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

9:04 PM

@leo $f(x)=\dfrac{e^{\frac1{d(K,x)-m}}}{e^{\frac1{d(K,x)-m}}+e^{-\frac1{d(K,x)}}}$ when $x\not\in K$ and $d(K,x)< m$ where $m=d\left(K,U^C\right)$

leo

@robjohn I see. But then the previous parts of the exercise are useless here

robjohn

@leo Oops, did I do this too quickly?

leo

@robjohn no. I think I should post this on main so you can understand

robjohn

@leo What am I missing?

leo

@robjohn no nothing. This works just find, the problem is that it seems like if the actual exercise I'm trying to solve is designed to proof the $C^\infty$ version of Urysohn Lemma by other approach

robjohn

9:11 PM

This only does it when one of the sets is compact...

leo

@robjohn yes

I have that

robjohn

@leo But it can be extended by looking at bigger and bigger compact sets in $\mathbb{R}^n$

Peter Tamaroff

@AymanHourieh Done!

Wanna see it?

Ayman Hourieh

@PeterTamaroff Sure.

Peter Tamaroff

@AymanHourieh What is the veredict?

Ayman Hourieh

9:23 PM

$h = \pi \neq 2\pi k$ can also be problematic. Looks good otherwise.

Peter Tamaroff

@AymanHourieh Hmm, yes, I mean that =D

@AymanHourieh I meant such that the function doesn't evaluate to zero.

So I should've said $h\neq k\pi$

Yeah

Ayman Hourieh

Yeah, you can even restrict it to $h < \frac{\pi}{2}$.

Peter Tamaroff

@AymanHourieh I have another exercise now, which I suspect is about $\limsup$, but the author uses other words.

Ayman Hourieh

@PeterTamaroff Upper limit? Outer limit?

Peter Tamaroff

He says "a number $x$ is called almost an upper bound of $A$ if there exists only a finite number of elements of $A$ for which $y\geq x$. Similarily we define almost a lower bound".

Suppose $A$ is an infinite bounded set. Prove the set $B$ of all almost upper bounds of $A$ is nonempty and bounded below.

We then define $\overline{\lim}A=\inf B$, and call it the limit superior of $A$.

Ed Gorcenski

9:29 PM

That's not really equivalent to lim sup

Ayman Hourieh

@PeterTamaroff Seems applicable to any set, not just sequences. First time I see it.

Peter Tamaroff

@EdGorcenski Sure?

This is Spivak talking!

Ed Gorcenski

Actually, I said that before your second line showed up. You closed your quotes so I thought you were done.

Peter Tamaroff

@EdGorcenski Oh, OK.

Ed Gorcenski

I still think that's strange... at first glance it suggests that the supremum of all subsequential limits of $A$ is less than some elements in $A$.

If... by limit superior he means $\limsup$

Ayman Hourieh

9:32 PM

@PeterTamaroff Actually, yes, you can prove that $\limsup$ is the only number with this property.

This is theorem 3.17 in Baby Rudin

Peter Tamaroff

@AymanHourieh With what property? The least of all the almost upper bounds of $A$?

@AymanHourieh Let me see.

Ed Gorcenski

Though, I'm thinking of increasing sequences... it makes much more sense for decreasing sequences

Peter Tamaroff

@AymanHourieh It is only the very worded definition that is a little baffling.

Ed Gorcenski

If you have a monotonically decreasing sequence, then every $a_n$ is an almost upper bound... the limit of any subsequence is smaller than that $a_n$

Of course, the sequence need not be monotonically decreasing... but I believe it may not be monotonically increasing.

Ayman Hourieh

@PeterTamaroff Baby Rudin also uses a non-constructive approach to $\limsup$. One can prove that both definitions are equivalent.

Ed Gorcenski

9:35 PM

I like Rudin's approach to limsup. I find his exclusive use of limsup in the ratio/root test definitions infuriating.

I mean it makes sense, it just seems to introduce an extra step

user19161

Rudin does not even call it limit superior. He calls it upper limit!

user19161

Rudin's a strange guy...

Ayman Hourieh

In Big Rudin, he uses the constructive approach to $\limsup$!

Ed Gorcenski

I like his treatment of it though; it makes more sense to look at $\limsup$ in a "nested" way

Ayman Hourieh

I really enjoy his books though. Still studying through Big Rudin.

user19161

9:40 PM

Grandfather Rudin is good too!

Ed Gorcenski

I didn't understand $\limsup$ at all the first time I encountered it

user19161

Hi @meand!

MeAndMath

@WillHunting Hello Willie!!!

user19161

I saw that again you were the victim of the moving target!

MeAndMath

@WillHunting yeah...

user19161

9:42 PM

@MeAndMath There are usually three victims. A wants to ping B but pings C instead. So the victims are ABC.

MeAndMath

@WillHunting Wanted help in something,but it is so simple that I'm embarrassed...

user19161

@MeAndMath I only know very simple stuff for now. I forgot everything!

MeAndMath

@WillHunting funny!

user19161

@MeAndMath You spelled "embarrass" correctly with double r and double s, very good!

Peter Tamaroff

@MeAndMath Hit me with your best shot.

MeAndMath

9:45 PM

@WillHunting A star for me!

Peter Tamaroff

@MeAndMath A smiley only. Don't get pretentious.

MeAndMath

Hit me with your best shot

Peter Tamaroff

@MeAndMath Why the '90s are awesome: Part I

@MeAndMath Did you see the Musical (or Movie) "Rock of Ages"?

Those are both interpreted there.

MeAndMath

@PeterTamaroff not yet

user19161

@meandmath You know, I usually don't know slang terms, but I just looked up Urban Dictionary and discovered the meaning of "willie"!

Peter Tamaroff

9:49 PM

@MeAndMath You should. I was so ecstatic I hugged the main character when the play finished.

user19161

@PeterTamaroff Pedro is so emo.

Peter Tamaroff

@WillHunting That's why we all crack up when we see "Willy Wong"!

@WillHunting Wait, what?

user19161

@PeterTamaroff You mean Willie.

MeAndMath

@WillHunting I don't know these terms either.That's why sometimes I may offend people without inttention

Peter Tamaroff

@WillHunting Potato Pohtato

user19161

9:51 PM

@PeterTamaroff Nothing, hug=emo.

rrampage

Hi everyone, I am a new user on the site. I would like advice on whether to post the link of a question I had asked some time back. It has not received any answers yet. I have read the chat room etiquette post about that. So pls advise.

MeAndMath

@WillHunting so,what's Willie?

Peter Tamaroff

@WillHunting Huh?

user19161

@PeterTamaroff Well, just short for "emotional", not a bad thing.

Peter Tamaroff

@WillHunting I know that it means, but I don't see how it relates here.

user19161

9:52 PM

@MeAndMath Erm, are you trolling me?

user19161

@PeterTamaroff I just make random associations all the time. Someone who hugs is emotional. QED.

anon

@rrampage go ahead; it's unlikely that chatters will be interested in your particular question however

MeAndMath

@WillHunting No,I am not.Something people must know:I'm not funny at all.

Peter Tamaroff

@rrampage Just ask.

user19161

@MeAndMath Erm, I am scared to write the meaning here. You may look it up yourself...

Peter Tamaroff

9:53 PM

@MeAndMath It means "dong".

user19161

@PeterTamaroff HAHAHA

rrampage

Integer weights used to cover all numbers from 1 to N with redundancies in case of breakage The problem is related to the famous Bachet's problem.

MeAndMath

@WillHunting Oh dear...

user19161

@MeAndMath HEHEHE

MeAndMath

@WillHunting such kind name...people are twisted...

user19161

9:55 PM

@MeAndMath Well, nothing bad, it's natural to think about such things...

anon

@rrampage What does it mean for a positive integer to measure other numbers?

rrampage

@anon , imagine measuring 10 kg with 2 weights of 5 and 15 kg. Its analogous to that. Basically, if a and b are weights, we can measure a+b and a-b as well

@anon think of it as a physical weighing balance.

MeAndMath

@WillHunting people only think about that...

user19161

@MeAndMath Well, and math too of course!

Peter Tamaroff

@rrampage So basically "weighing" means "writing as the sum of other integers."?

rrampage

10:03 PM

@anon yes

anon

I am not Peter..

user19161

OMG!

Jonas Teuwen

@PeterTamaroff Solved the drinking problem.

rrampage

@anon sorry.

Peter Tamaroff

@JonasTeuwen Do tell.

Jonas Teuwen

10:04 PM

It is solved.

user19161

@JonasTeuwen What is the solution?

user19161

Must be a whisky!

Jonas Teuwen

Beer.

MeAndMath

@JonasTeuwen Glad you're back!!!How are you?

Peter Tamaroff

@JonasTeuwen You should get some Artie Shaw with that.

Carioca, maybe.

►

Jonas Teuwen

10:11 PM

@MeAndMath Uh, well. Fuqdup?

leo

@robjohn, this is my problem

Peter Tamaroff

@JonasTeuwen Why?

Jonas Teuwen

@leo Yes, easy.

Even drunk.

MeAndMath

@JonasTeuwen excuse me,what's fuqdup?

Jonas Teuwen

@MeAndMath I don't know.

leo

10:14 PM

@JonasTeuwen 1.-4. are easy. prov 5. by using the others: don't know

Peter Tamaroff

@MeAndMath "Fuq'd up"

It is probably Klingon for "In trouble."

MeAndMath

@PeterTamaroff yeah ...probably..

Jonas Teuwen

Haha, I saw a cool letter from the tax service to one of my colleagues... Apparently somebody before him registered as "Master Yoda".

I wonder how he did that...

Peter Tamaroff

@JonasTeuwen He clearly used the force.

MeAndMath

@PeterTamaroff jIyajbe'

Jonas Teuwen

10:18 PM

"We need Master Yoda to sign this otherwise we cannot... wait wtf?"

Peter Tamaroff

@MeAndMath LAWL Babylon has a Klingon to English dicc.

MeAndMath

Babel fish could exist...

Peter Tamaroff

@MeAndMath ¿

MeAndMath

@PeterTamaroff you never read hitchhker's guide to the galaxy?

Peter Tamaroff

@MeAndMath No. Highways are good enough for me.

MeAndMath

10:23 PM

The Hitchhiker's Guide to the Galaxy

The Hitchhiker's Guide to the Galaxy is a science fiction comedy series created by Douglas Adams. Originally a radio comedy broadcast on BBC Radio 4 in 1978, it was later adapted to other formats, and over several years it gradually became an international multi-media phenomenon. Adaptations have included stage shows, a "trilogy" of five books published between 1979 and 1992, a sixth novel penned by Eoin Colfer in 2009, a 1981 TV series, a 1984 computer game, and three series of three-part comic book adaptations of the first three novels published by DC Comics between 1993 and 1996. T...

rrampage

@MeAndMath Hitchhiker's Guide is the single best piece of sci-fi comedy I have read. If it weren't for the chance of getting banned, I would answer every question with 42

MeAndMath

@rrampage AMAZING!

@rrampage what's the name of the poem no one likes to hear?

rrampage

@MeAndMath : Oh freddled gruntbuggly/thy micturations are to me/As plurdled gabbleblotchits on a lurgid bee.
Groop I implore thee, my foonting turlingdromes. And hooptiously drangle me with crinkly bindlewurdles,
Or I will rend thee in the gobberwarts with my blurglecruncheon, see if I don't!

remembered Vogons. searched wiki for it

MeAndMath

@rrampage hahahahahahahahahha!

rrampage

@MeAndMath have you read all the books in the series?

MeAndMath

10:29 PM

first only..:(

rrampage

@MeAndMath oh.. you must the read 2,3,4. Have read them. The 3rd book is especially good.

MeAndMath

@rrampage I will!

I did not have the time yet...

Peter Tamaroff

10:50 PM

@JonasTeuwen Could you help me?

Ed Gorcenski

11:25 PM

Wow, someone copied a comment of mine and tried to post it as an answer.

Classy

Peter Tamaroff

@EdGorcenski Dude.

@EdGorcenski I'm trying to prove something with a seemingly difficult idea.

Ed Gorcenski

Ok

What's that?

Peter Tamaroff

@EdGorcenski Well.

@EdGorcenski Recall what we just talked about the "almost upper bound".

Now, let $A$ be a bounded infinite subset of the reals.

Let $B$ be the set of all almost upper bounds of $A$. I want to prove that $B$ is nonempty and bounded below. The first one is easy.

Since $A$ is bounded, then there is a number $\mu$ such that $x\leq \mu$ when $x\in A$.

So $\mu$ will be in $B$ either because the last inequality holds vacuously, or because $\mu \in A$.

Ed Gorcenski

Thinking

It is beer o clock and I have not yet had a beer. This might take a second

Wait

$A$ is bounded

Peter Tamaroff

@EdGorcenski Yes.

Ed Gorcenski

11:36 PM

$B \subset A$. There is a $y \in \Bbb R$ such that $y \le a$ for all $a \in A$. Therefore, $y \le b$ for all $b \in B \subset A$.

actually, we have to refine that $B \subset A$ comment, because it's not true

Peter Tamaroff

@EdGorcenski I just proved $B$ contains all upper bounds of $A$.

Ed Gorcenski

but, it is true that by definition for all $b \in B$ there are infinitely many $a \in A$ such that $a < b$.

Peter Tamaroff

@EdGorcenski Yes, yes.

Ed Gorcenski

Right, so $B$ is bounded below.

There are infinitely many $a \in A$ that are lower bounds of $B$.

Peter Tamaroff

@EdGorcenski Well, I need to prove that there is at least one. How do you prove that?

Ed Gorcenski

11:39 PM

By the definition of an almost upper bound. $b$ is an almost upper bound if there are only finitely many $a$ such that $a \ge b$. Therefore, there are infinitely many $a'$ such that $a' < b$.

So if $B$ nonempty, then it is necessarily bounded from below.

GRRR

I love when the MathJax preview crashes my edit