@PeterTamaroff It does simplify the proof. Yes.

Peter Tamaroff

@AymanHourieh So, to disprove UC I have to find one $\epsilon>0$ such that for no $\delta >0$ will $|x-y|<\delta$ imply $|f(x)-f(y)|<\epsilon$, for some (fixed) $x$ and some (fixed) $y$?

MeAndMath

@GustavoBandeira

yunone

Does anyone have any silly ways to write the number $2$ in mathematical terms? (Like $\dim(\mathbb{P}_1)$, $[\mathbb{C}:\mathbb{R}]$, $|\mathbb{Z}/2\mathbb{Z}|$...)

Ayman Hourieh

@PeterTamaroff Exactly.

robjohn

@leo What is the question?

Ayman Hourieh

8:53 PM

@PeterTamaroff What you wrote is the logical negation of the definition of UC.

MeAndMath

@Gustavo "when I came to this world,I couldn't remember anything.Today I'm Gabriela,Gabriela I am...I was born so I grew so,And I'm still, I'll always be like this:
Gabriela, Gabriela always!"

Peter Tamaroff

@yunone $\{x:x\text{ is prime and } x \text{ is even}\}$?

yunone

@PeterTamaroff Thanks, that's another good one.

leo

2 hours ago, by leo

16 hours ago, by leo

Given a compact set $K$ of $\Bbb R^n$ and an open nhbhd $U$ of $K$, I'm trying to prove that there is a $C^\infty$ function $f:\Bbb R^n\to \Bbb R$ so that $f_{|K}\equiv 1$ and the support of $f$ is contained in $U$.

2 hours ago, by leo

15 hours ago, by leo

I've proved that given an open bounded set $A$, there exist a function $f:\Bbb R^n\to \Bbb R$ such that $f$ is $C^\infty$ and is positive over $A$. So I've tried take $g:\Bbb R^n\to \Bbb R$ of class $C^\infty$, positive over $U\setminus K$ which is open (and bounded without loose) and then consider $f=\chi_K+g$. This function is identicaly $1$ over $K$, but it can fail to be $C^\infty$

Peter Tamaroff

@yunone Any birthday coming soon?

leo

8:55 PM

@robjohn I think the idea is prove this without using Urysohn

robjohn

@leo Actually, you need to prove the $C^\infty$ version of Urysohn... not too hard.

yunone

@PeterTamaroff Ha, nope. I have a friend named Thu (pronounced "two") and I like to refer to her as these in written correspondence (fb, texts, etc.)

leo

@robjohn so that's what I am trying to do. In $\Bbb R^n$

Peter Tamaroff