Mathematics - 2016-11-25 (page 6 of 6)

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user227867

23:04

Meow.

user227867

@JessyCat Ideally one should know real analysis and complex analysis before tackling Fourier analysis.

Krijn

Shit, I will never be as cool as this guy: wstein.org

Mike Miller

that's true

Krijn

Professor of number theory and skateboards

Mike Miller

He is no longer faculty.

user227867

23:12

Sorry, but skateboards are not cool.

user227867

Yesterday, I got some blue socks to go with my blue watch and blue spectacles

Krijn

What are you, Yves Klein 2?

user227867

Nope.

user227867

I think I should watch Home Alone 1,2,3,4,5 again.

meow-mix

is it true that $F^{\mathsf{Ab}}_A \cong \mathbb{Z}^{\oplus \infty}$ for countably infinite set $A$?

usukidoll

23:16

I can't believe what I saw x. X

I was searching a question similar to what I have and the transcript from this chat showed my old messages haha

That was like 3 years ago

Adeek

yes

meow-mix

@Adeek to my statement?

Adeek

yeah @meow-mix

Fab

I like that notation

meow-mix

category of free abelian groups :P

and, $\mathbb{Z}^{\oplus\infty}$ is the set of all finite sequences of $\mathbb{Z}$

correct?

Adeek

no it is the category of fabulous groups !! @meow-mix

infinite sequences

meow-mix

23:20

not according to @Tobias

Adeek

oh wait it is the set where you have

it is the set of almost finite sequence

which means that it is 0 for almost all the sequence

meow-mix

what?

Mike Miller

The free group on $A$ is, as a set, the set of maps $A \to \Bbb Z$ such that all but finitely many elements of $A$ map to zero.

Addition is defined the obvious way.

Adeek

@meow-mix did you study infinite tuples ?

meow-mix

Adeek

23:23

It would have been easier to understand.. but yeah you can think of it as mike said

meow-mix

this is awfully confusing

why can't we say that $F^{\mathsf{Ab}_A} = \mathbb{Z}^\infty$?

Adeek

$\mathbb{Z}^{\infty}$ is another object

meow-mix

what is the difference between $\mathbb{Z}^{\infty}$ and $\mathbb{Z}^{\oplus\infty}$?

Adeek

it is defined the infinite sequence

$\mathbb{Z}^{\infty}$ isn't a group what would be an operation on it ?

meow-mix

addition component-wise?

Adeek

23:27

how do you define addition to make sense ?

@Krijn here ?

meow-mix

what?

Adeek

@meow-mix what is addittion of infinite sequence anyway ?

Krijn

@Adeek Yeah, w'up?

Adeek

@Krijn I think I have classified it would you like to hear my argument ?

meow-mix

$(a_1, a_2, \dots) + (b_1, b_2, \dots) = (a_1 + b_1, a_2 + b_2, \dots)$

Krijn

23:29

@Adeek Yeah, but I haven't got much time

So hand wavy is probably fine for now

Adeek

Okay

meow-mix

how would that not satisfy the group axioms?

Adeek

@Krijn so we computed the splitting field for $f(x) = x^4 - 2$ to be $\mathbb{Q}(i,(2)^{1/4})$ and we agreed that the $|Gal(K/\mathbb{Q})| = 8$. So our group must be of order 8.

Krijn

Yeah

Ted Shifrin

$x^4-2$, of course.

Krijn

23:31

Missed a 2 in $f$ but kay

Ohhhh, you @Ted

Ted Shifrin

And hello @Karim, @Krijn.

meow-mix

oh hey, its @ted

Adeek

hi @TedShifrin

Ted Shifrin

hi @meow

Adeek

ted could you answer meow question because I would like to finish this

meow-mix

23:32

@TedShifrin im having trouble understanding

Ted Shifrin

@meow: You're supposed to do projective geometry and you've gotten into category theory (which I hate).

Alessandro

Hi @Ted

meow-mix

its an algebra book

Ted Shifrin

heya @Alessandro

well, don't talk to me about that stuff, @meow

meow-mix

aluffis' specifically

Krijn

23:33

Lol

Ted Shifrin

I really think it is totally inappropriate for you at your level.

But you can do whatever you want ... just leave me out of it.

Adeek

an element of $\sigma_1$ is defined as $i \mapsto -i$ and fixes the other element. This $\sigma_1$ has order 2. $\sigma_2$ defined as fixing i and $(2)^{1/4} \mapsto (2)^{1/4}i$ we can see that $\sigma_2$ has order 4.

meow-mix

idk, somebody recommended it to me

Ted Shifrin

@Karim: Finding some of the intermediate fixed fields is cool ... have you figured them all out?

Adeek

If we denote $H = <\sigma_2>$ and $K = <\sigma_1>$ so then we can easily see that $Aut(K/\mathbb{Q}) = HK$

specifically $Aut(K/Q) = H \rtimes K$

meow-mix

23:35

you told me to study algebra, and i didn't have artin's, and i wont until i'm able to buy it, and so i asked for recommendations

Ted Shifrin

Leave me out of it, @meow, seriously.

Adeek

We can think of $H$ as $\mathbb{Z}_4$ and K as $\mathbb{Z}_2$ so we have $Aut(K/\mathbb{Q}) = \mathbb{Z}_4 \rtimes \mathbb{Z}_2$ so it is either $Z_4\times Z_2$ or $D_8$

Krijn

Okay

meow-mix

well, ok then

Adeek

so we have to check it is not a trivial semi direct product.

meow-mix

23:36

whatever you say.

Adeek

So after computing that I got it is not a trivial semi direct product as we can check that $\sigma_1 \sigma_2 \sigma_1^{-1}$ isn't the identity.

what do you guys think @TedShifrin and @Krijn ?

I computed the intermediate fields degrees but I thought for this problem I don't need it @TedShifrin ?

Jessy Cat

@JasperLoy well ideally, maybe, but it's not reality, and I am absolutely desperate to understand how to show the convergence for my Fourier Series on the intervals I need to show it on.

It's not that sophisticated Fourier Analysis. The book we use is used in undergraduate classes as well.

Krijn

It depends, I personally like arguments from Galois Theory where you can just see the results from drawing pictures of intermediate fields

@Adeek

Adeek

I should do that as well.. I will solve it this way as well to get more practice..

I still think little bit when I figure this pictures out so I need more practice.

I will also using how @TedShifrin suggested as well. But I think this way is correct that I did.

Jessy Cat

I just really want to understand how $\displaystyle \sum_{n=1}^N \frac{\left((-1)^{n+1}\cos(n\pi x_0)-(-1)^n\cos((n-1)\pi x_0)\right)}{2} =\frac12\left((-1)^{N+1}\cos(N\pi x_o)+1\right)$, but the guy won't explain it to me.

Adeek

23:40

@Krijn so how would you do it from the pictures and stuff ? do you just figure out the intermediate field extensions ?

and just use the galois correspodence to get the picture in term of the group and use that somehow ?

Krijn

@Adeek Yeah, it takes a while at first but after a while you can do it pretty fast

Adeek

oh ok I will try that.

I am cold it is pretty cold in my office here

Krijn

If you can't figure it out, this link will help (but it spoils a lot of the fun, SO DONT CHECK IT YET: math.jhu.edu/~vlorman/402s12/fieldlattice.gif)

Adeek