I am afraid the source is messier.

Sure, @MarcGato :)

I mean, there are a lot of unecessary things around.

I could maybe post it on the main.

Do it.

Phew

He really ran away

@Studentmath, I don't see why they both with saying $x>\epsilon n/2 + \epsilon^2 n/10$. The latter term can be absorbed into the first term by making $\epsilon$ slightly bigger.

You should ask such questions more often, @Studentmath. :P

haha

You mean slightly smaller, @Ted?

h

@MikeMiller Hello

The problem.

Thoughts.

Say.

See?

ok

oh, @Balarka didn't run away

hi @Mike

I think I meant slightly bigger, @Studentmath.

Prof. @Ted, than I'm afriad I don't get how..

@MikeMiller $L$ contains $\alpha$ s.t. $\alpha^n \in \Bbb Q$.

n

Thus, $L$ is the splitting fields of $X^n - \alpha^n$ as $L/\Bbb Q$ is galois. Thus $L$ contains the $n$-th root of unity.

@MikeMiller hmm?

$\epsilon' = \epsilon(1+\epsilon/5)$, right, @Studentmath?

Oh that was L

terminologies terminologies

Yes @Ted

no, $L$ is generated by some such $\alpha$, rather than containing all such

well, $\alpha \in L$, no?

that is precisely what I am saying

Oh, I see, it's just a matter of how we look at it. Proceed, @Studentmath :P

Good night all!

a generator of a field can't be outside the field!

night, @Marc

a generator of an extension can, @Balarka :P

that's not "precisely" what you're saying

@TedShifrin oh? examples.

don't be silly, @Balarka

Oh :P let me try to go over the first inequality again. I think they must leave $\epsilon$ in as it is important to the proof as is

$F(\alpha) = K$ ... $\alpha$ is outside $F$. :D

I am claiming $\alpha \in K$ rather than claiming $\alpha \in F$ @TedShifrin

=P That's why I was asking for examples

It's all a matter of language, @Balarka.

Teeeeed.

@MikeMiller Back to business.

I'm sick, @Anthony. Whatcha want?

I will try to post it on the mains, will be more readable there. I just can't wrap my head around this.

@Balarka Prove your claim

@TedShifrin It's not very hard, I don't think, I'm looking for a subset of $\mathbb{R}$^2 that projects to [0,1], but such that the projection map restricted to that subset is open, but not closed. What should I be thinking about?

Or I could just bother @MikeMiller.

your "thus" doesn't follow immediately from anything else so far

oh, projections are always open maps

mr @Pedro !

and I must say I'm being far more polysyllabic than I want to be

Just grunt, @Mike.

Mm.

They've changed the way names/avatars appear in here ...

@MikeMiller $X^n - \alpha^n$ has roots $\zeta^k \alpha$.

Projections are always open? But it wants one that isn't open later on! Hmm.

So

I think I must have stated my question incorrectly.

K/Q is galois, thus as K contains \alpha, K also contains the other roots

Thus K must contain n-th roots of unities

No, you were right, @Anthony.

I'm just informing you.

But is asks for a projection that isn't open later on.

k

What kinds of closed sets are you envisioning, @Anthony?

Projection maps are always open (in the product topology).

@MikeMiller however K is generated over Q by all the roots of unities L contains, thus $\zeta \in K$

k

@TedShifrin I dunno what I'm thinking then. Let me read this question a bit more.

"k" isn't even a grunt

did @Pedro ever say a word?

@MikeMiller Gak(L/K) fixes all the n-th roots of unities in L pointwise, by definition of galois groups. but i am still thinking how thats supposed to help us.

bah

ok, that's a grunt.

I'm terrified, @Ted

sigh posted, now all there's left is to hope and stare at this.

That's good, @Mike

Prof. @Ted, how long till the semester ends? Will you cover all of Sheldon's book?

Week 6 finishing now. 9 more ... I won't cover the whole book, no.

): @Ted

@MikeMiller you just said two words

I'm skipping various things along the way. Certainly don't need to do all possible discrete probability distributions. Will do binomial, Poisson, maybe hypergeometric.

Not to you, @Balarka, though you have now baited me into doing so

heh

Will you reach the law of big numbers, though?

YEs, definitely, @Studentmath.

With an idea of the proof, yes.

@Studentmath are you familiar with the strong law of small numbers?

Not sure.

@Mike, so why be you terrified?

google it @Studentmath =P

hahaha, yes just did that

You dare summon me, @Anthony?

@Ted Qual tomorrow.

Well, quit your smoking :P

Mike smokes?

That's news.

n

@Ted I really enjoyed that chapter and the proofs in it. Was a shame we had no work to hand in about it.

I don't think any mathematician is "super cool." :D

Alex is =P

I'm not a mathematician, @Ted

yes, you are.

I am tentatively an aspiring mathematician

tentatively? aw come on

Fortunately I am not. super cool

there's always the possibility I move to Portland and start a fried pickle stand, @Ted

Depends on how tomorrow goes

@Mike: Don't be silly.

Hey @Balarka, you could break your three-month hiatus from the mains and see how lovely I formulated my question there, while I walk the dog.

Most students don't pass quals first shot. Stop that.

@Studentmath no thanks

@Mike I heard they make tons of money there

You're right, @Ted, I hate fried pickles. I'll be a musician.

OK, time for me to go cook dinner. Do your best, @Mike, and it's no tragedy either way.

@TedShifrin Oh? So it isn't nessesary to pass quals on first shot?

no, @Balarka

did you?

Enjoy @Ted! @MikeM too close to Maths I'm afraid

how many shots are allowed?

depends who you ask

depends on the university ... some have an advanced oral exam, rather than quals on first-year material ... those one is expected to pass

of course he did @IceBoy

BTW, @Balarka, the word is necessary

and i am not being sarcastic, FYI

@TedShifrin hmm? what word?

@Studentmath I suppose I could just rent an apartment and listen to Iggy Pop until I run out of money.

You could start a pyramid business.

Rent the rented apartment for someone to rent.

rolls eyes ... bubye, guys :P

@MikeMiller Just get yourself a work at a burger shop

later

@Mike, if I don't see you later, get sleep and do your best.

and come up with something groundbreaking later on

like Zhang

Would rather listen to iggy pop, @Balarka

Or Eminem @Balarka

Okay I'm off to the dog. Be back later, maybe someone will answer..

\O_o/

@MikeMiller

ok no you're gone

lol

What can do I to provide more content for a letter of recommendation besides "this student was in my class and got an A and likes math a lot"

Hello @MikeMiller

So, how did it go?

h

Tell me good or bad.

how did what go

You know what.

algebra?

Errything.

algebra's tomorrow

Like the whole 12.

ya twit

Oh.

I didn't know.

yeah

So what about Analysis?

YEAH

I BET YOU DIDN'T

I'm not taking an analysis qual.

I'm lost.

Hi @TedShifrin

Hi @nabla

Huh?

@TedShifrin Hello Ted.

Hi mr @Pedro

So still mourning about probability?

=)

@PedroTamaroff I'm not sure why you're lost.

Sorta, also being sick, so I don't care :)

@MikeMiller I thought you did have analysis quals.,

Be lost no more: I don't.

You shall, though ....

Nope.

@Pedro: I've written up our exercise for the next problem set.

Nope?

Nope.

You really are selling fried pickles?

@TedShifrin Cool. =)

What are fried pickles?

I'ev enver heard of such

Oh god. How do you eat these things?

No, @Ted, last we spoke I decided to get an apartment and listen to Iggy Pop until I ran out of money.

Or Iggy Azela

I just don't get it. I think their inequality is wrong.

Don't say that @Studentmath

The Idiot is a classic album

Azela or inequality is wrong?

Oh.

Yeah it is.

@DanielFischer

Are you around?

Azalea's albums will never be called classics.

You can't know just yet.

None of her currently released albums.

Will ever be considered classics... by me.

Well, that's more probable

Complex Analysis: Given a function f(x) that is analytic in some domain D in the complex plane, if the conjugate function is conj( f(x) ) = c^2/f(x) then it is analytic. I read in a Churchill that this is true. Why does it follow that the conjugate must be analytic?

Churchill also did math?