Mathematics - 2016-12-01 (page 2 of 3)

Null

18:00

@robjohn can the prove i want to make be made with leading coefficients? or are leading coefficients only viable above 1?

Akiva Weinberger

@Alessandro I'm more-or-less familiar with them

Adeek

18:18

hi

Null

@NaCl hi

Adeek

java-devel

Hi!

Adeek

I don't understand the more generally par

part *

more generally even if H is not necessarily normal in G, the isomorphism bla bla

what is that

what isomorphism of E ?

Akiva Weinberger

Automorphisms maybe?

Null

18:21

@AkivaWeinberger eh, that only procastrinates the problem imo

that's like saying: what is surjective? maybe bijective?

Akiva Weinberger

Oh wait it's not that never mind

java-devel

I need to understand what is a smooth function.

Akiva Weinberger

They just mean the set of all such isomorphisms, I think, that go into the algebraic closure of $F$ and fix $F$

@java-devel Probably infinitely differentiable

Adeek

oh I see

I see @AkivaWeinberger

Akiva Weinberger

@Null No, that would be continuous

Null

18:25

@AkivaWeinberger i see

Adeek

@AkivaWeinberger so it is the set of all isomorphisms that go into algebraic closure of F containing the bigger field K ?

which fix F

java-devel

How's difference between smooth function and continous?

Mike Miller

one has derivatives, the other doesn't

Alessandro

smooth=infinitely differentiable

Adeek

smooth function is infinitely differentiable.

smooth function is very continous one could say

Null

18:28

@Alessandro so one prime example would be $f(x)=e^x$?

Adeek

the proof of fundmental theorem of galois theory is very long haha

Alessandro

yep, or any polynomial, sin, cos etc.

Akiva Weinberger

Polynomials, $e^x$, trig… non-smooth ones are like $\lvert x\rvert$

java-devel

Is exists a specific name for functions that f'(x)=\int f(x)dx=f(x)?

e^x is a such function

Akiva Weinberger

Yes, they're called $ce^x$ where $c$ is any constant

(This includes $f(x)=0$, by the way)

Null

18:29

ah, so differentable doesnt mean, that the derivative is $\not=$ 0?

Ted Shifrin

Greetings, @Alessandro, DogAteMy, Karim, @MikeM.

Alessandro

hi @Ted

Akiva Weinberger

Hi

java-devel

@AkivaWeinberger Hi

Alessandro

I want to prove that a nonprincipal ultrafilter can't have a countable filter basis @akiva

Akiva Weinberger

18:30

Huh, don't know, lemme think

Alessandro

I wrote a sketch of a proof somewhere, let me find it

Akiva Weinberger

Also, never heard of a filter basis, but I think I can guess?

Astyx

Hi everyone

Ted Shifrin

Salut, @Astyx

Adeek

hi @TedShifrin

Alessandro

18:32

I'm talking about filters over a set $X$ here, so basically filters in $(\mathcal{P}(X),\subseteq)$ if you're familiar with the definition for posets

A filter basis is a collection $B$ of subsets of $X$ such that for every $B_1,B_2\in B$ there is $B_3\in B$ such that $B_3\subseteq B_1\cap B_2$

if you have a collection $B$ of subsets of $X$ then the set of subsets of $X$ containing at least one element from $B$ is a filter iff $B$ is a filter basis

does that make sense?

Akiva Weinberger

OK, I see

So it's not a filter itself, but it generates a filter

Maks

Hi !

Alessandro

yes, in particular if $F$ is a principal filter then $\bigcap F$ is a filter basis for $F$

Astyx

I had a question : let $f:\Bbb R \to \Bbb R_+$ be $C^1$ and $\lim_{\infty} f'$ and $\lim_{\infty} f$ exist, let $$g: \begin{cases} [0, {\pi \over 2}] \to [-{\pi\over2}, {\pi \over 2}]\\x\mapsto \arctan\circ f\circ \tan(x)\end{cases}$$ continuously prolongated on $\pi\over2$. On what conditions does $g'$ have a limit at $\pi\over2$ ?

(I hope my terminology is right)

Akiva Weinberger

So to show that an ultrafilter has no countable filter basis, we need to show that for any countable basis, there's a set $A$ such that neither $A$ nor $X\setminus A$ contains any of the $B$?

Astyx

18:37

@Ted How are you ?

Ted Shifrin

Very strange notation, @Astyx. You mean $\lim\limits_{x\to\infty} f'(x)$, etc.

Alessandro

that's one possible way, there are a few characterizations of ultrafilters provided in the notes I'm reading

Astyx

@TedYou're right, I just don't remember how to write those (and I also forgot some underscores ...)

Alessandro

ok, so as I was saying a wrote a proof but I'm fairly sure there's a mistake somewhere

Ted Shifrin

So, @Astyx, if you write out the chain rule, the only hope will be for $\lim\limits_{x\to\infty} f'(x)=0$ ... indeed, in such a way that $f'(\tan x)\sec^2 x$ has a finite limit as $x\to \pi/2$.

Maks

18:42

Doesn anyone know how can I know if $ \int_{1}^{\infty} \dfrac {cos(x^2)} {x^2} $ converges or diverges ?
Because I wanted to prove it by comparison, but the theorem says both functions have to be positive, and that function is not always positive as $ cos(x^2) $ is not always positive

Alessandro

first a sanity check, by definition of nonprincipal filter we must have $\bigcap F\not\in F$, but if $F$ is an ultrafilter we actually have $\bigcap F=\emptyset$, right?

Ted Shifrin

@Maks: converges absolutely. So stick absolute values in there and it converges by comparison, as you said.

Astyx

@Ted But that isn't sufficient is it ?

Ted Shifrin

@Astyx: Why not?

@Maks: As with infinite series, if the improper integral converges absolutely, then it always converges without the absolute values.

Astyx

The derivative would be $$(1+\tan^2)f'\circ \tan\over 1+f^2\circ\tan$$ (If I'm not mistaken), thus you would need to "flatten down" $1+\tan^2$ right ?

Maks

18:45

@TedShifrin I compared it with $ 1/x^2 $ but I'm not sure if its bigger or smaller, so I'm still stuck

Ted Shifrin

That was what I wrote as $\sec^2x$, yes. @Astyx

@Maks: $|\cos(x^2)|\le 1$, of course.

Astyx

Yes sorry, we do not use $\sec$ in France ... my bad

Ted Shifrin

@Astyx: So, yes, you need $\lim\limits_{u\to\infty} (1+u^2)f'(u) = 0$.

So we need $\lim u^2f'(u)=0$, in particular.

i.e., $f'(u) = o(1/u^2)$ as $u\to\infty$.

Astyx

My question does seem a bit silly, but I was wondering wether there existed a compactifiation of $\Bbb R$ to a segment that preserves derivative's limits at infinities

Ted Shifrin

Well, we really need to be talking limits as $|x|\to\pi/2$, etc., then, @Astyx. You're asking for a differentiable version of Stone-Cech compactification (which allows all continuous functions to extend continuously). I have no clue.

The Stone-Cech compactification is horrendous all by itself.

Astyx

18:50

Okay right thanks for your help :)

Ted Shifrin

I guess you're assuming limits exist, so then provided limits agree in both directions on $\Bbb R$, you're compactifying with the one-point compactification (the circle).

Astyx

I'll look into that

Ted Shifrin

But, as our calculation shows, you need reasonably rapid decay of $f'$ as you to off to infinity ...

Astyx

Is that true for any compactification ?

I also had another question : Are $\left\lVert\cdot\right\rVert_1$ and $\left\lVert\cdot\right\rVert_\infty$ homotopic (on the vector space of real functions) ?

Ted Shifrin

Well, no, for example, Stone-Cech compactification of a topological space allows you to extend all continuous functions, but it is a hugely messy space. I have no idea whether one can do this with derivatives. I guess if you assume $f$ is $C^1$, you could make both $f$ and $f'$ extend continuously. But I have no idea ...

JeSuis

18:55

@TedShifrin Hi, always on my exercise. I have $f$ be a smooth real-valued function on an compact submanifold $M$ of $\Bbb{R}^n$. If $a<b$ and $f^{−1}([a, b])$ does not contain a critical point, then there exists a diffeomorphism of $M$ from $M^a$ to $M^b$.
We define $\nabla f$ by requiring $\langle \nabla_m f,v\rangle=df_m(u)$ for all $m\in M$ and all $u\in T_mM.$ the set of critical points of $f$ is closed

Ted Shifrin

@Astyx: Isn't a convex linear combination of norms always a norm?

Hi @PVAL: You saw my answer to your question a day ago?

Astyx

I didn't quite get what you just said ^^'

PVAL-inactive

@Ted Ya thanks

Ted Shifrin

JeSuis

Why it follows that $f^{−1}([a, b])$ is contained in an open subset of $U$ without critical points ?

Ted Shifrin

18:57

@JeSuis: You were supposed to work this out yesterday. :)

Alessandro

assuming what I wrote above @akiva if a filter has a countable basis $\{B_n\}_{n\in\mathbb{N}}$ one can assume wlog that $B_1\subseteq B_2\subseteq B_3\subseteq...$, but if $\bigcap F=\emptyset$ I must also have $\bigcap_n B_n=\emptyset$ which is a contradiction since this is $B_1$ and I can't have the empty set in a filter (or a filter basis)

Astyx

@TedShifrin And what does this imply concerning my question ?

Ted Shifrin

@JeSuis: If not, you'd have a sequence of critical points converging to a point of $M^a$ that would also have to be critical.

JeSuis

@TedShifrin Yes but I still don't see why...

Balarka Sen

@JeSuis Isn't this normality of the manifold?

$f^{-1}[a, b]$ and the critical set are both closed. So by normality they admit disjoint nbhds.

Ted Shifrin

18:58

@Astyx: Isn't that a homotopy?

Null

@Alessandro ah ;)

JeSuis

@TedShifrin arrf right

Ted Shifrin

@JeSuis: But what Balarka says avoids compactness, as I suggested you could :)

JeSuis

@BalarkaSen I don't know what normality does means

Ted Shifrin

Any two disjoint closed subsets can be separated by open sets. :)

Mike Miller

19:00

Is $f$ proper?

Ted Shifrin

Depending on your definition of manifold, this may take a little bit of work.

Astyx

@Ted Ah right of course ... and what about a homotopy that would go through all $\left\lVert\cdot\right\rVert_{p}$ for all $p\ge 1$ ?

JeSuis

@TedShifrin Ok thanks

Ted Shifrin

@MikeM: I think JeSuis stipulated compactness at some point in conversation with me.

@Astyx: That I very much doubt.

Null

@TedShifrin can you explain why open is called how it is? since open means every value except the ones we "write"

Ted Shifrin

19:01

Not true, @Null. What are you talking about?

Mike Miller

If $f$ is proper, this is easy. The critical set is closed, so $f(\text{Crit})$ is closed. Then the story is transported to $\Bbb R$: there's a small open neighborhood $(a-\varepsilon, b + \varepsilon)$ that contains no critical points, and you can just take its inverse image.

Astyx

`Ok, thank you very much ! @Ted

Null

@TedShifrin (0,1) means every value inbetween 0 and 1, but not 0 and 1. So why is it called open?

JeSuis

@MikeMiller I see, nice, thanks

Ted Shifrin

"Open" means that you can wiggle (move around a little) at every point of the set and stay in the set. What you're writing is notation special to $\Bbb R$.

Balarka Sen

19:03

@TedShifrin I don't know a simple way to do it, actually. The proof I have in mind is by giving it a Riemannian metric, which actually makes it into a metric space.

JeSuis

@TedShifrin what is strange is that I don't need bump function... ^^

Null

@TedShifrin ah, so open means, for every element i choose, there is a higher one (assuming ordering) that is still in the set?

Ted Shifrin

@JeSuis: I was going to make you use a bump function to extend $f$ to an open set in $\Bbb R^N$ and then project the usual gradient. But your way is more efficient and better.

@Null: Both lower and higher, yes.

Mike Miller

Seems like it should follow straightforwardly from second countable and locally Euclidean.

JeSuis

@TedShifrin it seems interesting too

Ted Shifrin

19:05

@MikeM: Probably using local compactness or something.

Mike Miller

In any case not very exciting.

Semiclassical

hi chat

Ted Shifrin

The usual way is to use Urysohn metrization @MikeM.

Balarka Sen

Actually maybe I can just use a bump function to separate those closed sets, and invoke Urysohn's lemma.

Ted Shifrin

Hi @Semiclassic

Balarka Sen

19:07

Err, yeah, Urysohn metrization is much easier. Whatever.

Semiclassical

Any interesting math today?

Mike Miller

n

Semiclassical

dang

Ted Shifrin

Of course, construction of bump functions really uses normality @Balarka :D

Null

@TedShifrin mmh, ok then let's rephrase: why is a closed intervall called that way. I mean literally. Has it something to do with algebraic "closed" or why?

JeSuis

19:08

@MikeMiller why f(Crit) is closed ?

Mike Miller

proper maps from locally compact to hausdorff are closed maps

Semiclassical

Here's a question, then.

Mike Miller

alternatively if you said M was compact, Crit is compact, f(Crit) is compact

Ted Shifrin

@Null: "closed" means that it contains all limit points of the set.

Balarka Sen

Fair, @Ted

PVAL-inactive

19:09

What is being jabbered about? You just do it in a closed ball in each chart using a Euclidean metric.

JeSuis

@MikeMiller cool

Semiclassical

Suppose I've got the 1-form $\frac{dx}{\partial_y F}=-\frac{dy}{\partial_x F}$ on a complex affine curve $F=0$.

Null

@TedShifrin so there is no nice to memorize analogy to the terms? (closed/open)

Astyx

@Null Studying

PVAL-inactive

A compact manifold is clearly covered by finitely many closed balls.

Ted Shifrin

19:09

@Null: The key thing is that the complement of every closed set is open and the complement of every open set is closed.

But they are notions very different in flavor.

PVAL-inactive

For non-compact you need a locally finite subcover using the usual nonsense and you use that

Semiclassical

I can view this as the Poincare residue of the 2-form $\frac{dx\wedge dy}{F}$ (correct?)

Mike Miller

yeah

Ted Shifrin

@Semiclassic: Assuming you're doing this all with holomorphic coordinates, functions, etc. ...

Semiclassical

yeah. I have in mind $F$ a polynomial in $x,y$.

Ted Shifrin

19:11

(Your question sounded real at the beginning.)

Semiclassical

Yeah.

Where I was going with this is that that's the only case I actually know re: Poincare residues.

Null

@Astyx what you mean with "studying"?

Alessandro

that you'll learn this stuff when you study some topology @null

Astyx

^

Semiclassical

So what would be the residue of a 2-form like $Q\frac{dx\wedge dy}{F}$?

Presumably I need some conditions on $Q$.

Ted Shifrin

19:13

@Semiclassic: The $Q$ is presumed holomorphic, and it goes along for the ride (restricting to $F=0$, of course).

Semiclassical

I figured as much. So it's just $Q|_{F=0}\frac{dx}{\partial_y F}$

Ted Shifrin

Yup.

(The $1$-form is likewise on $F=0$, so probably you just understand that and don't notate it.)

Semiclassical

Now, here's the case that creates problems for me.

Yeah, point.

Suppose I wanted to ascertain whether the one-form $y\,dx$ is a Poincare residue of some 2-form.

What I'd think I need to do is write that as $y (\partial_y F)\dfrac{dx}{\partial_y F}$

Ted Shifrin

Remember what you're doing is local, so there are various things to worry about. For starters, is $y\,dx$ even a well-defined holomorphic $1$-form on your $F=0$?

$x$ and $y$ only make sense on $\Bbb C^2\subset\Bbb P^2$.

Semiclassical

Well, the $F$ I have in mind in practice is $F(x,y)=(x^2+y^2+1)^2-t(x^3-3xy^2)$ with $t$ chosen so that $F$ isn't singular.

user379685

19:18

Hello could somone help me with x->0 1/x - 1/tgx ? WITHOUT LA'HOPITAL

Null

@BalarkaSen i finally understand why we can bound it by below ;)

Semiclassical

(The second part is chosen like that so that $F(x,y)$ is invariant under 120 degree rotations in the plane. not terribly relevant here)

Ted Shifrin

I'm worrying about things on the curve globally. Are you working on a compact Riemann surface or on a submanifold of $\Bbb C^2$?

Semiclassical

Do you mean, should I be working in terms of the projective curve rather than the affine curve?

Ted Shifrin

I'm raising that question, @Semiclassic, yes.

Semiclassical

19:20

I'm fine with doing that. That's actually where the issue becomes pressing

Ted Shifrin

So then you have to worry about whether things make sense globally (i.e., near infinity).

Semiclassical

In that case, $dx/\partial_y F$ homogenizes to $z\frac{z dx-x dz}{\partial_y F}$

Ted Shifrin

But your original meromorphic $2$-form $dx\wedge dy/F$ needs to be thought about at infinity, as well. It's not well-defined.

@user379685: What do you know that you can use?

Semiclassical

(I need that $z$ in order for it to be invariant under rescalings)

Going to need to head out for a bit.

later

Ted Shifrin

Bubye.

user379685

19:23

@TedShifrin hmmm try as elementary as possible

Ted Shifrin

That is not helpful, @user379685.

user379685

@TedShifrin sorry, i know that tgx/x=1 as x->0

Steve

How do I show that a function meets the stability criteria in order to be a subspace?

Tobias Kildetoft

@Steve I find that very hard to make sense of without further context

Akiva Weinberger

@Alessandro How did you get to the "without loss of generality" bit?

Steve

19:28

@TobiasKildetoft Suppose I have this set: ${a\cos(x)+b\sin(x)|a,b\in \mathbb{R}}$. How do I show that it's a subspace of $f(x)$? Are you Danish btw?

Tobias Kildetoft

@Steve I am

Steve

I imagine that I have to show that it meets the stability criteria for subspaces

Tobias Kildetoft

$f(x)$ is not itself a vector space (or space of any sort). Do you mean you consider the set of all functions $\mathbb{R}\to \mathbb{R}$ and that subset?

Steve

@TobiasKildetoft Du kan få opgaven på dansk, hvis du vil. Er ikke helt komfortabel med de matematiske termer på engelsk.

Tobias Kildetoft

@Steve OK

Aksel'sRose

19:32

ive noticed everyone uses mathJax in the chat but it looks exactly as typed. Am I missing a plug in that makes it readable or are all of you just practiced at reading it so you understand?

Tobias Kildetoft

@Aksel'sRose There is a link on the right

Aksel'sRose

ah thank you!

Steve

@TobiasKildetoft Er mængden $\{a\cos(x)+b\sin(x)|a,b\in \mathbb{R}\}$ et underrum i $C^0(\mathbb{R})$?

Ted Shifrin

@user379685: Your teacher likes sneaky tricks.

Tobias Kildetoft

@Steve Ok, so rather than all functions, it just considers the continuous ones, but that doesn't change how we need to solve it

user379685

19:34

@TedShifrin it has to be a sneaky trick :c

Tobias Kildetoft

(given that you know it is indeed a subset, which should be clear)

so you just need to show that it contains $0$, is closed under addition and under multiplication by real numbers

Ted Shifrin

Yes, substitute $x=2u$ and use trig identities to play around to see something.

Alessandro

ah, wait, @akiva I wrote it backward, it was supposed to read $B_1\supseteq B_2\supseteq B_3...$

Null

Is determine a different task than prove? (in analysis, which is proof-infested)

Ted Shifrin

I would need to see the context, @Null.

Tobias Kildetoft

19:35

@Null No, it is just a way to not reveal whether you are supposed to show the thing to be true or false

@Steve Where in Denmark are you at, btw?

Null

@TedShifrin determine a limit. We only have gotten the basic definitions of a limit, no laws we can build upon.

Steve

@TobiasKildetoft Copenhagen

Alessandro

I'm not sure about my statement on the intersection of the $B_n$ now @akiva

Tobias Kildetoft

@Steve Neat. I did my master's degree there

Balarka Sen

What's new

Steve

19:37

@TobiasKildetoft Cool, which university?

Ted Shifrin

@Null: You know your teacher better than I do. In an analysis course, it would ordinarily mean "determine, with proof."

Tobias Kildetoft

@Steve the :)

user379685

@TedShifrin have you found a solution? ;_;

Ted Shifrin

@user379685: Yes, I told you above what to do.

user379685

@TedShifrin sorry missed that

Ted Shifrin

19:39

Substitute $x=2u$, rewrite everything with a limit as $u\to 0$, and see what you find.

Alessandro

actually I think my "proof" doesn't work, I'll think about the "either $A$ or $X\setminus A$ belongs to $F$" characterization of ultrafilters you were suggesting earlier after dinner @akiva

Steve

@TobiasKildetoft nice :) I'm admittedly not comfortable in this topic yet, so how exactly am I supposed to show that the set contains 0

Tobias Kildetoft

@Steve Well, what would you need to set $a$ and $b$ to in order to get a function that is always $0$?

Steve

Oh, so just $0$ then. What about the addition and multiplication by real numbers? How do I show that's closed? @TobiasKildetoft

Null

eh, does $b^x$ grow faster then $x!$?

Ted Shifrin

19:45

No, @Null.

Tobias Kildetoft

@Steve Well, if you multiply $af(x) + bg(x)$ by a real number, what do you get?

Steve

A real value

Tobias Kildetoft

@Steve Let me try another way to say it: Try to expand $c\cdot (a\cdot f(x) + b\cdot g(x))$

Aksel'sRose

If $\alpha_1=(1\2)(\alpha+\alpha^*) and \alpha_2=(1\2i)(\alpha-\alpha^*)$ and I need to show they are selfadjoint, what would be the setup? I know that to be selfadjoint I need to show that $<\alpha(v),w>=<v,\alpha(w)>$ but my proof for the given alphas looks way to simple and incorrect....

Steve

Oh I get it now. I just have to show that $c(af(x)+bg(x))=caf(x)+cbg(x)$, right? @TobiasKildetoft

Ted Shifrin

19:50

@Aksel'sRose: Don't you know another definition for self-adjoint? Like $\alpha=\alpha^*$?

Tobias Kildetoft

@Steve right

Steve

Thanks

Aksel'sRose

@TedShifrin Yes, we havnt used that particular one very much.

Ted Shifrin

Try it.

The other way will work, too, of course, but it's more headache.

Aksel'sRose

on it

Alessandro

19:53

Hi @Balarka

Balarka Sen

Hi

Alessandro

We started talking about locally euclidean spaces today

Ted Shifrin

@Ali !!

Balarka Sen

Big fan of those

Ali Caglayan

@Ted !!

user379685

19:54

@TedShifrin could you guide me some more, i can't get it to work :l

Ali Caglayan

Your book came today

Ted Shifrin

Oh oh ... hides

@user379685: What did you do?

Alessandro

I'm trying to come up with a non Hausdorff one, but I don't want any hint

Ali Caglayan

How are you

also hi @BalarkaSen

Balarka Sen

Sure :)

Ted Shifrin

19:54

Doing fine, thanks, @Ali, and you?

Ali Caglayan

quite dandy

Balarka Sen

Hi @Ali

user379685

@TedShifrin ((tgx-1)+tgx/x)/2tgx

Ted Shifrin

Oh, @user379685. Better to write tangent in terms of $\sin$ and $\cos$ and use double-angle formulas for those.

Ali Caglayan

I went to an algebraic number theory lecture the other day

quite enjoyed it

Unique factorisation of prime ideals

Balarka Sen

19:56

Dedekind domains are good.

Ted Shifrin

Aha, @Ali ... Or failure thereof :P

Ali Caglayan

I think we are considering rings of integers so it worked

Ted Shifrin

So, Balarka, I see that you're fighting both mosquitoes and cockroaches. What phylum is next?

@Ali: Not always (I think).

Ali Caglayan

rings of integers of number fields?

Balarka Sen

Yes, those are Dedekind.

Ali Caglayan

19:57

hmm I would have to check, I just literally walked into a lecture

Ted Shifrin

But there are number fields where unique factorization fails, right ... ?

Tobias Kildetoft

@TedShifrin Not of ideals

Balarka Sen

Yep

Ted Shifrin

Oh, OK ... I've forgotten all this stuff.

sticks with calculus and geometry

Balarka Sen

Kunneth proved his version of FLT by passing to ideals instead, where everything worked.

Alessandro

19:59

That's completely unrelated but I was reminded of it by the mention of Dedekind, it took only 2 hours of the logic course today to get an additive group structure on the set of Dedekind's cuts of Q, we'll deal with multiplication next time...

Transcript for

$Mathematics$ Mathematics