Mathematics - 2014-07-17 (page 2 of 3)

robjohn

3:01 PM

@Vrouvrou with what?

Vrouvrou

on the definition of lim inf

robjohn

@Vrouvrou it is the infimum of a sequence as you look only at the tail $$\liminf_{n\to\infty}a_n=\lim_{n\to\infty}\inf_{k\ge n}a_k$$

Vrouvrou

how i can write this: $$\liminf_{|u|\rightarrow 0}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}\geq \alpha$$

using $\varepsilon $

robjohn

@Vrouvrou it is the same for functions.

Vrouvrou

lim inf of a function

for exemple in this paper

robjohn

3:07 PM

@Vrouvrou $$\lim_{x\to0}\inf_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}\ge\alpha$$

Vrouvrou

they say that

$\displaystyle\liminf_{|u|\rightarrow 0}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}\geq \alpha$ implies that $\forall \varepsilon>0, \exists \delta>0, |u|<\delta \Rightarrow \displaystyle\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2} \geq \alpha-\varepsilon$

robjohn

@Vrouvrou that is just what I have above...

Vrouvrou

how we delete the inf ?

robjohn

@Vrouvrou pardon? we don't delete the inf, we translate what it means.

Vrouvrou

in$\lim_{x\to0}\inf_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}\ge\alpha$

what i must do after that ?

robjohn

3:14 PM

the limit means that we can find an $x\gt0$ small enough so that for any $\epsilon$ we can make $$\inf_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}\ge\alpha-\epsilon$$

and that means that for all $|u|\le x$, we have $$\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}\ge\alpha-\epsilon$$

@Vrouvrou If the infimum of a function is greater than a given value then the function is greater than that given value

@Vrouvrou my $x$ is simply their $\delta$

Vrouvrou

we say that tere existe $x$ such that $\lim_{x\to0}\inf_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}=\liminf_{|u|\rightarrow 0}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}$ ?

robjohn

@Vrouvrou the $x$ is bound to the limit. It doesn't make sense to talk about it outside that scope.

@Vrouvrou that is the definition of lim inf

Vrouvrou

but how to pass from $\liminf_{|u|\rightarrow 0}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}$

to $\lim_{x\to0}\inf_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}$

i must say there exist $x$ no ?

robjohn

@Vrouvrou what do you mean how do you pass from one to the other? they are the same thing.

Vrouvrou

i want to write it correctly

i write directly $\liminf_{|u|\rightarrow 0}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}=\lim_{x\to0}\inf_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}$

but who is x ?

robjohn

3:26 PM

@Vrouvrou to use the limit, you do the same thing as you always have with a limit. say that there is an $x\gt0$ so that $\inf_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}$ is close to the limit value

@Vrouvrou $x$ is the variable in the limit...

Vrouvrou

what it means "is close to the limit"?

robjohn

Given an $\epsilon\gt0$, we can find an $x\gt0$ small enough so that $\left|\inf\limits_{|u|\le x}\frac{\tau F(t,u)-f(t,u)u}{a(t)|u|^2}-\alpha\right|\le\epsilon$

since the limit value is $\alpha$

Shisui

Is there a short way of proving that $x^2 + 1 > x$ for $x>0$?

robjohn

@Shisui note that $(x-1)^2\ge0$ and work from there

that shows that $x^2+1\ge2x$ with equality only when $x=1$

Shisui

Thanks.

$2x > x$ so it's just a matter of squeezing the two inequalities together thereafter.

Right? @robjohn

Vrouvrou

3:36 PM

@robjohn where i can found this definition of $\liminf$ ? please

robjohn

@Vrouvrou have you tried google?

Limit superior and limit inferior

In mathematics, the limit inferior (also called infimum limit, liminf, inferior limit, lower limit, or inner limit) and limit superior (also called supremum limit, limit supremum, limsup, superior limit, upper limit, or outer limit) of a sequence can be thought of as limiting (i.e., eventual and extreme) bounds on the sequence. The limit inferior and limit superior of a function can be thought of in a similar fashion (see limit of a function). The limit inferior and limit superior of a set are the infimum and supremum of the set's limit points, respectively. In general, when there are multi...

Vrouvrou

i found only for sequences

robjohn

@Vrouvrou did you read far enough into that link?

the whole thing is the same for functions and sequences anyway

Vrouvrou

yes yes i read it thank you

but please

Hippalectryon

@TedShifrin :D

Vrouvrou

3:42 PM

we have that $\liminf_{|u|\to0} G(t,u)\geq \alpha$ so by definition $\lim_{\delta\to 0}[\inf_{|u|\leq \delta} G(t,u)]\geq \alpha$

Ted Shifrin

Salut @Hippa ... Tu ne brûles plus? :)

Hippalectryon

@TedShifrin je ne suis pas chez moi et donc pas sur mon ordi

Ted Shifrin

Beaucoup mieux. Moi, je ne suis pas chez moi non plus.

Vrouvrou

@robjohn after i say that $\forall \varepsilon>0,\exists x>0 ~sutch that ~ inf G(t,u) \geq \alpha-\varepsilon $ ?-

Ted Shifrin

Hi @robjohn ...

robjohn

3:45 PM

@TedShifrin guten tag...

Ted Shifrin

Kein Umlaut :)

Daniel Fischer

Hi @TedShifrin. When will you be away on vacation?

Vrouvrou

@robjohn please

Ted Shifrin

Hi @DanielF ... i am typing in San Diego as we speak ... Just back from an hour walk.

Chris's sis

@r9m When I publish the book, I'll post here a question (or more from that book) every day. :D

Daniel Fischer

3:47 PM

@TedShifrin Ah. I forgot that you modern people don't need a computer to be online.

Vrouvrou

@robjohn please it is the last time just tell me if it's right please

Ted Shifrin

Teleportation. @DanielF.

r9m

@Chris'ssis :-) nice !

Ted Shifrin

I trust you're holding down the fort just fine, @DanielF.

Hi@Chris'ssis, @r9m

Chris's sis

@TedShifrin Hi Ted

robjohn

3:50 PM

@Vrouvrou $\lim_{\delta\to 0}[\inf_{|u|\leq \delta} G(t,u)]\geq \alpha$ means that for any $\epsilon\gt0$, there is a $\delta\gt0$ so that for all $x\le\delta$, we have $\left|\inf_{|u|\leq x} G(t,u)-\alpha\right|\le \epsilon$

r9m

@TedShifrin hello :D

Daniel Fischer

@TedShifrin Well ... at least the fort hasn't been force-closed yet.

Chris's sis

@r9m I'm cooking (better say) something for you ... :D

r9m

@Chris'ssis are you sure you ain't cookin me ?! :P

Ted Shifrin

Are you a good cook, @Chris'ssis?

Chris's sis

3:54 PM

@TedShifrin Yeap, I think so. :-)

Vrouvrou

@robjohn sorry an other question to obteain $\alpha-\varepsilon$ we must have $|G(t,u)-\alpha|\leq \varepsilon $ no ?

Ted Shifrin

Glad to hear it, @Chris'ssis

Chris's sis

:D

Ted Shifrin

No @Vrouvrou: No control on the upper bound needed.

r9m

Oh ! I'm a horrible cook :P ...

skullpatrol

3:56 PM

I have heard about math papers being written like this?

Vrouvrou

so why $-\varepsilon$ @robjohn ?

Chris's sis

@r9m :-)))))))

robjohn

@Vrouvrou okay... I have rewritten my last comment. Notice that $|x-\alpha|\le\epsilon$ implies that $x\ge\alpha-\epsilon$

Vrouvrou

4:11 PM

@robjohn so we have $\forall \varepsilon>0, \exists \delta>0, \forall x\leq \delta \inf_{|u|\x} G(t,u)\geq \alpha-\varepsilon$

@robjohn after that as $x\leq \delta$ we have that $|u|\leq \delta$

@robjohn and then $\forall\varepsilon>0, \exists \delta>0 \inf_{|u|\leq \delta} G(t,u) \geq \alpha-\varepsilon $

right ?

skullpatrol

@Vrouvrou Have you tried posting this question?

Vrouvrou

no

skullpatrol

Just a suggestion :-)

Vrouvrou

just @robjohn tel me if it's right and then i understand

@robjohn please

skullpatrol

Someone telling you does not really make you understand something;
you have to convince yourself that your way is right.

Vrouvrou

4:27 PM

i think that it's right

but i need someone told me that it's realy right

skullpatrol

closely analyze what he has done and convince yourself

r9m

@Chris'ssis :D $$\lim\limits_{n \to \infty} \dfrac{1}{\log n} \sum\limits_{k=1}^n \dfrac{1}{k}\tan \frac{\pi k}{2n+1} = \dfrac{2}{\pi}$$

Chris's sis

@r9m Nice :-)

skullpatrol

@Vrouvrou you can be that "somebody" who tells you that it is really right.

Chris's sis

@r9m Wait a second ...

r9m

4:37 PM

W|A says standard computation time exceeded for me :| .. but I guess the value of the limit is right :-) ..

Vrouvrou

but @robjohn know if it is right or not

r9m

@robjohn see this :D .. they claim the general inequality hold as well !! :|

skullpatrol

@Vrouvrou he would not intentionally tell you something wrong, but he would want you to closely analyze what he has done and convince yourself that it is right.

That is how Professors are :-)

Vrouvrou

@skullpatrol i just want to know if i write something correct that's all

my problem is on th wrting

skullpatrol

post a question

Vrouvrou

4:42 PM

?

skullpatrol

26 mins ago, by skullpatrol

@Vrouvrou Have you tried posting this question?

Vrouvrou

ok ok

Daniel Fischer

Looks like Jasper has a new account.

skullpatrol

@DanielFischer nice work champ :-)

caught'em in 12 minutes

r9m

@math110 hi .. :-)

Balarka Sen

4:53 PM

You're banished for a month.

@TedShifrin Go away.

r9m

@Balarka take a hike :P

Ted Shifrin

Ok.

skullpatrol

@BalarkaSen Are you looking for a spanking?

Balarka Sen

No, no, wait I was kidding.

skullpatrol

show some respect pal

Balarka Sen

4:55 PM

@TedShifrin I've been to learn comm. alg.

So whenever you mention Ok it reminds me of dedekind domains.

robjohn

@Vrouvrou That looks right. Just use the definition of a limit and an infimum...

Balarka Sen

Oh noes. Ted took my words literally.

Huy

How do I find $$\int_\Omega \nabla u \cdot \nabla \varphi \, dx = \int_\Omega f \varphi \, dx$$ with $\varphi \in C_0^\infty(\Omega)$ from $-\Delta u = f$? I don't really remember how to integrate by parts with a Laplace operator or a nabla.

robjohn

@r9m My counterexample shows that is not the case... right?

r9m

@robjohn right :-)

user1876508

4:59 PM

In the following problem, why is it that "1 = qd + r"?
If R is a Euclidean Domain and d is an element of minimal norm m, show that every element of minimal norm is a unit.

robjohn

@r9m I thought so... Is that the same estimate that you were trying to show?

r9m

@robjohn that estimate was given by Cesar Lupu in AoPS as well :| ... I got it from there initially ... now I see it here :|

robjohn

@r9m That is so strange, that that has been so widely spread.

r9m

@robjohn I've got no idea :|

robjohn

@r9m The version with $n=3$ is nice, but it fails for $n=4$

Jasper Loy

5:03 PM

Hello.

robjohn

@JasperLoy Hey there.. long time

Jasper Loy

I managed to do some edits to get 20 rep to chat.

robjohn

@JasperLoy only been an hour and already you have 13 points!

@JasperLoy that doesn't include the 9 from ELU

Daniel Fischer

19 mins ago, by Daniel Fischer

Looks like Jasper has a new account.

Welcome home.

r9m

@JasperLoy Hi :-) .. how is the 12 months - 12 books plan going ? :D

Jasper Loy

5:07 PM

@r9m Well, it has not started. It is supposed to start on Jan 1 in 2015.

r9m

@JasperLoy oh ! .. okay :-)

Jasper Loy

On SE, it is easy to start over. Just delete your account and create a new one. In life, it is difficult to start over.

I am very happy that my favourite country won the World Cup.

Anirudh

Hi.

I am maths graduate but I forgot everything about maths now, what do I do?

13 2

lol

5:18 PM

I am going to eat a burger now.

skullpatrol

bon appetit

Pedro Tamaroff

@JasperLoy You're from Malaysia right?

Shisui

5:37 PM

I've got a question on open and closed sets.

Chris's sis

@JasperLoy It's easier to start over than doing nothing. :-) In life there is no guarantee that things will be always fine ... (this is just an ilussion).

Pedro Tamaroff

@Shisui OK.

Shisui

@PedroTamaroff The set $[a,b]$ is closed, right?

Pedro Tamaroff

Under the usual topology? Yes.

Shisui

@PedroTamaroff What would you call the sets $[a,b)$ and $(a,b]$ ?

Pedro Tamaroff

5:40 PM

One usually calls them "half-closed" or "hald-open", but that has nothing to do with topology.

They are neither open nor closed sets.

Shisui

@PedroTamaroff Somebody I know said that the set $[0, +\infty)$ is closed. I don't believe them. Am I right in saying that it's half-closed/half-open ?

Jasper Loy

@PedroTamaroff Singapore.

Pedro Tamaroff

@Shisui No, that is closed.

Jasper Loy

@Chris'ssis You misspelled illusion.

Shisui

@PedroTamaroff Why?

Pedro Tamaroff

5:41 PM

So it begins, again.

Shisui

Is it just the way it's defined?

Chris's sis

@JasperLoy Yeah, it's illusion.

Pedro Tamaroff

@Shisui Because if you have a sequence of points with $x_n$ with $x_n\geqslant 0$; their limit $x\geqslant 0$.

There is a difference between [a,b) with b finite and [a,+\infty)

Shisui

@PedroTamaroff What's the difference?

I didn't understand what you wrote about the limit.

Pedro Tamaroff

@Shisui Well, one is bounded and the other isn't, say.

@Shisui Read it again.

If $x_n\geqslant 0$, $\lim x_n\geqslant 0$.

skullpatrol

5:48 PM

@PedroTamaroff somethings will never change

this time I'm going to handle it differently is what I said the time before...

Anthony

6:08 PM

Can someone help me with the answer on this question? I'mma little confused :/ math.stackexchange.com/questions/869710/…

Jasper Loy

I am very upset that I answered first but did not get any vote. Then amwhy answered and got upvoted and accepted.

Anthony

@JasperLoy Whatchu talkin bout

Jasper Loy

@Anthony math.stackexchange.com/questions/870121/… here

Anthony

I gave you an upvote <3

Jasper Loy

Yay!

Daniel Fischer

6:12 PM

@Anthony If you mean the comment you linked to, that is, if I don't misread the comment, a misreading of the answer to the other question. If the derivative of $f$ has a zero of order $n$ in $z_0$, then $f$ is $(n+1)$-to-one in a neighbourhood of $z_0$. Essentially, near $z_0$, $f$ is $a + b(z-z_0)^{n+1}$ then.

Anthony

@DanielFischer Oof I didn't mean to link to the comment.

Daniel Fischer

@Anthony Okay, which answer do you want explained?

Anthony

@DanielFischer I just meant the answer by Robert Israel, I didn't know why it is necessarily constant in some neighborhood of zero.

Huy

What is the meaning of the notion $L_\text{loc}^1(\Omega)$ again?

Daniel Fischer

@Anthony Since $f$ is not constant (well, that isn't stated, but we must require it), there is a neighbourhood $U$ of $z_0$ such that $f(z) \neq f(z_0)$ for all $z\in U\setminus\{z_0\}$.

Locally integrable functions in $\Omega$, @Huy.

So, @Anthony, if $r > 0$ is small enough, you have $f(z) \neq f(z_0)$ for all $z$ with $\lvert z-z_0\rvert = r$. Now consider the integral $$N(w) = \frac{1}{2\pi i} \int_{\lvert z-z_0\rvert = r} \frac{f'(z)}{f(z)-w}\,dz.$$

Huy

6:20 PM

@DanielFischer: Does locally integrable mean that restricted to any compact subset, the function is $L^1$?

Daniel Fischer

@Huy It should mean that every point has a neighbourhood $V$ so that the restriction to $V$ is integrable. If $\Omega$ is a subset of an $\mathbb{R}^n$ [more generally any locally compact (Hausdorff) space], the two conditions are equivalent.

Jasper Loy

math.stackexchange.com/questions/870141/… I consider my answer superior, yet it has no votes.

Huy

Thank you, yes, $\Omega \subset \mathbb{R}^n$.

Daniel Fischer

@Anthony On the one hand, as a parameter-dependent integral, the general theory of integration tells you that $N$ is a continuous function of $w$ on $\mathbb{C}\setminus f(\{ z : \lvert z-z_0\rvert = r\})$.

On the other, $N$ only attains integer values.

A continuous function that attains only integer values is constant on every connected subset of its domain. In particular, $N$ is constant in a small neighbourhood of $f(z_0)$.

Huy

@DanielFischer: Do you by any means understand the following definition?

"The linear map $$C_0^\infty(\Omega) \ni \varphi \mapsto - \int_\Omega u \, \frac{\partial \varphi}{\partial x_i} \, dx$$ defines the derivative of distributions (Distributionsableitung) $\frac{\partial u}{\partial x_i}, 1 \leq i \leq n$."

I tried googling it a bit but it seems most sources don't distinguish between derivative of distributions and weak derivative, however a weak derivative is defined just afterwards in the lecture notes.

What exactly is the Distributionsableitung now? Is it the expression $\frac{\partial \varphi}{\partial x_i}$? Is it the integral? Is it the map?

Daniel Fischer

6:30 PM

@Huy The map. And the correct denotation of the space is $C_c^\infty(\Omega)$ or $\mathscr{D}(\Omega)$, not $C_0^\infty(\Omega)$. A distribution on $\Omega$ is a continuous linear functional on $\mathscr{D}(\Omega)$. Its partial derivatives are also distributions. $$\frac{\partial T}{\partial x_i}[\varphi] := - T\left[\frac{\partial\varphi}{\partial x_i}\right].$$ In the case where the distribution $T$ is given by a locally integrable function, you get the integral in question as a

representation of the partial (distributional) derivative.

Huy

@DanielFischer: I think there are different interpretations of $C_0^\infty$, some use it to denote functions vanishing at infinity whereas others use it for compactly supported functions. In our lecture notes, I think the latter is being done throughout.

Or am I being completely wrong there? We never defined any distributions. This is a course about functional analysis, the chapter being Sobolev spaces.

Daniel Fischer

@Huy Unfortunately not uncommon, but entirely wrong. Since $C_0$ is the important space of continuous functions vanishing at infinity, it is unforgivable to use it to denote $C_c$, the space of compactly supported continuous functions.

Huy

@DanielFischer: Please tell that to my professor, Michael Struwe. :P

Daniel Fischer

@Huy Where would I find him?

Huy

math.ethz.ch/~struwe

Daniel Fischer

6:35 PM

@Huy Too far away.

Huy

I am sorry.

Jasper Loy

Only a world away, at most.

Daniel Fischer

Not your fault, @Huy.

Huy

@DanielFischer: So, the Distributionsableitung depends on $u \in L_\text{loc}^1(\Omega)$ which can be arbitrary?

How can I see if $\frac{\partial u}{\partial x_i}$ denotes the partial derivative or the Distributionsableitung? Do I have to figure it out from context?

Daniel Fischer

@Huy Every $u \in L^1_\text{loc}(\Omega)$ defines a distribution $$T_u \colon \varphi \mapsto \int_\Omega u(x)\varphi(x)\,dx.$$ That distribution has distributional derivatives $\frac{\partial T_u}{\partial x_i}$. These are often called "the distributional (partial) derivative of $u$ (with respect to $x_i$)". If the classical partial derivative exists and is locally integrable, then we have

$$\frac{\partial T_u}{\partial x_i} \left[\varphi\right] = \int_\Omega \frac{\partial u}{\partial x_i}(x)\varphi(x)\,dx,$$ where the $\frac{\partial u}{\partial x_i}$ in the integral denotes the classical partial derivative.

So, in case both derivatives exist, we can identify the two.

Thus, if both exist, it doesn't matter as which you interpret it, but if only the distributional derivative exists, there is no choice.

Huy

6:46 PM

@DanielFischer: In my lecture notes, it reads $\frac{\partial \varphi}{\partial x_i}$. Is that a mistake or how is it equivalent?

(in the integral even)

Daniel Fischer

@Huy If the partial derivative of $u$ exists and is locally integrable, then $$\int_\Omega \frac{\partial u}{\partial x_i}(x)\varphi(x)\,dx = - \int_\Omega u(x)\frac{\partial\varphi}{\partial x_i}(x)\,dx$$ by integration by parts. So then we have $$\frac{\partial T_u}{\partial x_i} = T_{\frac{\partial u}{\partial x_i}}.$$

Huy

@DanielFischer: How does the $u(x) \varphi(x)$ part vanish? I know that for solving Dirichlet's problem, we can assume w.l.o.g. $u = 0$ on $\partial \Omega$; is that why or is there something else?

Daniel Fischer

@Huy You mean the boundary integral? $\varphi$ has compact support in $\Omega$, so the integral over the boundary vanishes.

Huy

@DanielFischer: I'm sorry I'm not quite following. How does the integral over the boundary vanish exactly?

Daniel Fischer

@Huy You take a $U\subset \Omega$ with (sufficiently) smooth boundary such that $\operatorname{supp}\varphi \subset U$. Then (assuming that $u$ is regular enough) $$\int_\Omega \frac{\partial u}{\partial x_i}(x)\varphi(x)\,dx = \int_U \frac{\partial u}{\partial x_i}(x)\varphi(x)\,dx = \int_{\partial U} u(x)\varphi(x)\,dx - \int_U u(x)\frac{\partial\varphi}{\partial x_i}(x)\,dx.$$

Since $\varphi \equiv 0$ on $\partial U$ [on a neighbourhood of it, even], the boundary integral is zero.

Jasper Loy

7:06 PM

I want to do something to improve the community but it is against the law. How?

Daniel Fischer

What, @JasperLoy?

Jasper Loy

@DanielFischer I mean, in my country.

I can either do it and be jailed, or not do it.

Daniel Fischer

@JasperLoy Okay. But if it is against the law, there is a possibility that not all would consider it an improvement. So better think again.

Jasper Loy

I tell myself not to think about doing it, but it keeps bugging me.

@DanielFischer Yes, others consider it immoral, but I think otherwise.

skullpatrol

@DanielFischer

Jasper Loy

7:14 PM

@skullpatrol Do you have an answer to my question?

Huy

@DanielFischer: Thanks a lot.

skullpatrol

7:28 PM

@JasperLoy yes

Jasper Loy

@skullpatrol What is your answer?

skullpatrol

@JasperLoy ask here

Jasper Loy

7:45 PM

@skullpatrol But, do you agree with me that what is legal may not be right and what is illegal may not be wrong?

skullpatrol

@JasperLoy yes

laws are not perfect

even the laws of Physics change with a change of scale

Jasper Loy

@skullpatrol Well said. I heard about your head injury? Does it affect you now?

skullpatrol

I dunno what I would be like if it didn't happen.

Anthony

@DanielFischer I see! Thank you!

Jasper Loy

8:03 PM

@Anthony I see the cat.

blue

using hebrew for elements of fields

Anthony

@JasperLoy hohoho

Balarka Sen

@blue OK OK OK

Apparently, I am talking about direct products of finite copies of Dedekind domains.

blue

8:19 PM

wat

Balarka Sen

Dedekind domains.

blue

what about them

Balarka Sen

OK.

$\mathcal{O}_K$

@blue Our commutative algebra professor is freaking us hard. Dedekind domains are all we are speaking of now.

Anthony

@DanielFischer that just shows that it's not 1-1, right? The other answer shows that it is specifically looking like n to 1, correct?

Balarka Sen

OKs everywhere.

skullpatrol

8:22 PM

how many people in the class

Balarka Sen

~12

@blue I actually asked whether every Dedekind domains were PID. It turned out that if it was then Kummer would have made it to FLT.

Daniel Fischer

@Anthony That $N$ is constant in a neighbourhood of $f(z_0)$ means that $f$ is locally $n$-to-1 counting multiplicity. Then use that the zeros of $f'$ are isolated, so that in a small enough neighbourhood of $z_0$, there is no other zero of $f'$, whence the multiplicities there are all $1$. BUT

blue

@BalarkaSen yep

and there would be no such thing as class groups

Balarka Sen

@blue I know essentially 0 about class groups.

But, well, who cares about class groups.

It's the class number I want =P

We are just introduced to fractional ideals. I don't see the significance of them.

blue

it's nice to work with a group instead of just a monoid

Balarka Sen

8:28 PM

@blue well, then, why not the grothendiek group of the ideal instead?

oh wait. they are the same.

Daniel Fischer

@Anthony Much clearer is IMO to look at the power series expansion of $f$ around $z_0$. $$f(z) = f(z_0) = \sum_{k=n}^\infty a_k(z-z_0)^k = f(z_0) + (z-z_0)^n \underbrace{\sum_{m=0}^\infty a_{m+n}(z-z_0)^m}_{g(z)}.$$ Now, $g(z_0) = a_n \neq 0$, so in a neighbourhood of $z_0$, there is an $n$-th root of $g$, say $h$, i.e. $g(z) = h(z)^n$. Then $\alpha(z) = (z-z_0)\cdot h(z)$ satisfies $\alpha'(z_0) \neq 0$, hence is a biholomorphism of a neighbourhood of $z_0$ to a neighbourhood of $0$. And

Anthony

@DanielFischer Wait, where are we getting n-to-1 from in the first case?

Daniel Fischer

$$f(z) = f(z_0) + \alpha(z)^n.$$

@Anthony You mean for the argument principle, right? The assumption is that $f$ attains the value $f(z_0)$ with multiplicity $n$, so $N(f(z_0)) = n$.

Jasper Loy

@DanielFischer Your sentence is complete with the full stop.

Balarka Sen

@blue I think $\mathcal{O_k}$ is a Noetherian ring.

If it is, then it must be finitely generated and the additive part must also be so. Can we determine the rank of the free part of the group? Does it relate to $k/\Bbb Q$?

blue

8:33 PM

if k is an extension of Q then it's char=0 so O_k is char=0 so its underlying additive group has no torsion

and yes, [k:Q] = [O_k:Z], the first as vector spaces and the second is rank

Anthony

@DanielFischer I'm confused... Hnnggg. Why is it attaining the value $f(z_0)$ with multiplicity $n$? Because it's that degree of a polynomial?

Balarka Sen

@blue interesting. that means that the diagram has a fun property after all.

blue

@BalarkaSen it's more fun with e, f and g

you'll probs become friends with them in a bit

Balarka Sen

e, f and g?

blue

ramification index, residue degree, and ... g doesn't have a name

Balarka Sen

8:36 PM

Such names. Much big. So complicated. Wow.

Daniel Fischer

@Anthony No, it has nothing to do with polynomials. It's the order of the zero of $f'$. The function $f$ attains the value $w$ with multiplicity $n$ in $z_0$ if $f(z_0) = w$ and $f'(z_0) = \dotsc = f^{(n-1)}(z_0) = 0 \neq f^{(n)}(z_0)$.

Balarka Sen

@blue Is the ramification of alg. NT really have an intuition behind it's name? i.e., is there any relation with ramification in algebraic topology?

I don't get it.

blue

the relation is "etale" stuff that I do not get

Balarka Sen

yeah, i have heard a lot of hubbub about that.

etale morphisms. etale coverings. etale topology. etale cohomology. etale spaces.

who is this etale guy?

blue

it's a french word grothendieck used, IIRC

Shisui

8:45 PM

How would I prove this equality? $$ \sum_{k=0}^{l} \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}$$ The only hint I was given was to consider $$ (1+x)^n (1+x)^m = (1+x)^{n+m} $$

blue

symbolically or combinatorially?

Balarka Sen

@Shisui What have you tried?

Shisui

@blue Preferably symbolically. I've hardly done any combinatorics.

blue

@Shisui so, have you considered it?

what do we do with things that look like (blah + blah)^blah ?

Balarka Sen

Should be easy enough by that hint you are given, @Shisui

Think of the binomial formula.

Shisui

8:48 PM

@BalarkaSen @blue $$ \sum_{i}^{n} \binom{n}{i} x^{i} \sum_{j}^{m} \binom{m}{j} x^{j} = \sum_{l}^{n+m} \binom{n+m}{l} $$

Balarka Sen

Good. Hint : Compare coefs.

I'll let you think.

I think now one has to plug in $x = 1$ or something.

blue

@Shisui forgetting some x's on the RHS

Anthony

@DanielFischer Waaaaaat. I didn't know that. If there derivatives are zero, it's multiplicity goes up?

Jasper Loy

@BalarkaSen I thought you were going to delete your account?

Balarka Sen

@JasperLoy I didn't.

Shisui

8:50 PM

@blue Sorry! I forgot the $x^{m+n}$.

blue

@BalarkaSen no, equate coefficients of x^l

@Shisui x^l

Balarka Sen

2 mins ago, by Balarka Sen

Good. Hint : Compare coefs.

Just sprouting ideas.

One of them are bound to work.

Shisui

@blue Oops! Yes, that's what I mean! I was just writing down the last term of the RHS so I mixed the two up ^_^

Balarka Sen

@Shisui The sum on the right. $x^l$. $l$ is the index.

@blue Can we explicitly determine functions which are not in $\Bbb Q[[z]]$, upto power series expansion?

Daniel Fischer

@Anthony Yes, $a + (z-z_0)^m$ attains the value $a$ with multiplicity $m$ in $z_0$. And it is easily seen to be $m$-to-one in a punctured neighbourhood of $z_0$. Locally every non-constant holomorphic function looks like that for some $m$.

Balarka Sen

8:53 PM

I just saw some Q on MSE so that triggered up the question.

blue

@BalarkaSen I am not sure what you mean

Balarka Sen

@blue Take a function. $\exp(\exp(z)+z^2)$. How do we know that there exists a power series for this with rational coefs?

In general, is there any way to determine it?

blue

Q[[z]] is closed under +,-,x,/ and composition (wherever they are defined)

Balarka Sen

erm. yes. so?