Mathematics - 2013-03-05 (page 1 of 2)

Arkamis

00:58

Anyone around?

Link

I am?

Arkamis

01:27

Nevermind, I solved it :)

user19161

01:38

@anon You're doomed for life.

Peter Tamaroff

03:08

This is why this world is fucked up.

user19161

03:23

@PeterTamaroff OMG!

Peter Tamaroff

@JacobBlack Who teaches logic to these guys?

user19161

@PeterTamaroff Professor Pedro.

Peter Tamaroff

@JacobBlack Yeah, but not me.

Some Evil Pedro.

@JacobBlack This is just evil

IT is not the harlem shake.

Arkamis

wipes brow that was a hell of a problem.

Peter Tamaroff

@Arkamis Share, man,.

Arkamis

03:29

Sure

Let $C$ be a regular curve enclosing the distinct points $\omega_1, \omega_2, \ldots, \omega_n$ and let $p(\omega ) = (\omega - \omega_1)(\omega - \omega_2) \cdots (\omega - \omega_n)$. Suppose that $f(\omega)$ is analytic in a region that includes $C$. Show that
\begin{equation}\nonumber
P(z) = \frac{1}{2\pi i}\int_C \frac{f(\omega)}{p(\omega)} \cdot \frac{p(\omega) - p(z)}{\omega - z}\ d\omega
\end{equation}
is a polynomial of degree $n-1$, with $P(\omega_i) = f(\omega_i), i = 1, 2, \ldots, n$.

Peter Tamaroff

@Arkamis I see. I'll solve it when I learn CA!

Arkamis

It's actually not that hard

But it's very intimidating at first

Basically, you use the residue theorem to show that $P(z) = \text{Res}\left({\frac{f(\omega)}{p(\omega)} \cdot \frac{p(\omega) - p(z)}{\omega - z}}{;\ z}\right) + \sum_{j=0}^n \text{Res}\left({\frac{f(\omega)}{p(\omega)} \cdot \frac{p(\omega) - p(z)}{\omega - z}}{;\ \omega_j}\right)$

Peter Tamaroff

OK; whatever the residue theorem is!

Arkamis

Then the first term is zero

And every term in the sum evaluates to a polynomial of degree $n-1$, and vanishes when $z = \omega_j$, leaving only the $j$th term

The residue theorem is quite cool; it basically says that for some contour integrals of non-analytic functions, the integral is basically a multiple of $2\pi i$ times the $-1$th term in the function's Laurent expansion. Effectively, you get the value of the integral "for free" if you can do the Laurent expansion.

Peter Tamaroff

I see.

Is it tough to prove?

Arkamis

03:35

Not really

It's an exceptionally powerful tool

Peter Tamaroff

Feels like "Meet my little friend" scene from Scarface when you apply it, doesn't it?

PEW PEW PEW, problems!

Arkamis

Basically.

Also, I love using the word "annihilate" in any proof.

Peter Tamaroff

Explain.

Arkamis

ie "Therefore, evaluating the derivative at some $z_j$ annihilates every product in the sum except for when $k=j$"

I imagine taking terms and blowing them up.

Peter Tamaroff

Oh. I like "vanishes".

@Arkamis heh

Arkamis

03:40

Vanishes, to me, always has a notion of continuity

Like, as you approach something, it vanishes

Peter Tamaroff

I do like the annihilator method in ODEs, for example!

Arkamis

But we're not approaching, we're just evaluating

We're not setting the mic back in the stand; we're dropping the motherfucker

Peter Tamaroff

@Arkamis Hahhaha I get it.

Arkamis

Remember back when you were younger, learning basic algebra, and you would have to combine fractions or some shit, and then you get everything combined, and start canceling terms, and it felt so good

"(x-4)? SLASH THROUGH YOU. DIE!"

Peter Tamaroff

@Arkamis hahaha kinda.

Arkamis

03:45

Like the scene in Good Will Hunting, where the grad student gets all sad

Cuz Stellan Skaarsgard and Matt Damon are all like doing graph theory together as buddies

Peter Tamaroff

Can't get that one.

There are few math movies around.

@Arkamis You're an undergrad yes?

Arkamis

No, far from it.

Peter Tamaroff

@Arkamis Ah?

Arkamis

@PeterTamaroff Good Will Hunting is a spectacular movie; Robin Williams won an Academy Award for his role in it.

I'm working on a graduate degree, but I am a working professional.

Peter Tamaroff

@Arkamis Yes, I liked that one.

@Arkamis I hope I didn't offend you!

Arkamis

03:49

I finished my undergrad about 6 years ago, although I should have finished it more like 9 years ago.

Peter Tamaroff

@Arkamis What is a gradute degree?

Arkamis

A Master's degree or PhD

Peter Tamaroff

@Arkamis Oh, OK

Arkamis

I might eventually do a PhD, but I'm working on a Master's now.

Peter Tamaroff

I know the answer to this, but I am too darn lazy to type it down.

Arkamis

03:49

I work full time, so I can only really take one class.

Peter Tamaroff

@Arkamis What do you work in?

Arkamis

Mostly aeronautical and biomedical engineering.

tacos_tacos_tacos

@Arkamis curious about your experience working towards masters + working

Arkamis

My background is numerical analysis and aeronautics

@tacos_tacos_tacos It sucks, sometimes. But it's not that bad.

tacos_tacos_tacos

If you dont mind me asking, how long had it been since you had last gone to university

Arkamis

03:51

It all depends, if the class is a morning class, it doesn't affect my day very much; if it's a day-time class it really messes up my work week

Well, I finished my undergrad in 2007

And I started taking classes again in 2011.

@PeterTamaroff A lot of what I do has to do with developing novel methods for data analysis and modeling of dynamical systems

Some neural network stuff, some inverse methods

Also, mostly analysis of systems with parametric uncertainty

I've done a bit of work in polynomial chaos, which is a powerful computational tool for analyzing dynamical systems with a stochastic parameter (vs. a random variable that is the variable of integration)

Peter Tamaroff

Interesting.

Finally, gathered some forces

I'm off.

Arkamis

adios

Peter Tamaroff

@Arkamis Aha!

Bye byes.

Arkamis

04:06

This problem seems too easy.

Ah, because I did it wrong.

Or, maybe not.

BenjaLim

07:11

@anon Hmmm Olivier claims there is a problem in my answer here, is there really one?

anon

07:23

@BenjaLim I see no issue with the LHS or the reindexing, but upon looking at your answer a second time I no longer remember why $\sum_{i=1}^p e_{x^i}v\in V$ (we do not know that $V$ is an ideal in $K[G]$, and the summands are not generally in $V$ unless $uv=0$ by a stroke of luck) or why it is nonzero.

Alex Youcis

07:37

@anon Do you mind if I ask you a question about Q_p?

anon

go ahead

Alex Youcis

@anon Do you know any simple way to prove that $\text{Gal}(\mathbb{Q}_p/\mathbb{Q})$ is trivial? I know of one fairly tricky way which relies on describing the units of $\mathbb{Z}_p$ in a purely algebraic way.

anon

nope

Alex Youcis

Ok man, thanks.

BenjaLim

08:08

@anon The reason why it is in $V$ is because $V$ is a $K[G]$ - submodule

Dominic Michaelis

08:37

huhu

skullpatrol

09:36

hehe

Charlie

Haha

@skull Hello

skullpatrol

@Charlie Hoho gotta go...:(

Charlie

@skull bye... :(

Hi @dominic

Dominic Michaelis

Hi charlie

Charlie

@DominicMichaelis how are you?

Dominic Michaelis

09:48

i am fine and how are you ?

Charlie

Good

Dominic Michaelis

10:47

today is a bad day

user19161

11:03

@DominicMichaelis What happened?

Dominic Michaelis

no nice questions :(

user19161

Oh, I though something serious happened.

Dominic Michaelis

not yet :D

argh i hate when i miscalculate the complexity of a command

i wanted to compute the first 10 elements (?) of the iteration sequence math.stackexchange.com/questions/321331/…

but 4 gig ram aren't enough for it

user19161

@DominicMichaelis I have about 4 GB RAM on my desktop and 2 GB on my laptop.

user19161

Nowadays, everyone has much more RAM.

user19161

11:17

But I think 2 GB RAM is enough to run all 64-bit operating systems.

Dominic Michaelis

maybe the first 3 should be possible

now waiting for the fullsimplify :D

Dominic Michaelis

11:35

@AmWhy your comment should be as $A^n =0$ for $n>2$ shouldn't it ?

Gigili

Owow, it's unbelievable! The new moderator, Bill something is suspended. What a shame.

amWhy

@DominicMichaelis it is...

Dominic Michaelis

it wasn't at first ^^

Jayesh Badwaik

@DominicMichaelis You are doing something wrong. If you want to calculate first thousand terms or something, 100-200MB of program memory should be enough.

Dominic Michaelis

not if you calculate it analytical by a recursion

Jayesh Badwaik

11:48

Ohh, you meant analytically. I see. Even then, I feel first ten terms of the sequence should not be a big memory hog.

Dominic Michaelis

wait a minute i give you the third term

it takes a bit have to formate it

Jayesh Badwaik

okay no problem

Dominic Michaelis

mh i don't manage it but it should give the idea writelatex.s3.amazonaws.com/86058xgjlhq/page/…

thats a part of x_2

Carpediem

12:05

Hello

Dominic Michaelis

hi

Carpediem

Can someone help me with this? math.stackexchange.com/questions/321379/…

Dominic Michaelis

i can only see one subset

Carpediem

@DominicMichaelis What do you mean?

Dominic Michaelis

describe the subspace spanned by each of the following subsets of polynomials

Carpediem

12:08

Yes. There are others. but I would like you to show me how to do with the first one

The rest I will do on my own

@DominicMichaelis

Dominic Michaelis

done

Carpediem

Ah ok. I thought It was something more complex

Thank you very much @DominicMichaelis

Dominic Michaelis

i can make it more complicated if you like ;)

amWhy

12:29

@JayeshBadwaik Back to your real name (or at least the name I came to know and love) ;-)

Jayesh Badwaik

@amWhy Yes. :-P

amWhy

@JayeshBadwaik It reminds me...looks strikingly like an email address I need to use :-S

Jayesh Badwaik

@amWhy Yes, and quickly too. :-)

amWhy

@JayeshBadwaik You got it...I may need to take a nice long nap shortly...I stayed up all night! (6:30 am my time right now)

Jayesh Badwaik

@amWhy Ahh, I see. Good night then.

amWhy

12:33

@JayeshBadwaik Not quite yet...I've just, for some reason, kept going...kept delaying bedtime...until, lo and behold, the birds are chirping!

Jayesh Badwaik

@amWhy I actually feel more comfortable going to sleep when the birds are chirping. ;-)

BenjaLim

Any algebraists around? I have a question about Jacobson rings

amWhy

Yeah...me too sometimes!

Hello @Jacob Blue

user19161

@amWhy Yes, I am back to steelblue for now.

amWhy

@JacobBlack My favorite of all your colors!

user19161

12:39

@amWhy Yes, it seems to be my favourite as well.

BenjaLim

@JacobBlack Have you seen Mariano?

Dominic Michaelis

is the set of C^\infty functions an integral domain ?

user19161

The "blue" theme in gmail is a bit light, so I choose the "dusk" theme.

user19161

@BenjaLim No, if you have a question you can just post on the main site.

user58512

hi

skullpatrol

12:55

hi

user58512

@DominicMichaelis, I think the answer is no because of bump functions

analytic functions are though

Dominic Michaelis

i think as holomorphic is about as strong as analytic

or they should be the same or ?

because for holomorphic it is an integral domain

user58512

analytic is stronger than $C^\infty$

Dominic Michaelis

yeah but holomorphic is much much stronger than $C^\infty$

user58512

"but"?

I think we're in agreement :D

Dominic Michaelis

13:07

yeah :D

i already got an idea for a counterexample

user58512

a counter example to what?

Dominic Michaelis

that C^\infty is a integral domain

user58512

ah

I have one

Dominic Michaelis

don't tell it right now

if it is ok

i am eating and i want some minutes to get my idea on a sheet of paper

user58512

:)

skullpatrol

13:16

@Q__ who's going to show up next? R__ ;)

Q__

@skullpatrol Heh.

Jayesh Badwaik

@Q__ were you an IRC veteran before?

skullpatrol

Correction: U always follows Q :-D

Charlie

13:34

@skullpatrol good ;)

user58512

good morning @awllower

awllower

@user58512 Hello there!

user58512

is your leg ok now?

awllower

Much better!! :-)

skullpatrol

@Charlie How are you Chucky?

awllower

13:35

Thanks for your concern!

user58512

great - so you are recovering

awllower

Yes!

user58512

it is a lovely sunny day

Charlie

@skullpatrol I'm good thanks, and you ?

awllower

And for the moment I am trying to generalise this result

skullpatrol

13:36

@awllower Has the bleeding stopped?

awllower

Fortunately it has. :)

skullpatrol

@Charlie Fine thanks.

awllower

By use of some normal basis.

Charlie

@skullpatrol good :)

user58512

hm this result is not true?

so what do you mean to generalize it?

awllower

13:37

Oh, I mean, generalise the result by Qil'8

so that one can have a much clearer view toward this.

user58512

ah let me read it

awllower

But I am still stuck!

While I found another even more messy way to prove the same thing as Qil'8 did, I found presently no way of computing the stabiliser of that expression in the Galoisgroup.

user58512

i can't follow his proof! it uses some things I don'tknow

awllower

What things?
dedekind lemma?

user58512

yes but I felt a little lost before that too

awllower

13:42

Indeed the notations are a little messy...

user58512

I may try myself

awllower

And my proof uses some even more bizarre notations...

user58512

hehe

actually im a little confused

awllower

I will try to make it clearer.

user58512

in the problem statement are x,y arbitrary elements of $\mathbb C$?

awllower

13:43

Yes

user58512

I think they should be from K

awllower

No

user58512

ok

awllower

Yes they should come from K.

Sorry for the "fingerslap".

user58512

ah, :D

that makes more sense to me now

awllower

13:44

Good!

Jayesh Badwaik

what is a fingerslap? (in this context?)

user58512

we can easily set y=1

then we have $x^{\sigma_1}+x^{\sigma_2}+x^{\sigma_3}$ invariant under $C_3$ but not $S_3$

awllower

Fingerlap?

I am not sure about the usage, though.

@user58512 Indeed in some extensions the result is easier!

In fact I voted up the answer by Martin Brandenburg, before I knew what the question really was.
I regret for that...

I misunderstood the question at that time.

user58512

oops

I was thinking a little mistaken hmm :)

I'm thinking about $$K = \mathbb Q[X]/(X^3-2) = \mathbb Q(\sqrt[3]{2},\sqrt{3})$$

awllower

Oh

user58512

13:54

but in this case we have 6 automorphisms (3 for swapping $\omega^i \sqrt[3]{2}$ around, times 2 for swapping $\pm \sqrt{3}$ around)

so I think the form is invariant for $y \in \mathbb Q$

awllower

Indeed, I made that calculation earlier as well.

Quite misleading I found.

user58512

I should like to build a counter-example

for this $K$

awllower

oh

It could help by finding a normal basis firstly.

user58512

what is a normal basis?

awllower

Normal basis

In mathematics, a normal basis in field theory is a special kind of basis for Galois extensions of finite degree, characterised as forming a single orbit for the Galois group. The normal basis theorem states that any finite Galois extension of fields has a normal basis. In algebraic number theory the study of the more refined question of the existence of a normal integral basis is part of Galois module theory. In the case of finite fields, this means that each of the basis elements is related to any one of them by applying the p-th power mapping repeatedly, where p is the characteristic o...

user58512

13:59

hmm

awllower

Very interesting!
:D

user58512

I need to learn that theorem

awllower

The conclusion is like the followng:
$\sum_ia_{\phi_i^{-1}(m)}(b_{\phi_{i+1}^{-1}(n)}-b_{\phi_{\phi_j^{-1}(i)+1}^{-1}\phi_j^{-1}(n)})=0$ for every $m$ and $n$.

Quite ugly...

user19161

Ah, nowadays whether I answer a few seconds earlier or a few seconds later, I get less votes, lol.

user58512

I am confused :(

I should have wrote earlier $K = \mathbb Q[X]/(X^3-2) = \mathbb Q(\sqrt[3]{2},\sqrt{-3})$

awllower

14:11

It is by means of some calculations

Oh

user58512

what I don't get is, $\omega = \frac{-1 + \sqrt{3}}{2}$

wait

I just resolved it

was worrying that $\sqrt[3]{2} \mapsto \omega^2 \sqrt[3]{2}$ is same as the automorphism $\omega \sqrt[3]{2} \mapsto \omega^2 \sqrt[3]{2}$, but that's fine

awllower

Why $\sqrt[3]{2} \mapsto \omega^2 \sqrt[2]{3}$

Charlie

@jayesh :D

awllower

Maybe $\sqrt[3]{2} \mapsto \omega^2 \sqrt[3]{2}$?

user58512

oops

thanks

awllower

14:14

:)

user58512

imgur.com/a/TIEJg

a joke

awllower

Hm... Very difficult to understand...

user58512

its from the lord of the rings

awllower

And what is the second photo?

user58512

the ring gives you invisibility

user19161

14:27

Wow, my bleh got three stars.

Jayesh Badwaik

@awllower c'mon that lord of the rings is such an obvious one. :-)

skullpatrol

Why does the ring disappear?

awllower

I see. Haha

skullpatrol

You can't see the invisible.

awllower

I wrote a messy answer now.
Hope you like it! :D

user58512

14:30

I am amused by "by use of the same method, we can prove the same result"

awllower

Hm, on second thought, it appears to be quite strange now. Haha

user58512

I think the normal basis makes it much clearer

Dominic Michaelis

huhu

@jacobBlack sry for my terrible english

user19161

@DominicMichaelis Your English is better than my Chinese.

awllower

@DominicMichaelis Good to see you!

Maybe even better than my chinese. Haha

Dominic Michaelis

14:33

@awllower nice to see you too

awllower

:D

skullpatrol

We only see the gravatars...

awllower

:-)

Dominic Michaelis

maybe thats better ^^

peoplepower

Just so you all know, this is peoplepower's killer typing.

skullpatrol

14:34

...sort of like we're all wearing rings...

user19161

@peoplepower How does it kill?

awllower

My gravatar implies the class number formula for quadratic number fields: I think this is better than seeing me. :D

anorton

So guys... do you think I'd get in trouble if I appended the symbol "♦ " to my username? :P

user19161

@anorton Better not confuse others.

anorton

I had a feeling that would be the case... :) (It just clicked withe me that the ♦ symbol is a normal ASCII character)

Dominic Michaelis

14:38

are there unnormal ASCII characters ?

user58512

thats not ascii

user19161

Rather, abnormal.

user58512

it's black diamond U+25C6

fileformat.info/info/unicode/char/2666/index.htm

anorton

@user58512 Huh. I stand (well, actually sit) corrected.

awllower

I must depart to take care of my wounds now.
Later guys.

Dominic Michaelis

14:41

see you later

user58512

bye

Dominic Michaelis

i can'T enter the black diamond

◙

*

awllower

◆

Dominic Michaelis

its 2666 isn't it ?

awllower

I think it is the mode problem

Dominic Michaelis

14:44

*

awllower

And I think it is 25C6

Dominic Michaelis

nope doesn't change anything

♠

awllower

Hm...
I wish I could help...

anorton

@DominicMichaelis just copy and paste it...

skullpatrol

@JacobBlack When I saw this I thought about you calling yourself a banana ;-D

user58512

14:46

how do I produce a table in latex?

on this site

Dominic Michaelis

with \begin{array}{rlc} ...

maybe tabular works too

user58512

thanks, can you make ahve box lines?

Dominic Michaelis

jo

user58512

15:15

I need to get on with my group theory

awllower

To sleep.
Bye everyone.

Dominic Michaelis

bye

user58512

bye

Mats Granvik

Dr. Colin McLarty, Case Western University, Famous Math Theorem (Fermat's last theorem) Can be Proved Simply:
http://www.laboratoryequipment.com/news/2013/03/famous-math-theorem-can-be-proved-simply

user58512

bizarre misleading headline

McLartys project is showing that Wiles proof works in PA

Mats Granvik

15:22

Ok, I didn't know. PA = Peano Arithmetic I guess.

user58512

yeah, I think the original proof uses grothendeick universes (really big infinities)

and they wanted to see if that was actually necessary

it's just there's this whole thing about cranks claiming to have proved FLT using only high school math

so it kinda makes McLarty sound like one of those, if you didn't already know what he was doing

@MatsGranvik it's pretty intreresting beucase PA is very weak in some ways.. it can't prove some fast growing functions are really functions

oh..

> He hasn’t developed a proof for Fermat, but has shown that the theorem can be proved with much less set theory than Wiles used.

yeah, that's good

I didn't see that at first

Amzoti

Perhaps a moderator should have a look at this: math.stackexchange.com/review/low-quality-posts/50543

anorton

15:37

@user58512

$$\begin{array}{|c|c|c|}\hline
1 & 2 & 3 \\ \hline
2 & 3 & 1 \\ \hline
3 & 1 & 2 \\ \hline
\end{array}$$

(^ regarding box lines)

user58512

thank you, very nice

@Amzoti, oh it would be a good idea to flag it actually

that will specifically alert moderators

anon

@BenjaLim oh duh, they're both ideals

user58512

15:57

I don't understand fano plane, can you help me please

Sunny Marella

:need help with nonlinear dynamics.someone out there good with autonomous de's?

skullpatrol

@user58512 Do you know why Andrew Wiles chose Oxford over Cambridge?

Gustavo Bandeira

@SunnyMarella Just ask the question and pray for someone answering it.

Sunny Marella

Ok here goes.My system has 3 variables.Have been given a pair of 2d solution curves relating each of the variables to the other,X vs Y ,Y vs Z.From these curves,I need to arrive at a 3 system autonomous d.e w.r.t time.(To make it easier on the brain,one of the solution curves is a parabola opening to the right i.e a pair of implicits).**I pray now**

Gustavo Bandeira

16:13

@SunnyMarella btw, have you asked on the main site?

Sunny Marella

@gustavo,S.No one's answering there.Hope there's no rule for duplicating in a chat post.I'm super-desperate for a hint.Some headway...something.

user58512

@SunnyMarella, it's fine but I don't thik anyone knows here either

Sunny Marella

Tats a sham.Thght this site was for geeks :)

Gustavo Bandeira

@SunnyMarella Give your question a better polishing, using latex, etc.

Sunny Marella

@gustavo,wow!good idea,thanks man.will do.

Jayesh Badwaik

16:21

@SunnyMarella You have much better chance of getting the question answered on the site. Also, it is somewhat less clear what you have. You have three variables, and you have solution curves that pairwise relate the three variables, and you need to arrive at the set of differential equations that have the curves as a solution, is that correct?

Sunny Marella

@jayesh,yes and its got to be a 3 system diff.eq like a Lorenz system

Jayesh Badwaik

@SunnyMarella I understand. It would be difficult to help without actually knowing the equations.

Sunny Marella

okay,lets say X,Y,Z are the variables of the system. X vs Y is a parabola opening up to the right. Y vs Z is a sigmoidal curve,increasing for only +ve values of Y and Z.

Jayesh Badwaik

Okay.

user58512

16:39

hi @amWhy

Nimza

Hello-lo!

user58512

hey

Nimza

hi @user58512!

Hey guys, who is familiar with Betti numbers?

I wan't to find a space with my phone number as Betti numbers

2

user58512

haha

Jayesh Badwaik

@Nimza cool idea.

Nimza

16:49

@JayeshBadwaik :)))

@JayeshBadwaik Did you read about vector bundles (strenuously)?

Jayesh Badwaik

@Nimza no.

Nimza

@JayeshBadwaik so beautiful thing! I've downloaded lectures on it and watch it as some film :D

Jayesh Badwaik

@Nimza Haha. Cool! I know what they are. Do not know too much in detail.

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$Mathematics$ Mathematics