Mathematics - 2013-09-29 (page 3 of 4)

Kevin Driscoll

20:00

@masfenix So $e^{-\frac{1}{x}}$ doesn't work because $x e^{-\frac{1}{x}}$ is unbounded as $x \to -\infty$

Twink

So, it's not true that $1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+\frac{2}{6}-\frac{1}{6}+\cdots = \sum_{n=1}^{\infty}\frac{1}{n+1}$?

Pedro Tamaroff

@Alizter Just like you always do.

Alizter

@ped How do I always do?

Pedro Tamaroff

@Twink The right hand side is $+\infty$. Can you show the left hand side also is?

Twink

I mean, it's not so direct?

masfenix

20:01

so when you say decrease, you mean the function and the derivatives go to zero as x -> infinity? correct? But if you can find a polynomial $P(x)$ such that if you multiply this polynomial by this function (or its derivatives) then it doesn't go to zero then the function is not a Schwartz function, correct?

Twink

well if the left side is equal to the right side

Kevin Driscoll

@masfenix Correct. $e^{- x^2}$ is the prototypical example of a Schwartz function

Twink

then it is $+\infty$

Pedro Tamaroff

@Twink Well, how can you justify it?

@Alizter It is a real number, is it not? Or a complex number, say.

masfenix

Yeah so for my assignment, I have to prove that the function $\phi(x) = \exp{\frac{-1}{1-|x|^2}}$ belongs to $C_0^\infty$ for all $|x| < 1$ and is $0$ for $|x| \geq 1$.

Kevin Driscoll

20:03

Indeed, that is a Schwartz function because it is 0 everywhere outside 1

masfenix

Now so I think I have to show that this function vanishes outside its compact support.

Kevin Driscoll

infact it is an even more special kind of thing called a 'bump function' which are essentially Schwartz functions with compact support

Twink

$1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\frac{2}{5}-\frac{1}{5}+\frac{2}{6}-\frac{1}{6}+\cdots =(1-\frac{1}{2})+(\frac{2}{3}-\frac{1}{3})+(\frac{2}{4}-\frac{1}{4})+(\frac{2}{5}-\frac{1}{5})+(\frac{2}{6}-\frac{1}{6})+\cdots = \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots = \sum_{n=1}^{\infty}\frac{1}{n+1}$?

masfenix

so I am trying to show that first it is infinitely many times differentiable and then for each derivative, I need to show that it vanishes outside the compact set. Is my thinking correct, @KevinDriscoll

Alizter

@PedroTamaroff Apologies of my unclear portrayal of thoughts. Let me start from a basic level: How, using set theory (ZFC), would you represent an operation such as 1+2?

Pedro Tamaroff

20:05

@Twink Well, you must justify why introducing/removing parenthesis is legitimate.

Twink

that's what I'm asking you ¬¬!!

if it's possible or not

Kevin Driscoll

@masfenix The function $\phi$ that you gave is NOT 0 outside $\left|x\right| > 1$

You have to define the function to be $\phi$ inside $\pm 1$ and 0 outside

Pedro Tamaroff

@Twink There are criterions for the finitary case, IIRC. Apostol explains this.

I cannot recall the claim exactly.

masfenix

@kevin, that is exactly how it is stated in my book assignment problem

Pedro Tamaroff

One first defines what it means to "introduce parenthesis".

Twink

20:07

ok

I'm gonna look for it

Pedro Tamaroff

@Twink Let me find the section of the book.

Kevin Driscoll

@masfenix I mean if you just plug in $x =2$ you don't get 0

Pedro Tamaroff

I think it is in his "Mathematical Analysis".

Twink

I don't have that book

Pedro Tamaroff

@Alizter Via sets.

Twink

20:08

but I have some other good books

masfenix

@KevinDriscoll, its defined to be 0 for all $|x| \geq 1$.

Twink

which surely include the criterion

Alizter

@PedroTamaroff Would you care to demonstrate?

Fernando Martin

@Alizter: Do you know how to construct $\mathbb{N}$?

Kevin Driscoll

@masfenix Ah yes, precisely. The way you phrased it didn't seem to indicate that. It seemed like you were saying you wanted to show that $\exp{\frac{1}{1-\left|x\right|^2}}$ is 0 outside $\pm 1$

Pedro Tamaroff

20:10

@Alizter If $S$ is a set, define $S^+=S\cup \{S\}$.

Alizter

@FernandoMartin Yes

masfenix

Oh no, sorry so yeah I am trying to show that the derivatives are all polynomials * the original function.

Pedro Tamaroff

Then $0=\varnothing$. $1=0^+$; $2=1^+$, &c.

masfenix

and that it vanishes outside the compact set , but I am not sure how

Kevin Driscoll

@masfenix So yes, you know that away from $\pm 1$ the function and all teh derivatives exist and are continuous so you need to concentrate on the points $\pm 1$ and show that the function and all its derivatives go to $0$ there

masfenix

20:11

I see. okay, cool. I will work on that

Any hints? I found this which kinda answers my question math.stackexchange.com/questions/476195/…

Alizter

OK but what am I doing when I say 1+2?

masfenix

but I was wondering it there another way

Pedro Tamaroff

@Twink Here

Kevin Driscoll

In which case you have shown that the function that is $\phi$ side and $0$ outside is $C^{\infty}$

Twink

thank you

I thought it was in chinese xD

Kevin Driscoll

20:14

@masfenix Sorry, I don't think I could provide a proof that is sufficiently rigorous

Pedro Tamaroff

@Alizter We define $n+1=n^{+}$

And $n+m^{+}=(n+m)^{+}$.

@Alizter Do you know about the principle of recursive definition?

Alizter

yes

Pedro Tamaroff

@Alizter Well, that's how you define sum and product of natural numbers.

masfenix

@KevinDriscoll thankyou for your help

Kevin Driscoll

@masfenix no problem

Alizter

20:24

@ped what I don't get is that using your definition $2=1^+=\{\varnothing\}\cup\{\{\varnothing\}\}=\{\{\varnothing\}, \{\{\varnothing\}\}\}$

Twink

I'm gonna start a bounty for this question math.stackexchange.com/questions/507508/… in 6 hours

Pedro Tamaroff

@Alizter Dude, your set is $\{\varnothing,\{\varnothing\}\}$.

Not what you wrote there.

$A\cup B=\{x:x\in A\vee x\in B\}$.

Alizter

@PedroTamaroff OK that was confusing me

thanks for clearing that up

Fernando Martin

@Alizter: I like to think it as 2 = {0, 1}

Then $\mathbb{N}$ is well ordered with respect to $\in$

Pedro Tamaroff

@FernandoMartin It is that.

$0=\varnothing,1=\{\varnothing\}$!

Fernando Martin

20:28

I know!

Pedro Tamaroff

@FernandoMartin I know you know.

Kevin Driscoll

I know both of you know

Pedro Tamaroff

Why "I like to think it..."?

Kevin Driscoll

but I didn't know

Fernando Martin

But it's more natural (for me) to type $2=\{0,1\}$ than $\{\emptyset, \{\emptyset\}\}$

Pedro Tamaroff

20:28

@KevinDriscoll I know you know we know I know he knows.

Twink

Pedro I know you know Kevin knows you both know

Kevin Driscoll

Yes but did you all know that I didn't know that you all knew that I knew that both of them knew?

Alizter

@PedroTamaroff is the : in your set builder the same as |

Pedro Tamaroff

@KevinDriscoll I am cracking up right now.

Twink

I know that I know nothing

Kevin Driscoll

20:30

just fyi

@pedro If you can trick your opponent into anticipating your next statement by jumping 1 step ahead, you can then force them into a contradiction with that question

@alizter, I think so. Its 'such that'

Fernando Martin

Btw, the current definition of ordinals, which is exactly this thing (transitive sets well-ordered by inclusion) is due to von Neumann, when he was 19 years old!

19!

Kevin Driscoll

Ya, but when did he first bed a woman??? :-P

Pedro Tamaroff

@FernandoMartin You told me that, let me star it to make us all feel like shit.

Fernando Martin

He was quite a player IIRC.

Twink

@KevinDriscoll how do you know he wasn't gay?

Kevin Driscoll

20:33

I don't know why that makes you all feel bad. I wouldn't trade what i was doing at 19 for von Neumann's life

Alizter

@PedroTamaroff I am having trouble with $n+m^+=(n+m)^+$

Danny

the defintion of $lim sup_{n\rightarrow\infty} tn$ the supremum of all partial sums?

Ted Shifrin

@Pedro: I saw your roar. I wonder if I can switch to Russian in here :)

Fernando Martin

@KevinDriscoll: You must be really interesting then! I'd certainly would

Pedro Tamaroff

@TedShifrin Come againz?

Kevin Driscoll

20:34

@FernandoMartin ...... I'm the most interesting physics PhD student in the universe....

Twink

@TedShifrin I don't like Russia and you must know why

Pedro Tamaroff

@KevinDriscoll I wouldn't trade my life for another person's life, but I'd certainly enjoy some changes in favour of my math, I guess.

Ted Shifrin

Regarding my discussion with @Fernando earlier. But then thinking about typing our names.

Pedro Tamaroff

@TedShifrin You know Russian?

Danny

privet!

Pedro Tamaroff

20:35

@Twink Don't generalize.

Ted Shifrin

yes, @Twink, I don't either. But my heritage is still 75% Russian, 25% Polish.

Twink

I said Russia not Russians @PedroTamaroff

Danny

ted so u can speak russian

Kevin Driscoll

@TedShifrin I came up with 2 possible definitions for $$F.P. \int_a^b \frac{1}{\left| x - c\right|} dx$$ but I'm not sure if they are equivalent or consistent

Ted Shifrin

Yes, @Pedro, a bit. Studied it in college a little bit.

Ethan

20:37

@TedShifrin why

Pedro Tamaroff

@TedShifrin Cool. I have been told it is not easy.

Kevin Driscoll

I've heard Russian still uses CASE markers and word order is relatively free..... sounds like a mess

Twink

I don't like Putin and his stupid laws

that's it :D

Fernando Martin

@PedroTamaroff: Has Fava written something in Russian yet? :D

Kevin Driscoll

No one likes Putin, except for perhaps Putin himself

Danny

20:37

harasho

Pedro Tamaroff

@FernandoMartin Nah. Pity.

Ethan

I get all my news from the colbert report

Ted Shifrin

Not much anymore, @Danny. Even my once fluent French is rusty, same for German :(

Danny

bljath

lol

Kevin Driscoll

@Ethan at least then you know half of it is a joke

Ted Shifrin

20:38

6 cases in Russian, 5 in Latin, 4 in German

Ethan

lol

Ted Shifrin

@Kevin: $c \in (a,b)$, I presume?

Kevin Driscoll

@Ted indeed

@Ted I asked a question about it, but I think my definition is problematic

Fernando Martin

Dini's test is the bomb!

Pedro Tamaroff

@KevinDriscoll Cannot you remove the $\log$ singularity or something?

Fernando Martin

20:41

I guess you know it already @PedroTamaroff

Pedro Tamaroff

@FernandoMartin You mean for uniform convergence?

Fernando Martin

Yes

Pedro Tamaroff

It is a special case of a more convoluted test.

Fernando Martin

Really? How's it called?

Kevin Driscoll

@PedroTamaroff Ya thats what I did.... but I'm not sure that its justified in the same way that removing the $\frac{1}{x}$ singularity is for the traditional Hadamard Finite Part

Pedro Tamaroff

20:41

@FernandoMartin Dunno if it has a name.

Fernando Martin

What does it say then?

Kevin Driscoll

The HFP isn't just as hoc, its basically just meromorphic continution combined with contour integration. But I dont think that applies here.

Twink

Guys I'm gonna start a bounty for this question math.stackexchange.com/questions/507508/… in 6 hours, you may want to start formulating your answers :)

Pedro Tamaroff

@FernandoMartin I'd have to check.

Ted Shifrin

@Kevin: Seems to CPV doesn't exist for $\int_{-1}^1 dx/|x|$.

Kevin Driscoll

20:43

@Ted exactly

Ted Shifrin

I didn't say it should always exist! I just wanted to allow higher than first-order poles!

Danny

Dr.ted ,or anyone else can you look at this question and possible help? i dont really understand the answer i got

2

Q: closed,bounded not compact

Hi I was asked to prove that: if $S =\{ x \in \Bbb R : d(x,0) = 1 \}$ then $S$ is a closed and bounded set. The set $S$ contains only two points: $-1,1$,(it should not be a problem to prove that is it closed and bounded)but it gets really confusing when iam also asked to show that $S$ is not com...

Fernando Martin

@PedroTamaroff: Ok, no worries

Kevin Driscoll

@TedShifrin but if you can find the finite part of $\int_{-1}^{1} \frac{dx}{x^2} dx$ it seems like you should be able to do it in this case as well, because the singularity is somehow not as bad

Fernando Martin

I was just glad it popped out in my topology coursework, haha

Pedro Tamaroff

20:45

Ted Shifrin

I don't think you can there, either, @Kevin. Infinite area with no sign cancellation.

Kevin Driscoll

@TedShifrin This is the essence of Hadamard Regularization though

Pedro Tamaroff

@FernandoMartin There it is.

Fernando Martin

@PedroTamaroff: Thanks!

Kevin Driscoll

@TedShifrin Basically you treat $\frac{1}{x^2}$ as a generalized function that is equal to $\frac{1}{x^2}, \ x \neq 0$ and give it a value at $x=0$ so that the integral is well-defined

Ted Shifrin

20:49

@Danny: He's talking about metric spaces of sequences (but using superscript notation because you might want to do sequences in that metric space, which would mean sequences consisting of sequences). Maybe it helps to think of a sequence as an infinite-tuple of real numbers?

Kevin Driscoll

they explain it a bit on Math Overflow but I don't really know all that much

Danny

can we get it from the start ted

what we are considering are the supremum metric

Ted Shifrin

Hmmm @Kevin. Definitely out of the realm of Riemann integrals, so there! Let me look.

Danny

is the supremum metric right

Ted Shifrin

Yes. @Pedro can also help you with analysis, along with lots of others. :)

masfenix

20:53

Could someone help me understand this question: math.stackexchange.com/questions/476195/… Notice that in the main original question, the function is defined at $|x| < 1$ and 0 $|x| \geq 1$. But Peter in his answer changes these conditions

Are they talking about the same thing?

Kevin Driscoll

@Ted indeed. I've come to think this question isn't so trivial that I will appear stupid or lazy if I ask my advisor about it. Although I somehow doubt he can give em an answer that would satisfy me.

Ted Shifrin

This is like renormalization you physicists do, @Kevin. Definitely not CPV, but thinking about a Laurent series here and taking the constant term at $0$. I need to think more about the distributional derivative. But I also need to work on preparing my own grad course, so I'm tabling until I have more time.

Kevin Driscoll

@Ted Indeed. I promise that it isn't totally ad hoc in the problem I am working on. We didn't just arbitrarily decide that the integral should be interpreted in a regularized sense. It arises naturally in the derivation of the associated integral equation. Anyway, I understand. Will let you know if I figure it out.

masfenix

Hi, sorry guys did anyone get a chance to read my message or should I copy/paste it

Danny

what does he mean when he says $ e^{(k)}_ j=1$ if $j=k$

Twink

21:04

baila en la calle, de noche, baila en la calle, de dia

:D

Ethan

@robjohn can I talk to you?

Alizter

Ahh ZFC is messing with my head

robjohn

@Ethan sure

Ethan

in a separate chat?

for a second

Pedro Tamaroff

@Alizter Don't worry about ZFC now! =D

Kevin Driscoll

21:13

There are always more things to learn.

Alizter

@PedroTamaroff Thats what happens when you become curious when practicing limits. I should be studies in chaos theory. I diverged from my task so much.

Kevin Driscoll

@Alizter Shame on you! Nonlinear dynamics is so interesting

Pedro Tamaroff

@Alizter Chaos theory?

Alizter

@PedroTamaroff en.wikipedia.org/wiki/Chaos_theory

Fernando Martin

@PedroTamaroff: Know anything about adjunction spaces?

Pedro Tamaroff

21:17

@Alizter Very funny.

Fernando Martin

I'm studying them for my topology course. Really cool stuff.

Pedro Tamaroff

@FernandoMartin What are those?

Fernando Martin

Basically it's the way to formalize the "gluing" of spaces

Pedro Tamaroff

@FernandoMartin Right, guessed so.

Fernando Martin

Wedge products are a special case for instance.

Kevin Driscoll

21:21

@FernandoMartin be careful this almost sounds like.......... applied mathematics

Fernando Martin

Not products

Wedge unions

Don't know why I said products.

Pedro Tamaroff

@FernandoMartin Dunno what those are.

Fernando Martin

Hmm, I thought we talked about them back in Rosario but that might have been Germán now that I think of it.

Basically it's gluing spaces via identifying a single point in each of them

for instance the wedge union of two circles is "8"

Pedro Tamaroff

1700M CpS...... aaaaaaaaaaaaaaaaaaaaaaand its gone.

Karl Kronenfeld

@PedroTamaroff Yao to you too

Fernando Martin

21:24

hahahah

Pedro Tamaroff

@FernandoMartin Right.

Fernando Martin

Of course there's some detail "under the hood" to make everything formal but what's nice about it is that it's really geometrical

Pedro Tamaroff

@FernandoMartin I like the constructions like the cone over a space; the suspension of space and such. Quotient spaces, aye?

Kevin Driscoll

@FernandoMartin Is the direct product of two hilbert spaces an example as well?

Fernando Martin

The cone over a space is an example of adjunction space. I don't know what a suspension is.

@KevinDriscoll: Adjunction spaces are certain quotients of coproducts of spaces

When I say "gluing" I think of quotienting out so I don't think direct products fit here

Pedro Tamaroff

21:28

@FernandoMartin The cone over a space is a quotient too, $X\times[0,1]$ quote identifying $X\times \{0\}$ with a point.

Karl Kronenfeld

You're using a function to identify points of the domain with points of the target.

Pedro Tamaroff

The suspension is $X\times [0,1]$, and collapsing both endpoints, like two cones glued at $X$.

Fernando Martin

@PedroTamaroff: I see.

Pedro Tamaroff

@FernandoMartin Here.

The book I read about topology had something about it, never did the exercises though.

masfenix

I need to show that every function in the Schwartz space is in $L^p(R^n)$ any hints on how I can get started?

Pedro Tamaroff

21:30

@FernandoMartin Cool thing: Let $SX$ denote the suspension of $X$.

Then $SS^{k}\simeq S^{k+1}$.

=D

Fernando Martin

The thing about adjunction spaces is that not only they're quotients: they're pushouts

Pedro Tamaroff

@FernandoMartin Dunno what that is.

Alizter

It can be assumed but how to prove that $\lim_{u\to0}(1+u)^{1/u}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$

Danny

someone that can assist me with:

2

Q: closed,bounded not compact

Hi I was asked to prove that: if $S =\{ x \in \Bbb R : d(x,0) = 1 \}$ then $S$ is a closed and bounded set. The set $S$ contains only two points: $-1,1$,(it should not be a problem to prove that is it closed and bounded)but it gets really confusing when iam also asked to show that $S$ is not com...

real-analysis metric-spaces compactness

Fernando Martin

Some categorical construction; it's not really important

Kevin Driscoll

21:33

@masfenix well I suppose you need to use the definition of the $L^p$ norm and then show that it is bounded for any member of a Schwartz space

Fernando Martin

@PedroTamaroff: That result about the suspension of circles is really cool

Kevin Driscoll

@masfenix intuitively this should be true because all members of the Schwartz space decrease VERY quickly at infinity

masfenix

@KevinDriscoll I have no idea how to even begin showing that.

Fernando Martin

@PedroTamaroff: How did you prove it? I have a mapping $f:S^{k+1}\rightarrow SS^k$ that's continuous and bijective

Now since $S^{k+1}$ is compact

if I proved $SS^k$ is $T_2$ I'd be all set.

There was some proposition about when quotients are $T_2$

Pedro Tamaroff

@FernandoMartin I never said I proved it.

=D

Fernando Martin

21:38

Hahah

Karl Kronenfeld

@FernandoMartin On the cylinder, you need to be able to separate a face from nearby points using disjoint open sets.

Fernando Martin

Well, $SS^k$ must be $T_2$ since it's homeo to $S^{k+1}$...

Kevin Driscoll

@masfenix I'm not entirely sure.... you need to develop some kind of bound. For example for $L^p$ with $p>1$ then $\frac{1}{x^2 +1}$ is in that space (I think) @pedro will correct me if I am wrong

Fernando Martin

@KarlKronenfeld: Yes, I think that can be done. I was trying to do it as cleanly as possible.

Kevin Driscoll

@masfenix and then I guess the fact that any member of $S$ is $C^{\infty}$ and decreases faster than $\frac{1}{x^2 +1}$ also shows that members of $S$ are $L^p$ with $p>1$

Fernando Martin

21:40

By cleanly I mean without dealing with open sets and the like

Pedro Tamaroff

@KevinDriscoll Sure.

Karl Kronenfeld

@FernandoMartin Heh.. :)

Kevin Driscoll

@masfenix That is $\lim_{x \to \pm \infty} (x^2 +1) s(x) =0$ for $s(x) \in S$

masfenix

@KevinDriscoll so what is the interpretation of the limit, what is that telling us?

Kevin Driscoll

@masfenix Basically that the $L^p$ norm with $p>1$ can't diverge because it converges for $\frac{1}{x^2 +1}$ and $s(x)$ is 'smaller' than that function at infinity

Pedro Tamaroff

21:46

@masfenix That the function decreases damn fast.

Kevin Driscoll

@masfenix Sorry, I'm not a pure math person so I can't make this formal. Just trying to justify my intuition

Pedro Tamaroff

3000M CpS.

Fernando Martin

@PedroTamaroff: How on Earth...

Karl Kronenfeld

@PedroTamaroff What was it after 3 hours of play?

Fernando Martin

I've been playing for a day longer and I just have 1300M CpS

How did you do it?

Karl Kronenfeld

21:48

(I am trying to decide if my strategy actually is any good)

Kevin Driscoll

OH JESUS NO! I just figured it out

you people are playing COOKIE CLICKER

masfenix

@KevinDriscoll alright thanks, I am still pretty lost. Will work on it some more.

@PedroTamaroff could you help me formalize a proof for this? I want to show that every function in S is also in Lp

Fernando Martin

@KevinDriscoll: I'm deeply ashamed by my actions

Kevin Driscoll

@masfenix Sorry I can't be of more formal help.

@FernandoMartin I don't really understand the draw. I just worked on it a bit and then let it run in the background.....

Fernando Martin

Just wait for it...

Kevin Driscoll

21:54

@FernandoMartin No I mean this was weeks ago

Fernando Martin

Huh, well, don't know then

Kevin Driscoll

I let it run for a few days but...... I dunno it was just making more cookies

Fernando Martin

I was instantly hooked

Anything with achievements gets me hooked.

Pedro Tamaroff

@masfenix What is S?

Kevin Driscoll

@PedroTamaroff Schwartz Space

Pedro Tamaroff

21:55

@KevinDriscoll That tells me nothing.

Kevin Driscoll

@PedroTamaroff All functions that are $C^{\infty}$ and both the function and all derivatives decrease faster than any power of $x$

Pedro Tamaroff

@KevinDriscoll Darn.

So what is the fuzz?

Kevin Driscoll

@PedroTamaroff Its the natural space for doing Fourier Transforms because it leaves the space invariant

@Pedro 'fuzz' ??

masfenix

@PedroTamaroff I want to show that the every function is in Lp (or in other words, I want to show that Schwartz space is a subspace of Lp)

Transcript for

$Mathematics$ Mathematics