Mathematics - 2013-11-30 (page 1 of 3)

Don Larynx

12:00 AM

@Akshaj: HINT Rewrite the sequence.

Pedro Tamaroff

@DanielFischer And how do I obtain that compact nbhd? By local compactness?

Daniel Fischer

@PedroTamaroff Yes.

Pedro Tamaroff

@DanielFischer How do I know I can fit it inside $U_{i+1}$?

Daniel Fischer

One of the first things one proves about locally compact spaces is that every point has a neighbourhood basis consisting of compact sets, or isn't it anymore?

Sometimes, that is even the definition.

Pedro Tamaroff

@DanielFischer Ah, OK. I'm not following a course, so bear with me =)

Anyhow, got it. I basically mimic the construction for complete metric spaces.

Daniel Fischer

12:08 AM

Of course, effectively it's the same proof as $\text{compact}\Rightarrow\text{normal}$.

90intuition

Okay, it now works. So again, if you want to give it a fair judge. Try out how this looks with little bit changed chatjax script: http://jsfiddle.net/4cqVD/13/
ˊx^2+y_1+z_12^34ˊ
ˊexp(2pi*i/3)ˊ
ˊsin^-1(x)ˊ
ˊ<= => -> >=ˊ
ˊ(a,b]={x in RR | a <= x <= b}ˊ

2

Pedro Tamaroff

@DanielFischer Aha. Well, I have a checklist of things the prove.

Daniel Fischer

@PedroTamaroff Yes, it's basically the same proof. Compactness gives you what the shrinking diameter gives you for complete metric spaces, a nonempty intersection.

Pedro Tamaroff

@DanielFischer ;)

I should review for my midterm tomorrow, though. =O

@EnjoysMath Dude, WTF?

Enjoys Math

idk

Seth

12:12 AM

Who is flagging stupid things?

iKlsR

@EnjoysMath ಠ_ಠ

kalina

hello maths

Seth

Like I've always said.. If you want to start a party, cast a stupid flag.

kalina

please don't flag silly things

fire will be involved

Pedro Tamaroff

Damn, three people?

Seth

12:13 AM

4

Pedro Tamaroff

Are you flag hunters or something?

kalina

nope

Pedro Tamaroff

So you came here with popcorn in your hands?

kalina

when silly people flag silly things, we get notified

Pedro Tamaroff

Expected some action?

kalina

12:14 AM

and have to decide whether or not the saying of silly things should be silly banned

meh...

--^ that.

still...

@kalina Right, we all do.

kalina

good

now it's been a while since a silly flag came out of this silly channel

Pedro Tamaroff

12:14 AM

Still, you can stay in your room.

kalina

so I silly invalidated it

don't tell me what to do

Seth

but it got silly approved.

anon

@mick I am hungry and tired and interest is waning, so I am leaving out a lot of details.

kalina

@Seth well that's silly

Pedro Tamaroff

@kalina Just saying, you can approve and disapprove over there, without coming here and ranting about something silly.

Seth

12:15 AM

@kalina IKR

kalina

@PedroTamaroff silly context is silly everything

Seth

Ooh, I like the classy orange notifications here

kalina

@Seth rule of stack exchange: no matter how good your theme is, you still prefer somebody elses'

Braiam

....

Seth

@kalina lol.

Pedro Tamaroff

12:16 AM

@DanielFischer I looked at Rudin's book. Seems nicer than his little book.

Seth

Well since the world isn't ending, I think I'll come back when it is. Cya folks.

kalina

@PedroTamaroff I don't know how you math types roll, but in the real world, people prefer context before making assumptions

still, enjoy being silly

just do it quietly

Pedro Tamaroff

Sigh.

iKlsR

haha. I just come for the lectures.

Danny

dr pedro!!

Pedro Tamaroff

12:18 AM

@Danny No, no. Long run before Doctor.

Daniel Fischer

@PedroTamaroff The "little book" is Principles of Mathematical Analysis? That's also nice, but RCA and FA are masterpieces.

Pedro Tamaroff

@DanielFischer Yes. I feel like he put more of him in RCA, maybe he felt PMA was more of a simpler job, who knows.

Kasper

@iKlsR how did you type those symbols ?

Danny

pedro you are young and very promising , you should go for it!

iKlsR

@Kasper just google ascii emoticons.

Danny

12:20 AM

dont dissapoint me pedro!

Kasper

@iKlsR so you copy paste it ?

iKlsR

@Kasper unicode really, so If you have the time, you can combine and make your own. fileformat.info/info/unicode/char/0ca0/index.htm

@Kasper Most times, yes.

@Kasper Have fun.. gist.github.com/endolith/157796

Kasper

It's funny because I recently wrote a script that is able to type that exact symbol by keyboard. I just used that as an example:
https://github.com/sanisoft/jQuery-auto-correct/issues/2

Karl Kronenfeld

Speaking of people from other channels flooding in, what happened to twink?

Pedro Tamaroff

@KarlKronenfeld Dunno.

Maybe he just find a good tutor IRL?

@leo Heya.

=O

leo

12:28 AM

@PedroTamaroff Hello!

How are you?

Pedro Tamaroff

@leo Fuzzy. You?

Kasper

@iKlsR So if you are interested in typing those kind of unicode smileys by keyboard, you could use this: jsfiddle.net/4SWy6/19 bookmarklet.

leo

@PedroTamaroff I think I've got it. The last thing IO was stuck on. Fuzzy about what?

Kasper

ఠ_ఠ

Pedro Tamaroff

@leo Dunno, light headed.

You're studying Galois theory right?

leo

12:31 AM

@PedroTamaroff yep

@Kasper lol

Kasper

@90intuition It doesn't work.

Pedro Tamaroff

@leo I was reading about Galois' life some days ago. He was quite a maverick.

A rebel.

90intuition

@Kasper Did I fail again ? :P

Kasper

Trying to implement asciimath as well for my forum. It's quite a nice language.

leo

@PedroTamaroff Indeed. When I started uni I read El Elegido de los Dioses by Leopold Infield. Was pretty good.

Kasper

12:38 AM

I needed to restart (to turn the usual chatjax off): Now it works:
ˊ(1/2,pi/2]={x in RR | 1/2 <= x <= pi/2}ˊ

How do I write those high comma's ?

90intuition

Cool !

leo

My professor said many things of the course (probably not in its modern form), Galois said they were true in a letter he wrote the day before his dead.

90intuition

Just type two times comma's. That is part of the script.

Kasper

works. ˊ<< x,y >> in RRˊ

90intuition

Who is Gone ? =>

Pedro Tamaroff

12:44 AM

@90intuition Bill Dubuque.

90intuition

8 golden badges

and 1 point haha

Pedro Tamaroff

@90intuition Well, he had more, but now he's banned.

So his points are set to $1$ by default.

90intuition

Ah okay. Why is he banned ? Man he got lots of answers.

And zero questions. Why on earth would you ban someone like that ?

You could better ban persons like me, that only ask questions :P

He has some strange way of writing, all variables upright.

Wow this is some real drama.

Don Larynx

@Charlie Hi @Charlie

Charlie

@DonLarynx hi don

Don Larynx

12:53 AM

@Pedro I considered your series $1+q+p+q^2+p^2+\cdots$ with $q=\frac 1 3$ and $p=\frac 1 2$. I have no idea why $\liminf_{n\to\infty} \sqrt[n]{a_n} = \sqrt[2n]{\frac{1}{3^n}} = \frac{1}{\sqrt{3}}$. For Erdos sake man that doesnt make sense.

Why the $2n$? That came out of nowhere.

Pedro Tamaroff

@DonLarynx Consider all subsequential limits of the sequence. What is the smallest one?

That is, of the sequence $a_n^{1/n}$ with $a_{2n}=q^n$ and $a_{2n+1}=p^n$.

Don Larynx

oh

@Pedro: Are we dealing with sequences of real numbers in the extended real system?

@Pedro: Yes.

Pedro Tamaroff

@DonLarynx There's no need to do that here.

Don Larynx

I answered my own question a bit too soon, sorry.

robjohn

@DonLarynx what else would it be?

Don Larynx

1:08 AM

@robjohn: After reviewing the definition of upper and lower limits I considered the question.

I only thought it was the real numbers

forgetting infinity is not a real number...

anyways, did you guys know the brother of Niels Bohr, Harald Bohr, was a famous mathematician AND a soccer player? That's insane.

@robjohn too late to edit, but Rudin didn't properly define divergence before upper and lower limits.

Pedro Tamaroff

Did 5 meters in GIRP. Meh godz.

robjohn

@PedroTamaroff GIRP?

Pedro Tamaroff

foddy.net/GIRP.html

Don Larynx

Bad Pedro, bad. That's such an easy distraction lol.

How can I find a closed form of the series?

Do I just group the terms by 3's?

and then find the pattern for $a_{n}, a_{n+1}, a_{n+2}$...

?

leo

Proof by diagram!!

@DonLarynx isn't it just a rearrange of $\sum 2^{-n}$?

Don Larynx

1:25 AM

Maybe, but that changes the limit.

Pedro Tamaroff

@DonLarynx You-re flipping $\frac{1}{2^n}$ as @leo said.

@DonLarynx No, it doesn't.

leo

@DonLarynx nope

as Pedro said. Now, why not? Because the series converges absolutely.

Daniel Rust

good evening

Don Larynx

Sorry. I mean it changes the sum.

Pedro Tamaroff

I must go. Too tired. Bye.

leo

1:29 AM

@PedroTamaroff Get some rest. Sleep well

Don Larynx

@leo: to be more specific.

leo

@DonLarynx uh?

Don Larynx

I am looking for a closed form of that type

Benja

@DanielRust Hey!

@DanielRust I saw your question on TQFTs

always been wondering what the big fuss was about

leo

@DonLarynx say, are you looking for a closed form of that alternating series?

Daniel Rust

1:31 AM

@Benja they're interesting!

and fun to play with

leo

@DonLarynx or a closed form of one of its rearrangements?

Benja

I am interested in AG, is there anything in TQFT I might be interested in @DanielRust

Daniel Rust

AG?

alg geom?

Benja

yea

Daniel Rust

Hmm I dunno

They seem to be very well rooted in category theory

which is of course the playground of AG

but beyond that I don't know

Benja

1:33 AM

right. Are there links between representation theory and TQFT?

Because I am also interested in representation theory

Daniel Rust

well given the link I mentioned with Frobenius algebras, and their link to representation theory of finite groups, I'd say so

Benja

frobenius algebras?

man in rep theory they keep coming up with more and more algebras

Daniel Rust

haha

yeah tell me about it.. I'm a topologist

Don Larynx

@leo: Yes.

Benja

frobenius algebras, cherednik algebras, weyl algebras, gan-ginzburg algebras to name the few I've seen

Don Larynx

1:35 AM

@leo its rearrangements

Benja

@DanielRust You like topology?

Daniel Rust

A friend of mine works on quantuum groups.... but he lies! he told me the definition and they're really algebras

Yeah, i'm working in an area of topological dynamical systems atm

but using a lot of algebraic topology

Benja

right.

Daniel Rust

haha, what a coincidence :P

the main invariant i'm interested in is Cech cohomology

Benja

@DanielRust ah that's section 4 of hartshorne chap 3

Daniel Rust

1:38 AM

I think the way alegebraic geometers approach Cech cohomology is more abstract than how algebraic topologists approach it though

Benja

yea. Sometimes this AG makes me sick because everything is so highfalutin

@TedShifrin hey.

Ted Shifrin

Hi, @benja

Daniel Rust

I don't use much more beyond the inverse limit definition and then automatically turn everything in to a direct limit of abelian groups

Benja

you must be good at computing direct limits then.

Daniel Rust

not as good as I should be...

Hey @ted

Benja

1:39 AM

I can get away with "if every term in the limit is constant then so is the limit" :D

Ted Shifrin

Hi @DanielRust

Daniel Rust

@TedShifrin How's life? I haven't been on chat in a while

Ted Shifrin

Doing well, thanks. Thanksgiving dinner down, final exams next.

Daniel Rust

Did you take part in any black friday madness?

agent154

Anybody know of any good resources for keeping tabs on developments in cryptography and its applications? The only name that comes to mind is Bruce Schneier. Aside from obviously learning about what's going on, I'd like to maybe stumble upon something I can do further research on for my Honours Thesis in a couple of years.

Ted Shifrin

1:43 AM

Nah, Daniel :)

Daniel Rust

@agent154 arxiv.org/list/cs.CR/recent

@Ted do you know much about TQFTs by any chance?

Ted Shifrin

Nope, sorry.

Daniel Rust

np

just looking for some applications to mention in a talk i'm giving next week

Ted Shifrin

go for it :)

Alexander Gruber

@Benja geez, everybody's on my case this week.

Bitrex

1:54 AM

Question: for the inverse Laplace transform of a function of the form $X(\frac{s}{b} + a)$ can you apply the shifting and scaling theorems to get $x(t)$ in terms of $X(s)$

a and b are constants

agent154

@DanielRust Thanks a bunch

Enjoys Math

2:07 AM

Hey, back from the ban

I drew a penis and it was offensive, because private parts

specifically :-D bleep O=

Daniel Rust

@EnjoysMath I'm like totally not talking to you atm :P

unaccept my answer will you....

:D

Enjoys Math

where's your answer?

be specific!

this is math

Daniel Rust

lies!

this is sparta

Enjoys Math

If I did that it was probably because the other user spent like 5x longer on theirs, you're welcome to edit yours and ping me for a revote

Daniel Rust

some question about tensor products

Enjoys Math

2:12 AM

lol

Daniel Rust

can't remember where it is

Enjoys Math

lemme see

Daniel Rust

haha i don't mind

the other answer was much more detailed

I basically gave a bare bones answer

Enjoys Math

Oh, this one @DanielRust

uh math.stackexchange.com/questions/579456/…

Daniel Rust

haha yeah

that's the one

Enjoys Math

2:17 AM

The user has 500 rating, you have 9000, and they wrote a book

Daniel Rust

Yeah I agree

Enjoys Math

you wrote a page, dey wrote a book

Though your answer really did help me, I wish we cold select two or something

you got 6 points for that, I'd say it's win-win

6 words 6 points, seems fair :D

Daniel Rust

hehe :D

Enjoys Math

6 is the smallest order of a non-cyclic group!

right?

Daniel Rust

klein 4 group?

anon

2:19 AM

S3 is the smallest non abelian group

Enjoys Math

ah, ic

eXtremiity

What's the definition of an e-neighborhood of infinity in C^*?

Daniel Rust

is C^* the one-point compactification of the complex plane? (ie the riemann sphere)

Bitrex

$\frac{1}{2}Ae^{-\sqrt{\alpha}x}\left(\mathrm{erf}\left(\frac{2\sqrt{\alpha}t - x}{2\sqrt{t}}\right)+ e^{2\sqrt{\alpha}x}\mathrm{erfc}\left(\frac{2\sqrt{\alpha}t + x}{2\sqrt{t}}\right) + 1\right)$

blech!

eXtremiity

I am not sure. The question is as stated in my first message. However, C^* is the set of all complex numbers excluding the point z=0.

Daniel Rust

2:30 AM

Well it probably means the complement of a compact region in C.

the larger the compact region, the smaller the epsilon

eXtremiity

Ok, thanks.

usukidoll

2:46 AM

yo

anyone out there know how to do laplace with complex roots in the denominator?

it's like you have to take partial fractions, but the denominator has complex roots in it

anon

the same formulas apply as if it were real numbers

but the justification involves complex analysis

so you might be more comfortable with just doing partial fractions with linears and quadratic irreducibles

then, along with completing the square, linearity and shifting, you can do the inverses, with the prototypical ones being 1/(s+a) and 1/(s^2+a^2) and s/(s^2+a^2) and 1/(s^2-a^2) and s/(s^2+a^2)^2 etc.

usukidoll

grr no $ $ latex :(

$M_n$

ok my chatjax is on

can't be me

Benja

@AlexanderGruber What da ya mean?

Zeezee

3:10 AM

hello everyone, I want know why line is closed in product of two lower limit topology? it is not a problem that I can put it in site

anon

why not?

anyway, hint: $\Bbb R\times\{0\}$ is the complement of the union of appropriate "infinite squares" above the line and appropriate "infinite rectangles" below the line

Karl Kronenfeld

@AlexanderGruber Don't worry, AG really stands for amazing guy, Benja is just jealous.

Benja

3:26 AM

@KarlKronenfeld ?

Karl Kronenfeld

@Benja It's a joke about how the acronym AG is also his initials.

Benja

@KarlKronenfeld ah ok.

@KarlKronenfeld I posted an answer here

do you know if we need the algebraically closed hypothesis in the second proof?

@KarlKronenfeld Also I am a little unsure about the cutting with hyperplanes business

Karl Kronenfeld

3:42 AM

@Benja I don't know whether the hypothesis is necessary. I have to go, but right now I don't see why you would not be able to cut like that.

anon

haha, u=C?

Bitrex

Yah, but I forgot the boundary conditions. :[

looks like it's $u_t - u_{xx} + u = C_1$ with $u(0, t) = C_2$ and $\lim_{x \to \infty}u(x,t) = 0$

user60887

4:05 AM

hey guys

Bitrex

On the interval $[0, \infty)$

Vrouvrou

hello

someone of you have seen the saddle point theorem on Rabinowiz's book ?

???????

Antonio Vargas

4:38 AM

There is some serious confusion here: math.stackexchange.com/questions/586370/…

anon

troll thread

Antonio Vargas

That was my reaction too.

Muhammad Raja

Morning all

I am trying to solve this question but not able to find any helpful material. It involves factorial with multiplications,

8!/5! x 7!/7!10!

I tried crossing 8 and 5 and 7 with 7 but it's not giving me right answer

finished!

4:45 AM

I would ask a question on site, but my internet is being just too slow :( as when I post the question, I get time out error

usukidoll

turns out doing the complex root method was a bad idea..just partial fractions and put Ax+B whenever you see a non factorable equation EWOO

dailup?

dialup?

Muhammad Raja

Nope, it's a 3G but in a developing country :(

well at last I am able to post it :D math.stackexchange.com/questions/586438/…

masfenix

5:11 AM

Hey guys, when my prof says let $\Omega \subset \mathbb{R}^n$ be bounded, what does this mean exactly?

i know when a function is bounded, it means that $|f(x)| < C$ for all x.

Sanchez

It means that there is some $C$, $||x|| \leq C$ for all $x \in \Omega$.

masfenix

is that equivalent to saying this

Bounded set (topological vector space)

In functional analysis and related areas of mathematics, a set in a topological vector space is called bounded or von Neumann bounded, if every neighborhood of the zero vector can be inflated to include the set. Conversely a set which is not bounded is called unbounded. Bounded sets are a natural way to define a locally convex polar topologies on the vector spaces in a dual pair, as the polar of a bounded set is an absolutely convex and absorbing set. The concept was first introduced by John von Neumann and Andrey Kolmogorov in 1935. Definition Given a topological vector space (X,&t...

Mike

5:28 AM

@anon Why do you consider that post on square roots a troll post? It read to me like someone who was googling things they had no idea about.

I guess I can see it because he's being condescending in the comments

anon

I said thread not post :)

Mike

I meant question, which is what I assumed you meant

anon

so yes, I had to take into consideration the comments, the question itself is far away from justifying a claim of troll thread

Mike

Haha whoops I just looked at the comments

"you must be one of the Internet trolls. Good laugh, kid – jack 55 mins ago"

Neeevermind, I'm with you

Astrum

8:13 AM

what does the statement $|h(x)| \le M$ mean in the context of limits? O.o

robjohn

@Astrum $\limsup\limits_{x\to a}h(x)\le M$?

@Astrum $\liminf\limits_{x\to a}h(x)\ge -M$?

for any $a$

Astrum

@robjohn my text hasn't mentioned any of that, so I don't really know

I can write out the full question to give you more context, but I can't see what it means

robjohn

@Astrum The question is kind of broad. I don't know what they are looking for.

@Astrum might be helpful

Astrum

Prove: if $\lim_{x\rightarrow 0} g(x) = 0 $ and $|h(x)| \le M$ for all $x$, then $\lim _{x \rightarrow 0} g(x)h(x) = 0$

Spivak has never mentioned anything about this mysterious $M$ value

@robjohn any idea?

TheNotMe

8:32 AM

Can anyone tell me if the language $L = \{ w \mid |w|_a = 2^n + 273 \}, \Sigma = \{a,b\}$ is regular?

Alexander Gruber

@Astrum it's just some value

Astrum

@AlexanderGruber weird, so it's nothing fancy

Alexander Gruber

you can read it like "$|h(x)|$ is bounded above"

Ethan

$h(x)=O(1)$

robjohn

8:52 AM

@Astrum Use the definition of limit and everything is pretty straight forward.

Astrum

@robjohn yeah, except for this new problem I'm having some trouble on. It might be worth another post when I wake up later

robjohn

$\lim\limits_{x\to0}g(x)=0\iff\forall\epsilon\gt0,\exists\delta\gt0:|x|\le\delta‌\implies|g(x)|\le\epsilon$

Astrum

@robjohn yeah, but this new problem is trickier

robjohn

For $h(x)g(x)$ you use the $\delta$ that would give an error of $\epsilon/M$ with $g$

Astrum

9:10 AM

prove that if $\lim _{x \rightarrow 0} f(x) \le 0$ and $\lim _{x \rightarrow 0} g(x)$ does not exist, then $\lim _{x\rightarrow 0} f(x)g(x)$ does not exist

I meant $\lim {x\rightarrow 0}f(x) \ne 0$

robjohn

@Astrum that is not true. If the $\lim\limits_{x\to0}f(x)=0$ and $\lim\limits_{x\to0}g(x)$ does not exist, $\lim\limits_{x\to0}f(x)g(x)$ might exist

@Astrum Ah, then it is true.

Astrum

@robjohn I'm assuming that we set up our definitions, and then we have something like $|(f\cdot g)(x)-L\cdot m| < \epsilon$

robjohn

I would start by assuming that $\lim\limits_{x\to0}f(x)\ne0$ and $\lim\limits_{x\to0}f(x)g(x)$ both exist, then prove that $\lim\limits_{x\to0}g(x)$ exists. (contrapositive)

Astrum

9:26 AM

so we'd just be restating the proof that $\lim _{x\rightarrow a} [f(x)g(x)]=(\lim _{x\rightarrow a} f(x)) (\lim _{x\rightarrow a} g(x))$

that proof was kinda of ugly, if I recall correctly

Chris's sis

Greetings the great one! (I also have a great question I sent to some professors)

Compute elementarily $$\sum_{n=1}^{\infty}\frac{1}{n^2}\left(\zeta(2)-\frac{1}{1^2}-\frac{1}{2^2}-... -\frac{1}{n^2}\right)$$

Astrum

this chat is like a problem help line XDDDDD

Chris's sis

(well, some professors that are so willing to send them questions - no response in this case)

One told me this morning "it's not possible to do it elementarily" and I replied "Really?".

Anyway.

It's lovely!

hehe, I have such a craving for attending calculus questions today. I hope to finish a lot of proofs.

(brb -- I'm around -- beginning to work on some questions)

Transcript for

$Mathematics$ Mathematics