Mathematics - 2012-09-08 (page 3 of 4)

@anon OK - thanks :)

I have been accused of similar - but with less humour :)

John Junior

hi @WillHunting

Do you think 200 rep bounty is not high enough to make people want to answer a (very easy) question?

user19161

@JohnJunior Hey! (remembers not to call you bro as instructed)

user19161

@Matt Where? Are you thinking of giving one?

John Junior

9:06 PM

@WillHunting Thanks for remembering :-D

@WillHunting No, I put 200 on this one. I really want it answered.

And I don't mind offering more.

Looks as if I cannot increase a bounty once set : (

user19161

@Matt Not many people deal with knots, so I hope you get an answer. I guess it must be related to your thesis.

Well, this guy knows about knots and can probably answer my question.

user19161

Also, although tb is not here, I am here if you need me. :-)

@WillHunting Sorry, no one can replace tb for me.

But thanks : )

user19161

9:13 PM

That is certainly true considering the fact there is physical proximity between you two as well.

I can assure you that not: I'm straight and have a girlfriend : )

user19161

Hehe, I thought you were gay a few times.

user19161

I am straight and have no girlfriend.

Yeah, not so easy to find one, is it : /

@WillHunting Why? Just curious.

user19161

@Matt Oh, from some of the things you say in chat or the way you say it. Of course, straight people can fantasize over someone of the same sex too sometimes.

9:16 PM

I struggle to find a girlfriend too ...

user19161

@JohnSenior Is that some silly joke? :)

... but if I did, my wife might just complain :)

: P

although I do know married guys who have girlfriends too ...

@WillHunting I thought this about you.

user19161

9:18 PM

@GustavoBandeira I don't tell lies here.

user19161

@JohnSenior That is not good.

What you mean?

user19161

@GustavoBandeira Oh, I mean that I already said I am straight which means I am, that's all.

user19161

But I can't draw a long straight line without a ruler though.

@WillHunting It means you tell lies IRL...

user19161

9:20 PM

@GustavoBandeira Nope, I try not to very hard at least.

@WillHunting True. And I'm afflicted with quite some imagination... : )

user19161

@Matt I recently fantasized over...

...Justin Bieber?

user19161

Nope, it's a secret!

Tease!

user19161

9:21 PM

I don't want to freak anyone out here!

@Matt ^ deliberately trying to sound gay now.

These guys are weird....

: D

@WillHunting Heh, someone in chat here?

I'll stop asking, it's ok.

user19161

@Matt Er,... :-)

: )

@JohnSenior Do you also think I'm gay? (just curious)

9:26 PM

I once threw my computer out of my window.

@Matt ...

user19161

@JonasTeuwen Why?

@Matt I think. =D

Because it was a piece of shit, that's why.

@GustavoBandeira : D

9:26 PM

@JonasTeuwen =D

@Matt no - never thought you might be - not thought that about anyone I've met here, actuially

@JonasTeuwen Next time buy a mac...

@Matt It was a mac.

As if.

I bought a new one. Was the best thing I ever did.

9:27 PM

Yo! Take that, you miserable fanboy!!! @Matt

Throw my computer out of the window.

@GustavoBandeira : D

The new one was a mac too.

user19161

All the computers I have now are somehow free.

I first tried to bite it, but that did not help.

9:27 PM

@JonasTeuwen Eating macs is bad for your health

It had some bad hardware connection somewhere which randomly caused it to crash.

@N3buchadnezzar Locking up computers when you want to work do too!

@JonasTeuwen For that you have 3 years of apple care so that they fix it for "free".

I will not throw my iPad out of the window, it is sweet.

user19161

I think this Ubuntu Unity is not too bad after all, I am currently on Ubuntu.

@Matt I want it NOW.

9:28 PM

Ipaid to much

Not in whatever weeks.

@N3buchadnezzar Ipaidnotcare.

Oh I see.

@JonasTeuwen My desktop has been performing perfectly again after I reinstalled Linux onto a SATA drive instead of SSD

@JonasTeuwen But they're usually quite quick no? Bring it in and they fix it within 2 days.

So throw out of window, confirm is it completely dead, make sure it is and go to the store!

@JohnSenior Ah! :-).

user19161

9:29 PM

@JohnSenior What did you install?

Or just go to the store.

SSDs needs some additional commands to work well.

Heh!

@Matt But then I will still be frustrated.

Mint this time - just out of curiosity

9:30 PM

@JonasTeuwen You fancy gadgets dude.

@JonasTeuwen *Imagines Jonas throwing his deceased relatives out the window.

And the miserable thing will still be there when I am home.

It seems that there are two kinds of math books: The more "procedural" kind, in which the authors tell you that something is true, and there's another kind of books in which the authors tell you why it's true. Do you guys know some related question on this?

@Matt What? Why?

@JonasTeuwen I think I still don't get it.

@JonasTeuwen SSD = solid state disk, I assume?

user19161

9:30 PM

@GustavoBandeira No, there is only one kind and they tell you both but to various degrees of quality.

I considered that too, when I replaced my disk a few months ago.

@BrianM.Scott Are you there? Be there, be there!

@rob Hey!

@Matt Yes.

SSDs need TRIM to perform optimally.

9:32 PM

@PeterTamaroff Sort of half here.

SSD is fancy stuff.

But these...

@BrianM.Scott I think I'm done.

Nevermind.

@Matt Yes, but very nice for portable battery devices!

@JonasTeuwen Buy one of these to throw your computers when they're sucking. When the shredder starts to suck, throw it through the window too!

9:32 PM

@PeterTamaroff With the proof?

@GustavoBandeira That's a mean shredder.

I guess I can't use "they're sucking"

user19161

@PeterTamaroff I hope you are not done with your life.

Can you also put small children in it?

@WillHunting What ya mean?

9:33 PM

@JonasTeuwen Yes. But good old hard drives have been produced for a long time and hence are probably more reliable.

So I thought. And bought a normal drive.

@Matt Yes, then you choose for reliability.

With the nice side effect that it's huge and cheap.

user19161

@GustavoBandeira I mean what I mean. Every decent book will give you theorems and their proofs.

@Matt I don't need much space. Just to mirror my repositories and OS + tools.

user19161

@JonasTeuwen Do you have a lot of stuff in your repos? :-)

9:34 PM

@WillHunting Not so much, is about ~800MB now.

If $q_n\in(x,x+2^{-n})$ and $p_n\in(x-2^{-n},x)$ then it must be that $|p_n-x|<2^{-n}$ and $|q_n-x|<2^{-n}$ so, $$\eqalign{
& f\left( {{p_n}} \right) < f\left( x \right) < f\left( {{q_n}} \right) \cr
& {p_n} < f\left( x \right) < {q_n} \cr
& {p_n} - x < f\left( x \right) - x < {q_n} - x \cr
& - {2^{ - n}} < f\left( x \right) - x < {2^{ - n}} \cr
& \left| {f\left( x \right) - x} \right| < {2^{ - n}}\forall n \Rightarrow f\left( x \right) - x = 0 \Rightarrow f\left( x \right) = x \cr} $$

Code + documents.

user19161

You know what? I usually have zero files in my thumbdrive.

@JonasTeuwen Yes. Because I have zero patience. If my computer lets me down just for a second I already think of throwing it out the window. So I could never be a windows user. I'd have to buy a new computer every week.

@Matt Ah, so you do understand my computer throwing!

9:35 PM

@JonasTeuwen Heh. And where do you store all the pron?

There is a Dutch computer throwing championship. I should join.

@JonasTeuwen Sure.

@Matt I do not!

I think I would beat them if it runs windows and they make me work on it!

Otherwise not so much.

I'd hand in my notice.

cats are lovely

user19161

9:36 PM

What kind of cats?

@PeterTamaroff Looks okay. A little more roundabout than I’d have written, but that’s no real problem

All cats.

@JonasTeuwen Finland has a mobile phone throwing competition too

@BrianM.Scott What does roundabout mean?

@JohnSenior Ah! If it is a smartphone, I'd be happy to throw it.

user19161

9:37 PM

Dogs are lovely too.

@PeterTamaroff That it could probably have been stated a little more directly.

@WillHunting Stood is the past [something] of stay, right?

@Matt More cats, less people.

user19161

@GustavoBandeira No, of stand.

@JonasTeuwen That would be nice.

9:37 PM

We should trade some cats for hoomins. With the Cat Aliens.

@BrianM.Scott Well, maybe I have the need to be more detailed, just in case =P How would you do it?

Yes.

@JonasTeuwen The only problem of the cats is that they crap a lot. =/

user19161

We need a cat lady.

Pretty good deal, just in weight, they get 50% discount.

9:38 PM

so many cats ... so few recipes

2

@GustavoBandeira So do humans. Also verbally.

@JohnSenior 8-).

user19161

I have no idea what we are talking about now...

Well, humans are taught to crap in the toilet.

I hate what cats do to my garden - although I did take a nice pic of a big cat at a zoo today

Some days ago, I've seen a method to teach that to the cats.

9:39 PM

@JohnSenior Come to think of zoos: I really wish this sort of animal concentration camp did not exist.

@BrianM.Scott I have as a theorem that if $x,y$ are such that for some $a$ and every $n\in N$, $$y\leq x \leq y+\frac{a}{n}$$ then $x=y$, IIRC (that is from Apostol)

@PeterTamaroff For each $n\in\Bbb N$, $x-2^{-n}<p_n=f(p_n)<f(x)<f(q_n)=q_n<x+2^{-n}$.

@JohnSenior If you would have a cat yourself, no cat would do that to your garden! 8-).

@BrianM.Scott Yes.

Especially if it is a mean one.

9:39 PM

@BrianM.Scott Now subtract $x$

The cat would be like "No crappin' in my garden.".

@JonasTeuwen Or a dog.

@Matt I worry about that too - but in some cases they probably help preserve endangered animals too

user19161

Isn't it expensive to keep a pet?

@JohnSenior When I was abroad, instead of asking for the receipt (after buying a sandwich) I asked for the recipe... cashier was like "DAFUQ dude"

9:41 PM

@PeterTamaroff wonderful!

@PeterTamaroff I would take you back and have you make them.

user19161

@PeterTamaroff WTF!

"The best way to learn it is to do it."

"WORK!"

@JohnSenior It's not hoomins' rights to "preserve" species.

@PeterTamaroff So you have $-2^{-n}<f(x)-x<2^{-n}$, and if you want to be really niceyou can multiply by $-1$ and combine to write $|f(x)-x|<2^{-n}$.

9:41 PM

Is it possible to simplify $(1+x+O(x))^x$?

@WillHunting No.

@Matt its not our right to destroy them either - but we do

@JonasTeuwen LAWL

@Argon Sure.

You are so easily amused.

user19161

@PeterTamaroff Is LAWL the new LOL?

9:42 PM

@WillHunting It is all about inflection.

@JohnSenior Yes, including other hoomins. I don't care about that. But after we have already destroyed them we should not torture the last remaining individuals.

user19161

I imagine a world where I am the only male and the females have no choice but me.

@Matt Total annihilation minimalizes pain.

: D

Very benthamish thus to destroy it all!

9:43 PM

@WillHunting They would became lesbians. =D

user19161

@JonasTeuwen minimizes

@WillHunting ... not with a bang, but a whimper.

@WillHunting They would die!

@PeterTamaroff What would it become? How do I handle to $O(x)$?

: D

9:44 PM

You would have to provide the food!

You are so funny!

user19161

@JonasTeuwen They would die of pleasure!

Who is?

@WillHunting Of terror.

All of you.

This chat room gives me the lolz.

Pleasurable terror?

user19161

9:45 PM

Pleasure rhymes with terror.

@Argon Use $a^x =e^{x \log a}$. Then use the expansion of $e^x$

Their eyes will be filled with terror.

Charles Mingus!

And their heads with despair.

@PeterTamaroff Thanks, I will try that.

user19161

9:45 PM

Horrified terrified petrified stupefied by you

@WillHunting ...and your children will be next!

@JonasTeuwen You haven't heard of Russel Edgington, have you?

It's the name of the guy who compiled a handbook on how to teach cats to poop in the toilet, look: openculture.com/2011/11/…

I am going to sleep. Good night!

user19161

@Matt Good night. I will see you in your dreams.

9:46 PM

@WillHunting Erotic.

: D

user19161

@PeterTamaroff Sweet!

user19161

9:48 PM

@PeterTamaroff Like you!

user19161

Hello @iuli. Welcome to chat!

@PeterTamaroff If there is somebody gay here... it must be you. You are talking about it all the time.

There is this Dutch saying: "Waar het hoofd van vol is loopt de mond van over.".

@JonasTeuwen Translate please.

user19161

@JonasTeuwen Like I said, everyone is gay. The only difference is the concentration.

9:50 PM

@JonasTeuwen That is not true!

Where the head is full of the mouth pours over of.

Or something like that.

So that means your head is full of gay.

Denial is futile.

You're gay. (that's a period)

@JonasTeuwen HAHAHHAHAHAHHA

►

user19161

@PeterTamaroff Now that is like a Benja laugh.

WTF - Google translate just gave me "Where the head is full on the mouth speaketh. "

user19161

@JohnSenior WTF!

9:52 PM

and changed the translation when I deleted a couple of spurious punctuation marks at the end

"Where the head is full of running the mouth speaks."

Google gave that

user19161

The mouth speaks from the heart.

user19161

There there, my lovely version.

@PeterTamaroff try adding a full-stop and double quotes at the end (with no matching quotes at the start)

here

@PeterTamaroff Hi there.

9:54 PM

@PeterTamaroff My expansion seems quite unsuccessful.

@robjohn Seen this?

@Argon I think Apostol treats that. Let me see.

@PeterTamaroff You keep on referencing him; I should really get this book!

@Argon Maybe you should.

user19161

@Argon He will keep referencing Mendelson Topology too!

@WillHunting You know why?

user19161

9:57 PM

@PeterTamaroff Because you read it of course.

@WillHunting I was hoping Mendelson had been forgotten now :)

@WillHunting No.

user19161

@PeterTamaroff Why? It is your course text?

@WillHunting Because fuck you, that's why.

user19161

@PeterTamaroff :-)

9:58 PM

@PeterTamaroff Whoa!

Prrrretty hungry.

user19161

@JonasTeuwen It is midnight for you.

Yea.

It is bro, it is.

user19161

You should get a midnight snack.

I am a werewolf, howlin for food.

10:04 PM

@PeterTamaroff thanks :-)

@robjohn You think you can solve it?

@PeterTamaroff check it out

@robjohn You missed the $2n+1$! Oh, dear!

@PeterTamaroff really? darn

user19161

@robjohn 2 min and 2 votes already!

user19161

10:07 PM

Just now Iuli came to chat. I was wondering whether it was luli or iuli.

@robjohn I need to find noncosntant $f$ with $|f(x)-f(y)|\leq |x-y|$

@PeterTamaroff $f(x) = x$.

I'm trying to show $\sin x$ does the work.

@JonasTeuwen That one is too easy!

So?

@PeterTamaroff Mean value theorem and note $|\cos| \leq 1$.

@JonasTeuwen Dunno. I did think about that.

@JonasTeuwen No MVT allowed this time.

10:15 PM

What the heck.

That's stupid, I'm off.

@JonasTeuwen I haven't got to derivatives yet! =D

@JonasTeuwen last night I was mumbling about discontinuous $f$ such that $f(x+y)=f(x)+f(y)$ ...

@JonasTeuwen Maybe you like this one more.

... now wondering if that is the same thing as the existence of a discontinuous/unbounded linear functional on a normed space

@JonasTeuwen Suppose that $$f(y)-f(x)\leq (y-x)^2$$ Show that $f$ must be constant.

10:18 PM

@PeterTamaroff Well. Also easy.

Divide by.

So if it was differentiable then the derivative is $0$ everywhere.

@JonasTeuwen Spivak says: Divide the interval $[x,y]$ into $n$ equal parts.

Yea, also works.

@JohnSenior Can’t be quite the same: a discontinuous function satisfying the Cauchy functional equation does not in general satisfy $f(ax)=af(x)$.

@BrianM.Scott For $a$ rational?

@BrianM.Scott Why does Spivak suggest to divide the interval into $n$ equal parts?

10:22 PM

@JonasTeuwen It’s true for rational $a$, but not in general.

@PeterTamaroff That should be better

@BrianM.Scott yep - that would be true

@BrianM.Scott Yes, but that is clear :-). You would take a vector space over $\mathbf Q$!

@BrianM.Scott: thanks for noticing :-) Now I think I have it right.

@BrianM.Scott I was wrong about the linear functional bit - but would it be possible to create a discontinuous $f$ just by taking a Hamel basis and defining $f$ pretty much randomly for each element of the basis?

10:25 PM

@JohnSenior That’s pretty much the only way to do it.

@BrianM.Scott Thanks

@JohnSenior Go Go Old John!!

@robjohn Slick.

@PeterTamaroff what???

@JohnSenior " That’s pretty much the only way to do it."

10:26 PM

@PeterTamaroff ah - ok

@BrianM.Scott I don't understand this $$|f(x)-f(y)|\leq (x-y)^2$$ thingy

I have to prove that $f$ is constant by dividing the interval $[x,y]$ into $n$ parts,

Let me think about it for a moment.

Just divide this in small intervals, if you take $n$ then $(x - y)$ will be at most $\frac1n$ apart.

And to the left you use the triangle inequality.

I'm now drawing arbitrary non constant functions to see how it fails.

If it is differentiable it is easy right?

10:31 PM

@JonasTeuwen Well, yes.

Yes. So for equal division into $n$ pieces you get.

$|f(x)-f(y)|\le\sum_k|f(x_k)-f(x_{k+1})|\le n\left(\frac{|x-y|}n\right)^2=\frac1n|x-y|$.

@JonasTeuwen So take $(y-x)/n$ in length

Yea. Easy I said right?

But I am very tired.

Try to do this one: if $f(nx) \to 0$ as $n \to \infty$ for some $x$, then $f(x) \to 0$.

Now I need to add some assumptions...

Continuity!

Maybe some $x$ needs to be from a discrete countable set...

Let $f : \mathbf { R} \to \mathbf{R}$ be real. Assume that for all $\alpha > 0$ we have that $f(n \alpha) \to 0$. Show that $f(x) \to 0$.

@PeterTamaroff ^ is correct.

10:42 PM

@BrianM.Scott Are you taking each $x_k$ in a different subinterval?

No, those are the endpoints of the subintervals.

@BrianM.Scott $x_k=x+\frac{k}{n}(y-x)$?

@PeterTamaroff The idea is...

Yes.

that if you split them up your $|x - y|$ will be "small".

LIMITS.

Done.

10:43 PM

@JonasTeuwen Split up what?

The interval from $x$ to $y$.

@JonasTeuwen Oh, wait. $$|f(x_{k-1})-f(x_{k})|\leq (x_{k-1}-x_{k})^2$$!!!

...?

But $x_{k}-x_{k-1}=(y-x)/n$.

Yea... so?

Yes.

That's the bloody idea, Sir.

10:47 PM

@JonasTeuwen Well, I wasn't seeing that.

@PeterTamaroff That's normal I think.

@Peter: The point is that the squares of the lengths shrink faster than the number of subintervals grows.

@BrianM.Scott Oh, OK. I thought that the square had to shrink fast, yes.

Is $\alpha =2$ kind of a limit point for $$|f(x)-f(y)\leq |x-y|^\alpha$$ or does $\alpha>1$ suffice for $f$ to be constant?

@JonasTeuwen Wait, how do yuo get $\leq 1/n$? I'm getting $$|f(x)-f(y)|\leq \frac{(y-x)^2}{n}$$

Don't troll me man.

4 mins ago, by Brian M. Scott

@Peter: The point is that the squares of the lengths shrink faster than the number of subintervals grows.

@JonasTeuwen I'm not trolling!

$$\displaylines{
\left| {f\left( x \right) - f\left( y \right)} \right| = \left| {\sum\limits_{k = 1}^n {f\left( {{x_{k - 1}}} \right) - f\left( {{x_k}} \right)} } \right| \cr
\leqslant \sum\limits_{k = 1}^n {\left| {f\left( {{x_{k - 1}}} \right) - f\left( {{x_k}} \right)} \right|} \cr
\leqslant \sum\limits_{k = 1}^n {{{\left( {{x_{k - 1}} - {x_k}} \right)}^2}} \cr
\leqslant \sum\limits_{k = 1}^n {\frac{{{{\left( {y - x} \right)}^2}}}{{{n^2}}}} = \sum\limits_{k = 1}^n {\frac{{{{\left( {y - x} \right)}^2}}}{{{n^2}}}} \cr} $$

Where $x_k=x+\dfrac{k}n (y-x)$

10:55 PM

@PeterTamaroff You have $x - y = (x_1 - y_1) + (x_2 - y_2)$.

@PeterTamaroff $=\frac1n(y-x)^2$

Add some points.

@BrianM.Scott Yes, I forgot to write that.

So all in all it is $$\left| {f\left( x \right) - f\left( y \right)} \right| \leqslant \frac{{{{\left( {y - x} \right)}^2}}}{n}$$