Mathematics - 2012-10-29 (page 4 of 5)

user19161

20:00

Hey @argon, how's your day?

Argon

@JasperLoy Fine.

School made it annoying!

user19161

@PeterTamaroff Oh man!

Peter Tamaroff

@JasperLoy I'm drinking ginger tea!

Ed Gorcenski

user19161

@PeterTamaroff You weren't here. There were 9000 flags in chat just now.

Ed Gorcenski

20:02

"Oops"

Peter Tamaroff

@JasperLoy Why?

user19161

@PeterTamaroff Well, the usual reasons. People got upset by our language.

Argon

@PeterTamaroff No Yerba?

:)

Peter Tamaroff

@JasperLoy What a bunch of fartheads.

@Argon Hehe, no.

user19161

So guys, this is my advice. It is best to avoid the f word altogether in this room.

Peter Tamaroff

20:05

Fuck that.

No really.

Argon

hahaahaha

Peter Tamaroff

People should know that "bad" words are just convention.

Being "offended" by such words is just a matter of personal taste. Saying "shit" is no worse than saying "moron". It is simply arbitrary.

Ed Gorcenski

This is that crane, formerly

Charlie

@JasperLoy what word?FUCK?

user19161

No, but really guys, you know what I mean. Let's not attract more unwanted attention.

Charlie

20:08

►

LIsten carefuly!

user19161

Barbara Streisand, a great singer.

Charlie

@Argon it always does...

user19161

I like her With One Look.

Charlie

"Think for one fucking second...shut the fuck up!"

user19161

She is shy in real life but not so on stage, applies to many singers I guess.

user19161

20:11

I guess I feel the same sometimes. When I sing, I forget all the fears and the music takes over. But that is only after you get enough stage experience.

Charlie

@JasperLoy then SING!!!!

user19161

So, those of you who get to meet me next time, I will sing for you!

Charlie

BTW, Good night , sleep tight @JayeshBadwaik!!!

user19161

...

Charlie

in portuguese , mean words sound more offensive...

agressive...

user19161

20:19

@anon Since stupidity also relies on neurotransmitters, it is probably finite as well.

Charlie

@JasperLoy human creativity to create stupidy is incredibly immense

anon

since stupidity is a measure of negative space, it is infinity minus a finite quantity pertaining to mental ability, thus it is infinite

user19161

@anon claps hand

anon

usually infinities for the same cardinal number are not further comparable, but you know how partial orders are obtained from subset containment

Charlie

:D

@EdGorcenski were you who advised me to "think less and write more?"

Ed Gorcenski

20:24

Yes

Has it helped?

Charlie

@EdGorcenski Thanks a lot!

@EdGorcenski it's helping.little by little

user19161

Thinking helps one write, and writing helps one think.

user19161

Mind affects body, and body affects mind.

Charlie

@JasperLoy man..I just think TOO MUCH

user19161

@Charlie Yes, you think too much of XXX.

Charlie

20:26

@JasperLoy math

user19161

These days, I usually think of ... . Hmm, that is my secret.

Charlie

@JasperLoy huh

Ed Gorcenski

You all have too many secrets.

Charlie

too many

Ed Gorcenski

Secretly, I hope the hurricane cancels class tomorrow.

Charlie

20:28

@EdGorcenski what hurricane?where do you live?

user19161

@EdGorcenski Is your life in danger?

Charlie

Gorcenski..Gorcenski i like to say:Gorcenski

Argon

@Charlie I like to say Marilia.

user19161

@Argon Shh, you are not supposed to say her name...

Charlie

@Argon the R sounds different

Argon

20:31

M

@Charlie I can roll R's

if needed

user19161

@Argon I like to say anon.

Charlie

@Argon translate.google.com.br/#pt/pt/mar%C3%ADlia

listen

@JasperLoy anon?like danon?

user19161

@Charlie I don't know. I just call anon "ah-non".

Argon

@Charlie I can say that!

Yay

Charlie

@Argon yay!! it's good to hear it, isn't it?i always feel weird when someone call me by my name

Argon

20:33

@Charlie Would you rather "Charlie?"

Charlie

@Argon it's not that i don't like my name, it's a poetical name, i love it.but it feels ... i don't know how to explain.

user19161

I am off to bed, I will see all of you in my dreams.

Argon

@JasperLoy I will see you there...!

Charlie

@JasperLoy bye bye.have nightmares with me :P

Argon

:)

user45170

20:35

I have an urgent question:

Charlie

Marília de Dirceu

Marília de Dirceu () is a poetry book written by Luso-Brazilian Neoclassic poet Tomás António Gonzaga. It is divided in three parts — all of them published in different years. The first part, published in 1792, has 33 "lyres" (or poems), and they tell mostly about Gonzaga's (using the pen name Dirceu on the book) love by a woman named Marília (who was, in real life, a girlfriend of his, Maria Doroteia Joaquina de Seixas). The second part, published in 1799, was written when Gonzaga was serving time in Ilha das Cobras because of his involvement with the unsuccessful Minas Conspiracy. Its 38...

Argon

@Charlie That book was probably dedicated to you

Charlie

@Argon it's a sad stupid love story

Argon

@Charlie About you?

user45170

I have an antisymmetric matrix. In one of my linear algebra scripts it says that, now every unitary matrix can be written as e^A. Does anyone have a proof for that?

Charlie

20:37

@Argon the true story!

Argon

:)

Ed Gorcenski

@Charlie Hurricane Sandy

@JasperLoy Virginia

anon

@user45170 take the matrix logarithm of $UU^*=I$

Ed Gorcenski

@Charlie There's like, a huge storm wrecking the east coast right now: cnn.com

Charlie

@Argon she was what?15 he was more than 30, they fell in love, blahblah and he had problems with government, had to go away to Moçambique, met a daughter of a farm owner, married her and forgot the love promises

Ed Gorcenski

20:40

@JasperLoy No, my life is not in danger. At least, not any more danger than normal, given my life as a secret agent.

anon

3/4 million without power, wowzer

Argon

@Charlie Weird.

Charlie

@Argon he was a douchebag

@EdGorcenski Oh my God!!!

Ed Gorcenski

The storm is actually three storms merging together: a hurricane, a cold front coming from the west, and a very cold front coming from Canada

The mixture of the three is expected to last for a week. It's predicted to be bad enough that it may actually affect the US Presidential Elections

Charlie

@EdGorcenski O.o

Argon

20:42

@EdGorcenski It is miserable today...

anon

yes, they were recommending early voting, in case you didn't get a chance later

Ed Gorcenski

Not every state allows early voting. Also, as a northeast native, people in the northeast are exceptionally stubborn

user45170

@anon, I never used the matrix logarithm, but do you mean the following: Let A be the log of UU*=I, that means that e^A=UU*=I

anon

no

Ed Gorcenski

This is a live video from NYC: video.msnbc.msn.com/nbcnews.com/49598904#49598904

That crane is about to come crashing down.

anon

20:44

@user45170 you do not want e^A=I, you want e^A=U with A antisymmetric.

Charlie

@Argon is it very cold today?

i feel guilty now...

Argon

@Charlie Cold, windy, rainy, cloudy...

Awful

Charlie

@Argon oh...

Ed Gorcenski

@Argon where are you?

user45170

@anon, may you can explain me what it means to take the logarithm of UU* ?

anon

20:46

google matrix logarithm

Argon

@EdGorcenski Toronto

anon

for one thing, it's an inverse to the exponential function on matrices

almost anyway

Argon

Bleh - Pedro

anon

my mobile is picking up all sorts of strange static

user45170

I just read the wiki article. It says that, if A is the log of B => matrix eponential of B = A

Charlie

20:48

@Argon hehehe

Dedalus

Hmm, is it true that any ideal I in a polynomial ring K[x] over a field K has minimal generators of any given cardinality?

anon

@user45170 and $\log (UU^*)=\log U+\log U^*=\log I$ implies $A=-A^*$, where $A=\log U$

Charlie

@Argon do you have Fb?

Argon

@Charlie No

Charlie

@Argon oh..

Argon

20:51

I never really needed it

Dedalus

Like (x,x^2+x) is a minimal generating set of cardinality 2 for (x)

(in that no proper subset generates it)

anon

doesn't {x} generate the same space as {x,x^2+x}?

Charlie

@Argon okay okay good for you!it's because there's a very funny page called:"I fucking love science".Thought you might like it

Dedalus

x^2,x^3+x,x^3-x^2 one of cardinality 3

oh , sorry

I meant {x^2,x^2+x}

anon

same thing, generated by {x} as well

Argon

20:53

:)

anon

every ideal in K[x] is principal

Dedalus

{x^2,x^2+x} is a minimal set of generators for x in that no proper subset generates it

anon

oh I see

Charlie

@Argon heavy metals

Argon

Hahahahahahaa!

Dedalus

20:53

and the same with {x^2,x^3+x,x^3-x^2} , should be a minimal for x of cardinality 3

user45170

@anon, thank you, that is really short. A must be a unique matrix, right?

Dedalus

So can we find for an arbitrary non-zero ideal I in K[x] minimal generators for any positive integer?

Charlie

@Argon noble metals

anon

@user45170 no, it is never unique. at least with complex numbers.

Argon

@Charlie Is it a thing to dress up as metals??? :)

anon

20:55

much like the $e^{2\pi i n}=1$ for each $n\in\Bbb Z$.

Charlie

@Argon nah..don't think so.

Argon

S'funny

user45170

@anon, so the proof holds for complex and for real numbers, but A is only in the complex case uniquely determined. But why only with complex numbers?

Charlie

@Argon check this out

Argon

@Charlie Is this where your Argon jokes are from??

Charlie

20:56

@Argon no.Argon jokes comes from myself.Totally new!

Argon

Hahaaa

anon

@Dedalus I think so. Let $I=(p(x))$. Given $n$, let $\{p,e_1,\cdots,e_{n-1}\}$ be linearly independent over $K[x]$, and then set $I=(e_1-p,e_2-e_1,\cdots,e_{n-1}-e_{n-2},p-e_{n-1})$. This is of course just a generalization of the stuff you were writing.

Argon

Yay

Dedalus

Right, that seems reasonable

anon

@user45170 Wait, what? I said it is never unique with complex numbers. I told you why by giving you the scalar explanation: because $\exp(2\pi i n I)=I$ for ever integer $n$.

Dedalus

20:58

thanks

Charlie

@Argon why yay?

Argon

@Charlie Just yay.

yay

Charlie

@Argon YAY!

Argon

@Charlie YYYYAAAYYYY!!!!

Charlie

@Argon hehehe

@Argon you know what's funny?

Argon

21:02

@Charlie What's funny

Charlie

@Argon to observe people studying mathematics!

Argon

:)

:)))))))) - Marilia

Charlie

@Argon in my study room , in my uni, it's really funny.It's the most quiet place on earth

Argon

@Charlie So you just sit around and laugh at people :)?

Charlie

@Argon no, i study but when i look around...it's fascinating

Dedalus

21:04

wait anon: How do you know that e_i-p is in I?

Or wait, you just pick the e_i linearly independent from I to start with I guess

Argon

@Charlie Never tried!

Charlie

@Argon some with their heads down, some sleep, some work hard, writing , some just stare at what their are doing...the faces!!! you can see when someone is not understanding!

Argon

Hahahahahaa

Charlie

@Argon it's a peaceful place!

Argon

@Charlie Except before exams?

Charlie

21:07

@Argon even before exams.

Argon

@Charlie Wow! No one is going nuts?

Charlie

@Argon no one is going nuts.We are nuts

anon

@Dedalus that's a good point, I don't think this works. Instead, consider $\{e_1,\cdots,e_n\}$ a minimal generating set for $K[x]$ itself, and write $I=(pe_1,\cdots,pe_n)$. Maybe that will work.

Argon

@Charlie Hahaha, very true!

Charlie

@Argon and i love to walk and listen to what they say!

the mathematical stuff!

the professors talking about their stuff

Argon

21:09

:)

Charlie

@OldJohn HEEEEEEEYYY!

Dedalus

@anon So then we need to show that we can find, for every n, a minimal generating set for $K[x]$ of any given cardinality

Old John

@Charlie Hi - how are things?

Charlie

@OldJohn good, good, good!

@Argon go to uni..it's a good thing!

anon

@Dedalus right. we should be able to use $\{x-1,x^2-x,\cdots,x^{n-1}-x^{n-2},1-x^{n-1}\}$

Argon

21:11

@Charlie I will be there soon!

Dedalus

@anon Can't we just leave out say x^2-x from that set? Since it is just x*(x-1)?

Charlie

@Argon yeah! there was a time that i couldn't take anymore school..I needed to go to a good place, where i could deal only with math

anon

drats, I'm thinking K-linear independence

Argon

@Charlie School annoys me a lot. A lot.

Charlie

@Argon you have 2 years left?

Dedalus

21:13

I tried some induction earlier, but didn't get too far with it

Argon

@Charlie This year and next

Charlie

@Argon hmm..you're 16, right? oh yea

Argon

Yup

Yay

Helmut

Can someone tell me what it means for a kernel to be a finite limit? I know it is a categorical limit, but why finite?

Dedalus

Helmut: The indexing category is finite.

Charlie

21:16

@Dedalus put @ before helmut

Helmut

Since the kernel diagram is finite then, the indexing limit guy is finite?

Dedalus

@Helmut The indexing category is finite, yes.

Helmut

what is the indexing cat for the kernel? isn't it isomorphic to the standard kernel diagram?

with the terminal object being the kernel

Dedalus

@anon Did you get anywhere with the cardinality argument? I am completely stuck

Peter Tamaroff

WHAT IF TELESCOPY IS INDUCTION IS DISGUISE?

@anon You there?

jdoe

21:20

it's not JUST induction

Peter Tamaroff

@jdoe "JUST"

Charlie

@PeterTamaroff you are induction in disguise, oh yes you are....

anon

@dedalus nah @peter hardly a disguise

Peter Tamaroff

@jdoe What is your objection?

@anon Just joking. Man. Primitive roots of unity.

Dedalus

Too bad :/

Charlie

21:24

@PeterTamaroff did i tell you i like your answers?

Peter Tamaroff

@Charlie Nope.

user45170

@anon, The same result should also hold for symmetric matrices, right? I am not quite sure, but if A symmetric => e^A positiv definit, and every positiv matrix T=e^A, whereas A*=A. Is it here also possible to use the logarithm ?

Charlie

@PeterTamaroff i do

@PeterTamaroff you helped me to study for a real analysis test,man!3 months ago

i passed

feel good about it

anon

@user45170 show that every eigenpair (a,v) to the matrix A corresponds to an eigenpair (e^a,v) of the matrix e^A, and clearly e^a is positive. I think there's also some book-keeping to do on the sidelines

Peter Tamaroff

@Charlie =)

Charlie

21:30

@PeterTamaroff :D

user45170

I already proved that e^A is positive deifnit

anon

wait so what is the question?

Helmut

@Dedalus what is the indexing category of the kernel?

user45170

Every positive matrix can also be written as e^A, with A*=A. Is it possible to use the logarithm here also?

jdoe

log(e^M) = M?

then again what's the logarithm of a matrix..

Peter Tamaroff

21:36

@jdoe Why not infinite series =P

Dedalus

@Helmut I consists of two objects, 1 and 2, with two morphisms besides the identity.

Peter Tamaroff

@anon Man. Can you tell the basics about primitive roots of unity?

Old John

@PeterTamaroff what do you need to know about them? - in some ways they are really simple, but they can have some really interesting developments :)

Peter Tamaroff

@OldJohn They smell groupish.

...and cyclish.

Argon

What's a root of unity?

Dedalus

21:41

@anon (Sorry if I'm disturbing you, tell me if that is the case) : Maybe some induction argument could be made to work? It is true for n=1, and if we assume it is true for n, let I be any ideal and take a set of n minimal generators {x_1,...,x_n}. Now, {x_1,...,x_{n-1}} generate a proper ideal of I, and thus have a set of n minimal generators

Old John

they do indeed form a group (or several groups - depending on which roots of unity you look at)

Peter Tamaroff

@Argon A complx number $\omega$ such that $\omega^k=1$.

Argon

@PeterTamaroff Oh right :)

Peter Tamaroff

We actually call that $k$-th root of unity.

Argon

De Moivre's theorem

Old John

21:42

if you just look at the n'th roots of unity for a fixed n, then they are a cyclic group

Peter Tamaroff

@OldJohn Yes, under multiplication, right?

Old John

but if you look at the group of all n'th roots of unity for all n, they are a group, but not cyclic

Argon

Scary

Old John

yep - multiplication

anon

I speculate cyclotomic polynomials categorify something, although I'm not sure what.

Old John

21:43

in the complex plane they are equally spaced around the unit circle starting at 1

Dedalus

maybe we can show that if we take those n generators and x_n, they generate I and no proper subset generates it

Old John

useful property is that the sum of all the n'th roots (for fixed n) is 0

Peter Tamaroff

@OldJohn I just proved that, but it is actually $\omega^n=1$ then $$\sum_{k=0}^{n-1}\omega^k=0$$

For $\omega \neq 1$.

Old John

@PeterTamaroff you need some assumptions there - like ...

(you just said it)

Peter Tamaroff

@OldJohn =P

Old John

21:46

actually you need more assumptions than that

Peter Tamaroff

@OldJohn ORLY?

I needed no more nor less.

Old John

unless you are assuming that n is a prime

anon

no you don't need that

even if $\omega$ doesn't generate the whole group, it will generate a subgroup but with multiplicity, and you will still get 0

Old John

yep - anon is spot on (as always!)

Peter Tamaroff

@OldJohn How is that sum useful?

anon

21:47

character theory

gauss sums, discrete fourier transforms

Peter Tamaroff

@anon Dirichlet's characters?

user45170

@jdoe, where is the relation between log(e^M)=M and my problem?

anon

@PeterTamaroff as particular cases yes

Peter Tamaroff

@anon Like $\sum \omega^{k^2}$?

Old John

@PeterTamaroff Yep

There are some really neat things that can be done with Gauss sums - most depend on that sum being zero (I think)

and some good examples in Ireland and Rosen's book

jdoe

21:50

quadratic reciprocity!

Peter Tamaroff

Darn. So many stuff to learn.

Old John

@PeterTamaroff and the more you learn - the more you find out that there is out there that you don't knowe

Argon

@PeterTamaroff You'll be fine; you are a genius anyways

Old John

think of your knnowledge as an expanding ink-blot - the bigger it gets, the more stuff there is that is just beyond the horizon :)

Peter Tamaroff

@anon I need to find all $n$th primitive roots of unity for $n=2,3,5,9$

anon

21:52

cool. do it by hand. work with Z/nZ's for simplicity.

Peter Tamaroff

Any hints to make things less tortuous?

user45170

@jdoe, quadratic reciprocity?

Old John

@PeterTamaroff just look for equally spaced points round the unit circle starting at 1

Peter Tamaroff

@anon Integers $\pmod n$?

anon

you think $n<10$ is tortuous? heh.

Peter Tamaroff

21:52

Why?

@anon =D

jdoe

@user45170, sorry I don't mean about your problem

Peter Tamaroff

@anon Is there any general formula for the number of primitive roots of unity for each $n$?

anon

the group of nth roots of unity is isomorphic to Z/nZ, with exp(2 pi i / n) mapped to 1 and 1 mapped to 0

Old John

if you want really tortuous, try doing arithmetic in the algebraic integers of the 23rd cyclotomic field :(

anon

@PeterTamaroff $\varphi(n)$

Peter Tamaroff

21:53

@anon Hmmmm

jdoe

what happens in the 23rd field?

Old John

@jdoe it is the first one in which unique factorisation fails

Kummer found it the hard way

Peter Tamaroff

@anon My definition is that $\omega$ is primitive $n$ if $\{1,\omega,\dots,\omega^n\}$ is the set of all $n$th roots of unity. In some sense, it generates the set, yes?

jdoe

oh no!

anon

sorry, I was thinking of the generators of $U(n)$ when I said phi(phi(n))

Old John

21:55

@anon too many phi's?

Peter Tamaroff

@anon I think I'll read about these bastards in Apostol's book of NT.

anon

don't worry, he rescued UF by being an idealist

3

jdoe

hahaha

Old John

@anon Brilliant!

I once read (in Edwards book on FLT) the actual calculations he did - they were hideous in the extreme

user45170

@jdoe, may you explain what you meant by log(e^M)=M

jdoe

21:57

@user45170, I thought that solved the problem bt I guess I was wrong

anon

see ya I'm going to class

jdoe

bye

Peter Tamaroff

@anon You mean you're lecturing.

Transcript for

$Mathematics$ Mathematics