Mathematics - 2017-08-02 (page 2 of 4)

Secret

08:04

[Random] The above discussion inspired the following weird idea: Suppose $a,b,c\neq 0$ are zero divisors such that $ac=0$ and $bc=0$. Now there's the following relation:

$a+a+\cdots + a = b$
Then by right distributivity

$ac+ac+\cdots + ac = bc\implies 0+0+\cdots + 0 = 0$
Not sure whether there is something special about such rings

Alessandro Codenotti

Take $\Bbb Z/p\Bbb Z$ with the usual addition and trivial multiplication $xy=0$ for all $x$ and $y$, it should have characteristic $p$ by daminark's definition

Daminark

Rip

Secret

hmm, actually, if $ac=g\neq 0$, the presence of zero divisors can limit the characteristic of a commutative ring since $a + a + \cdots + a = b$, then $ac + ac + \cdots + ac = bc \implies g+ g + \cdots + g = 0$ so the characteristic of such rings will be $< (n-1)$

so, zero divisors kinda behave like as if you have a fragment of zero floating around in your structure which given the correct conditions, becomes zero

Secret

08:34

Attempted sketch proof: A Schauder basis is countable and for a Banach space that has a Schauder basis, all elements can be expressed uniquely in terms of a countable linear combination of the basis, so any open set in the norm topology must contain at least one member of the Schauder basis, meaning that the basis forms a countable dense subset and hence separable. For $\ell^{\infty}$ it is not separable thus it cannot have a Schauder basis.

As for why $\ell^{\infty}$ is not separable, I currently have no idea as I cannot find any countable bounded sequence in it that can fall outside of a…

Daminark

That's not right about the Schauder basis

Think about $\ell^2$

A Schauder basis is given by $(1,0,0,\ldots),(0,1,0,\ldots),\ldots$

But those vectors are all in the unit ball

So the Schauder basis itself won't work, but what can?

Remember that you wan't thinks to become dense but still countable, and you have a Schauder basis and know that any vector is converged to (is that a word? Well it is now)

Secret

all unit vectors in $\ell^2$ should converge towards the spherical surface of the unit ball, because of how the norm is defined?

Daminark

I mean, each of those vectors is norm 1

As for $\ell^{\infty}$ not being separable

Consider the set of sequences of $0$ and $1$

Each of those is in $\ell^{\infty}$ (and not necessarily in $\ell^p$ for finite $p$), and the distance between any two of them is 1

But they've all got norm 1 and there are uncountably many

Or I mean no need for norm 1 here

Anyway I've gotta pack real quick so I'll let you dwell on that for some time, it's sneaky

Secret

08:52

ok

Alessandro Codenotti

going on holiday? @Dami

Daminark

Nope, I've got this geometry/topology RTG for the next few days

There are some nifty talks

Alessandro Codenotti

"Matt Gursky - How to hear the topological and geometric properties of a manifold " that one seems super interesting

The professor of my PDE course went through some stuff related to this during the last 2 classes since we were done with the syllabus and he wanted to show us something interesting

Balarka Sen

presumably it's about spectral theory of Laplacians on the manifold

how to hear the shape of a drum etc etc

Alessandro Codenotti

It is, according to the abstract

@mathiu_lady I don't know if you saw my message earlier but I think that should work as a counterexample

Leaky Nun

09:11

Is there a power series with radius of convergence 0?

Alessandro Codenotti

$\sum\limits_{n=0}^\infty x^nn!$

2

Any series where the coefficents grow fast enough to have $\limsup \sqrt[n]{|a_n|}=\infty$ will do

mathiu_lady

@AlessandroCodenotti yeah I just saw it and also it may be one of counterexample : Let R= Z_4 (the set of all integers modulo 4) and take the subring {0,2} of Z_4 then the subring is of characteristic 2 but does not contain Z_2 or F_2

Alessandro Codenotti

isn't $\varphi(0)=0$, $\varphi(1)=2$ a field isomorphism from $\Bbb F_2$ to your ring?

Secret

[Chemistry] Just submitted draft 1 of the proposal for review. Currently re-tidying all the data folders and need to once again redo the numbering of the files as it once again become so messy with letters and caps everywhere after some time of sequence breaking

Meanwhile, trying to think of a way to code a rotation matrix along some given axis which is determined by the structure of the molecule

Alessandro Codenotti

ah, of course not, that ring has no identity

2

nevermind, I was thinking about it as a group

mathiu_lady

09:20

@AlessandroCodenotti yeah

Leaky Nun

@AlessandroCodenotti heh, thanks

Kirill

09:43

I have questions to the proof. Could someone help?

Leaky Nun

@Kirill just aks; don't aks to aks

Kirill

@LeakyNun ok. I need to present the proof, thats why. There are more than one question there.

Leaky Nun

@Kirill what is it above?

@LucasHenrique bom dia

Lucas Henrique

bom dia! :)

Kirill

@LeakyNun pretty simple for you I think. Given $\Omega \subset \mathbb{R}^n$, that is open, $f: \Omega \in \mathbb{R}^m$ and $f$ differentiable in $x$ prove that $f$ is continuous in $x$ and partially differentiable in $x$.

Leaky Nun

09:48

about*

Kirill

so, step by step: first part proves that $f$ is continuous:

Leaky Nun

you mean $f:\Omega \to \Bbb R^m$?

Kirill

oh, sure, $ \subset$, not $\in$, sorry

nonono

$f: \Omega\to\mathbb{R}^m$

Leaky Nun

continue :)

Kirill

The proof says: let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a linear map, that satisfys the definition of a differentiable function. With $r(h)$ (our remainder from yesterday), so $r(h) = f(x+h)-f(x)-T(h)$ is true that $$\Vert f(x+h) -f(x) \Vert = \Vert r(h) + T(h) \Vert \le \Vert r(h) \Vert + \Vert T(h) \Vert \le \Vert r(h) \Vert+ C\Vert h\Vert $$, where $C$ is some from $h$ independent constant.

Leaky Nun

09:57

what is the language?

Kirill

ok, we can find such a $T$, as $f$ is differentiable. Why has been $r(h)$ set in this way?

@LeakyNun originally German

Said exaggeratedly, I do not understand, why $r(h)$ not equals $\sqrt{2}$, or $r(h) = f^2(x)$ or something else. Why exactly this term, used in the proof?

Leaky Nun

18 hours ago, by Kirill

Does somebody know this guy: $\lim_{h \to 0, h \ne 0}\frac{r(h)}{h}$ and his brother $\lim_{h \to 0 h \ne 0}\frac{\mid r(h) \mid}{\mid h \mid}$? They come from the formula $f(x+h) = f(x) + f'(x)h + r(h)$. What are they?

Kirill

@LeakyNun if so, why $f'(x)h$ equals $T(h)$?

Leaky Nun

because it is defined that way?

Kirill

@LeakyNun it is said, that $T$ is a linear map, that is unique for every $x \in \Omega$. But I do not see, why it should be exactly the first derivative of $f$, and not of some other function $g$, multiplied by the $h$.

Leaky Nun

10:05

I'm also thinking about this

Kirill

But said generally, you think that it is just a transformed formula from yesterday, wih respect to $r(h)$, right, @LeakyNun?

Leaky Nun

I've no idea

hast du der originell Beweis?

Kirill

ok. So, further I would like to ask, what does the term on the left side of the equation in the center mean. He says, "let $r$ be defined this way. Then I take the difference of $f(x+h) - f(x)$ and take the norm of it." What for?

@LeakyNun klar, wenn das für Dich einfacher wird.

Leaky Nun

@Kirill is it a soft-copy or hard-copy?

ich verstehen einfacher wurde

Kirill

@LeakyNun tell me what these two mean

Leaky Nun

10:12

was ist 5.1?

Kirill

that is the part proving that $f$ is continuous

5.1 is:

@LeakyNun $\lim_{h \to 0, h \ne 0}\frac{\Vert f(x+h) - f(x) - T(h)\Vert}{\Vert h \Vert}=0$

I mean all-in-all that is a proof that $f$ is continuous. I understand only one step here: triangle unequality. All of ther others: mystery at the moment.

so, $1.$ Let $r$ be the term I do not understand. What does the difference on the left of the unequality means? Why there is a norm there?

$2.$ Why is $\Vert T(h) \Vert \le C \Vert h \Vert$?

$3.$ What is $C$? Why that should be independet from $h$?

$4.$ What does the last equation mean? How it follows from it that $f$ is continuous?

Leaky Nun

oh!

$x$ is nicht eine Variable, aber eine Konstante

dafur $f'(x)$ ist eine Konstante

Kirill

@LeakyNun that is a certain point in Omega, that we look at

Leaky Nun

@Kirill yes, that is why $f'(x)$ is a constant

Kirill

@LeakyNun maybe English?

Leaky Nun

10:22

so $T(x) = f'(x)h$

Kirill

:)

Leaky Nun

and then $\|T(x)\| = \|f'(x)\| \|h\| = C \|h\|$

Kirill

$\le$!

Leaky Nun

where $C=\|f'(x)\|$

@Kirill you don't speak German, or is my German too bad?

14 mins ago, by Kirill

@LeakyNun klar, wenn das für Dich einfacher wird.

wait, you clearly speak German, so it's the latter

Kirill

@LeakyNun ich kann freilich und gern Deutsch reden, es geht ja darum, dass man mathematische Aussagen möglichst präzise ausdruckt. Es klappt bei dir in dem Englischen besser, als in Deutschem

Leaky Nun

10:26

so you speak German and I speak English? :-P

is $h$ a vector or a scalar?

Kirill

I speak about math clearly better in German, but English is a good exercise, too. I will have my master in English anyway.

Leaky Nun

well, in your English rendition of the proof, you left out the very important "welche 5.1 erfullt"

AlwaysNeedHelp

Does any one here know anything about finite groups and GAP?

Leaky Nun

@AlwaysNeedHelp maybe you could tell me what conjugacy ... and r... triplets are

Kirill

@LeakyNun that is the definition of the differetiable function that says, "if such a $T$ exists and 5.1 is true, then the function is differentiable". I wrote that $T$ is the one from the definition :) $h$ should be a vector, as $x+h$ is undefined then

Leaky Nun

10:30

but $x$ is a scalar

AlwaysNeedHelp

@LeakyNun I'm afraid [email protected] would be more help.

Kirill

@LeakyNun $x$ is a vector too, we are in $\Omega \subset \mathbb{R}^n$

and $x \in \Omega$

Leaky Nun

wait

if $x$ is a vector, then what does partial differentiability mean?

@AlwaysNeedHelp done

Kirill

@LeakyNun every partial derivative of $f(x)$ exists

Leaky Nun

what does it even mean

Kirill

10:32

$f(x) \in \mathbb{R}^m$

Leaky Nun

partial derivative is when you have another variable, right

Astyx

Should it not be "every partial derivative of $f$ exists" ?

Kirill

e.g. $(x_1,x_2,x_3) \mapsto (x_1 sinx_2, cos x_2, x_1 + x_2)$

so, $\partial_1f_1(x) = \sin x_2, \partial_2f_2(x) = -\sin x_2, \ldots $

SteamyRoot

If $f = (f_1, f_2, \dots, f_k)$ and $f_i = f_i(x_1, x_2, \dots, x_k)$, then $f$ is partial differentiable if $\frac{\partial f_i}{\partial x_j}$ exists for every $i$ and $j$

Leaky Nun

oh ok

Astyx

10:36

Really ?

Leaky Nun

13 mins ago, by Leaky Nun

and then $\|T(x)\| = \|f'(x)\| \|h\| = C \|h\|$

Is this correct?

Astyx

Does partial differentiable on a basis imply partial differentiable ?

Kirill

or, $f$ is partially differentiable in $\Omega$, if $\partial_if(x)$ exists for all $x \in \Omega$ and all $i, 1 \le i \le n$.

SteamyRoot

@Astyx I have no idea what you mean with "on a basis" o.O

Kirill

@LeakyNun according to the proof that is not correct, as it should be $\le$ there. According to my question - ok, the triangle unequality is perfect, but why there is such a $C$, and $C$ doesn't depend on $h$?

Astyx

10:40

@SteamyRoot Like, if you have a basis $(e_1,\dots, e_n)$ of $\Bbb R^n$ and $f : \Bbb R^n \to \Bbb R$, does "$\forall i {\partial f\over \partial e_i}$ exists" imply "$\forall x\in \Bbb R^n {\partial f\over \partial x}$ exists" ?

Leaky Nun

@Kirill because $C = \|f'(x)\|$?

SteamyRoot

How do you define $\frac{\partial f}{\partial e_i}$ ?

Astyx

$${\partial f\over \partial e_i}(x) = \lim_{t\to 0} {f(x+te_i) - f(x)\over t}$$

Kirill

he says $\Vert T(x) \Vert \le C \Vert h \Vert$. So, $\Vert T(x) \Vert \le C$. @LeakyNun

SteamyRoot

Oh, a directional derivative

Leaky Nun

10:43

@Kirill I'm out of this :)

@Astyx do you have any idea?

Astyx

About what ?

Kirill

@LeakyNun but have you understood the proof on your own?

SteamyRoot

Ew, that's one abusive notation :P

Leaky Nun

@Kirill I haven't

Astyx

@SteamyRoot Was that for me ?

SteamyRoot

10:44

Yup

Leaky Nun

48 mins ago, by Kirill

The proof says: let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a linear map, that satisfys the definition of a differentiable function. With $r(h)$ (our remainder from yesterday), so $r(h) = f(x+h)-f(x)-T(h)$ is true that $$\Vert f(x+h) -f(x) \Vert = \Vert r(h) + T(h) \Vert \le \Vert r(h) \Vert + \Vert T(h) \Vert \le \Vert r(h) \Vert+ C\Vert h\Vert $$, where $C$ is some from $h$ independent constant.

@Astyx help him understand the proof

Astyx

How do you define $\partial f\over \partial e_i$ then ?

SteamyRoot

I define it the same way, I denote it differently

Kirill

@LeakyNun we normally say that the function is continuous, so that small differences of function arguments imply small differences of function values. But I do not understand, why that proof shows that $f$ is continuous.

Leaky Nun

@Kirill well, continuous would be $\lim\limits_{h\to0}f(x+h)=f(x)$

Astyx

10:45

Oh right. Yeah I've been taught to abuse notations a lot when dealing with multivariable functions

Kirill

@LeakyNun true. But he says it follows from the big equation. I do not see the implication.

SteamyRoot

Anyway, if the $e_i$ and $x_i$ match, then obviously it's true. But in general, I doubt it... I'll have to think a bit :P

Leaky Nun

@Kirill I have no idea

please ask @Astyx

SteamyRoot

@Astyx I think functions like $f(x,y) = \frac{xy}{x^2 + y^2}$ may pose problems

Not sure, though, analysis isn't exactly my forte :P

Astyx

@Kirill Any linear map $T$ on a finite dimensionnal normed vector space $(E, \Vert\cdot\Vert_E) \to (F, \Vert\cdot\Vert_F)$ is Lipschitz (ie $\exists C \in \Bbb R_+ \forall x\in E, \Vert T(x)\Vert_F \le C\Vert x\Vert_E$)

Kirill

10:50

@Astyx hi!

Astyx

@SteamyRoot Yeah, that's the kind of examples I was afraid of :p But I guess we simply don't have the same terminology

Hi @Kirill

user84215

The subscription price of ANNALS OF MATHEMATICS is $345 for an individual (for just one year).

Astyx

I need to go soon though

as in, now

user84215

I am not rich enough to read mathematical journals.

Mike Miller

sci-hub.io

user84215

11:00

It is illegal.

Kirill

@aminliverpool caomparing to the Wolfram and MatLab annual licenses, 345 is like a cheesburger by McDonalds.

@aminliverpool there are a lot of libraries with such papers, as much as free works online

user84215

Selling science is not a good job.

Kirill

sure, but respecting the job of others is a good job

user84215

What should I do if I can not find the libraries having those journals?

AlwaysNeedHelp

not learn mathematics.

Clearly mathematics should only studied by the exceedingly wealthy.

That is what the bureaucrats want.

Kirill

11:11

@aminliverpool you can search better and use evtl Google for this

Astyx

Lol, you have all you need to study mathematics on the Internet for free if you have the will to

user84215

that is, mathematics is for rich people?

Astyx

And legally ofc

Kirill

@aminliverpool my bicycle keys cost 40 Dollars each, are bicycles for rich people?

user84215

But some articles are not published for free.

Kirill

11:13

no, they are just too specific to be cheep

Mike Miller

Virtually every paper that appears in the Annals first appeared as a preprint on the arXiv.

Kirill

@Astyx are you gone, or, here?

Astyx

Both

On my phone

Kirill

wow, like an electron

two conditions at once

@Astyx I am not sure about the Lipschitz argument. It seems plausible, but we want to show that $f$ is continuous, and using Lipschitz-continuity for proving continuity seems for me to be wrong. I can mistake though.

Astyx

But here I'm only talking about linear maps - which I believe you know to be continuous in finite dimension ?

Kirill

11:23

@Astyx yeap. I do not know why, but I know the fact. The discussion with Leaky-Nun was about this certain proof. There were several certain points that were not clear.

Astyx

Linear maps are continuous iff they are lipschitz

It's simple enough to prove with some topology

What points aren't clear ?

Kirill

Is there any way to repost the message? Do you speak German?, @Astyx

Astyx

I do not know and not well enough

Kirill

Astyx

Yeah I saw that message

Kirill

11:31

1. Why he defines $r(h)$ that way?

Astyx

That's the non linear component of the function

Kirill

Why exactly that way, not as $f^2(x)$ or $\sqrt{1+x}$?

what do you mean?

Astyx

What do you know about the differential of a function ? How is it defined ?

Kirill

nothing

@Astyx we are not told about differentials at this course

Astyx

Like , how is T defined ?

Kirill

11:34

@Astyx T is defined as a linear map, going from the same space into the same space as $f$. It is said that $T$ is unique for each $x$ of the domain.

I have struggled to get something about differentials, but they say it is too early for them.

Astyx

Yeah but what makes it unique ?

Kirill

@Astyx I don't know.

Astyx

Huh that's a problem

Kirill

for every $x$ there is correspoding linear map

Astyx

Unfortunately I don't quite have the time to explain it

Kirill

11:37

the proof I do not understand and that I try to understand shows (lectro says) that this $T$ is unique

So, it shows him, not me

And I try to understand the proof to see why it shows it

Astyx

It's because it's he unique one such that $\Vert f (x+h)-f (x)-T (h)\Vert=o (\Vert h \Vert)$ (hope that renders correctly)

Kirill

so, we can stop at this point, I can ask you later, or I can ask other things, @Astyx. What would be the best for you?

Astyx

You can still ask, I won't guarantee I'll be there to answer

Someone else might

Kirill

@Astyx I do not know Landau good. But I hope you mean that the numerator goes to 0 faster as the denominator

Astyx

If I understand what you mean correctly, yes

Kirill

11:44

it still doesn't tell me why $T$ is unique. So, what was the main idea to define $r(h)$ as such, and not as $\sqrt{2}$? @Astyx

Astyx

Because then you can bound it upward by the norm of h in your triangular inequality (by definition)

Why T is unique is another proof

It doesn't really matter here that T is unique

Kirill

We have $A \le$ triangle unequality $\le B$ in the central unequation. What is $A$ and $B$? He constructs $r$ that I do not understand. Then he takes $A$ that I do not understand, too. What is $A$?

$A$ doesn't look like a numerator of the differential quotient, it is just a difference. Why he takes a difference, not a sum, a product, etc?

Astyx

I can't answer on my phone sorry

Perhaps tomorrow same time ?

Kirill

are you French?

Astyx

Or in 7 hours if you prefer

I am

Kirill

11:51

at some evening would be better, in 7 hours is perfect. I will have a course tommorow at the daytime.

Astyx

Right, I'll try and be there

Kirill

Asking because if you were in Canada, the evening is defined differently there

Astyx

True

Kirill

ok, so ~ 21.00 if you can.

Astyx

Do you not have holidays at this time in Germany ?

Agreed

Kirill

11:54

@Astyx I have, but the exam is in 6 days, and tomorrow is a whole-day-repeating course for the whole module.

Akiva Weinberger

@TedShifrin Do you think there's a geometric interpretation of the determinant of a complex matrix ($\Bbb C^{m\times n}$)?

Secret

uh, you mean nxn, else rectangular matrices have no determinant

Akiva Weinberger

Right yes I do mean that

Leaky Nun

12:19

@AkivaWeinberger The "hyper-volume" formed by the "hyper-parallelepiped"?

Daminark

Hey everyone!

@AkivaWeinberger product of eigenvalues

Leaky Nun

At least that's from 3B1B's essence of linear algebra

BAYMAX

Hi

Actually I saw a research paper published recently can I ask the author to share that paper or is that not decent or do i have to purchase it , i am confused.

parvin

12:34

@BAYMAX what is the name of it?

SteamyRoot

@BAYMAX See if either the author has put the article on his/her personal website, or look if there's a preprint available on arxiv

parvin

hello guys. there is a math problem i'm solving, can some one tell me if the word "minimal" is a strict comparison or is just normal comparison? (i mean < or <= ?)

Akiva Weinberger

Normal I think

parvin

@BAYMAX I may be able to find the pdf online if u want

@AkivaWeinberger do you have a bit of time can i send the question?

with solution

BAYMAX

thanks @parvin @SteamyRoot !

I got it in arxiv

:)

How we can know the latest papers published like say "what are the papers published today or this month in a specific area like say differential equations" ?

any idea

@SteamyRoot

Akiva Weinberger

12:39

I might need to run in the middle

parvin

@BAYMAX also check this one if needed anytime : gen.lib.rus.ec

BAYMAX

ok

parvin

Akiva Weinberger

Looks like a stars and bars sort of problem

user84215

What is gen.lib.rus.ec ? why is copyright violated here repeatedly?

parvin

12:46

is the second one uploaded?

user84215

None of you believes in copyright?

Akiva Weinberger

So I guess thirteen stars and twelve places between the stars in which we can place bars?

parvin

@aminliverpool because we are living in iran or some other places in which people have no access to the shop or bank whatever and in addition to that it's not my website ...

@aminliverpool do you believe in free world wide education?

@AkivaWeinberger yes I can see that but why? why did he choose 1 n 1 n 2 n3 instead of 2 n 2 n 3 n 4?

is that bcuz it's "minimal" not minimum?

user84215

@parvin I can buy your required books and journals and post them to you; if that is your only reason.

Leaky Nun

@parvin so that the y are positive

but i would have used 2,2,3,4 and natural numbers :)

parvin

12:51

@aminliverpool that is...and I guess there are people wanting paper books so amazon still have the books. they buy it. some may want the ebook...I dunno..

@LeakyNun i see...

user84215

They also can buy ebooks

parvin

should I trust the book's solutions?

@aminliverpool you are right the poor author should get what ever the income is, and it's sort of a theft ...

Leaky Nun

@parvin your book is correct

user84215

Please don't advertise websites of hackers.

user84215

or maybe better to say of thieves.

parvin

12:57

ok..I thought he may be in need

user84215

Stealing from rich people also, I think, is not good.

SteamyRoot

He might, just as a huge amount of academics pretty much need it. But, it's still illegal, and should probably not be advertised here.

user84215

^

SteamyRoot

Though, on the other hand, when @aminliverpool says "None of you believes in copyright?", my answer is I don't believe in the copyright enforced by academic publishers, no.

Daminark

One book sale is usually enough to get a profit and lawyer fees are expensive

Fargle

13:03

I just learned something that I had completely forgotten, and that is very obvious.

Daminark

Enlighten us

Fargle

Namely, that the graph of a function $f : U \rightarrow \Bbb R^m$ (where $U \subset \Bbb R^n$) is always homeomorphic to $U$.

SteamyRoot

For $f$ continuous, I take it?

Fargle

Er, yes.

I'm tired and careless.

Very very not true for $f$ discontinuous--e.g. the indicator function on $\Bbb Q \subset \Bbb R$

SteamyRoot

You can do it for arbitrary topological spaces, by the way.

user84215

13:12

I think we should not force publishers and authors to do their jobs in such a way that we think it is right.

SteamyRoot

What do you even mean by that?

It's basically the authors being unhappy about how the publishers treat them.

user84215

They have right to sell their products in such ways that they want

user84215

Authors can publish their works for free

SteamyRoot

Yeah, in a perfect world that all makes sense.

Daminark

Has there been talk of forcing? It's mostly grumbling as I see it

SteamyRoot

13:16

Also, first of all, academics are paid by the university to, amongst other things, publish.

Daminark

Though I do think that authors are given less control than they ought so... It's sketch

SteamyRoot

If I only throw my papers on ArXiV and don't publish them in a journal, my university will never award me a PhD.

user84215

We can produce free products to reduce the price of books.

SteamyRoot

You want to publish in journals that are aimed towards people in your area. You preferrably want it to be a "good" journal too, whatever your standards of good may be (impact factor, etc)

Good journals are often owned by big publishing companies, who charge way too much money for anyone wanting to read your paper; or they allow you to publish your paper as "open access" for a measly 2000 - 3000 dollars.

Also, as soon you publish your article, you hand over the entire copyright of your article to this publisher

user84215

You can publish your preprint of it in ArXIV.

SteamyRoot

13:20

Sometimes. It depends on the publisher.

But, unlike a journal, ArXiV is not peer-reviewed.

A "good" journal also entails that papers appearing are "good".

user84215

We should create a good free journal by ourselves.

SteamyRoot

So academics prefer to look in journals aimed towards their research interests, because they know the papers will be good.

And hence, if you want many people to read your good article, you need to get it published in a good journal

and so the circle continues and continues...

@aminliverpool Free is impossible. Cheap/fair is possible, but it takes a lot of time and effort.

And that's what's happening right now, very slowly, step by step.

user84215

Also writing free books

SteamyRoot

Also free food for the entire world overnight, while you're at it?

user84215

No. Science must be free for all.

SteamyRoot

13:24

Yeah, well, a lot of people agree.

But things aren't going to change overnight.

Abcd

Short question: Is $\dfrac{\sin2n\pi}{(2n+1)} = \dfrac {\sin \pi}{(2n+1)} $ ? where n is integer

SteamyRoot

Is $n$ an integer?

Abcd

@SteamyRoot Yes

SteamyRoot

In that case, you can just cancel the denominators.

and $\sin(k\pi) = 0$ for all $k \in \mathbb{Z}$

Abcd

@SteamyRoot So are they equal? Only then would I be allowed do cancel the denominators

My question is whether the two expressions are equal or not

SteamyRoot

13:30

Well, you already know the denominators are equal.

So all you should care about is whether the numerators are equal.

Regardless, both expressions are just $0$, so yes, they're equal.

Abcd

@SteamyRoot True. Thanks :)

user84215

Let's write free books all together like Wikibooks.

Abcd

@SteamyRoot Well I made a big typing mistake.

Is $\sin\dfrac{2n\pi}{(2n+1)} = \sin \dfrac {\pi}{(2n+1)} $ ?

Leaky Nun

$\pi - \dfrac{\pi}{2n+1} = \dfrac{2n\pi}{2n+1}$

so yes

Abcd

@LeakyNun Oh. Thanks.

Akiva Weinberger

14:31

@aminliverpool I wonder why Wikipedia is so much higher quality than Wikibooks

3

Probably something about how articles are structured in comparison to books

So here's a question. Is it possible to find a family of (overlapping) convex sets $A_1,A_2,\dots,A_n$ such that the union is $[0,1]^2$, each one has a diameter less than $1$, and $\bigcup_{i=m}^nA_i$ is always convex for $m\le n$?

Leaky Nun

@AkivaWeinberger probably because the former is more well-known

so more editors would contribute

DukeZhou

15:14

Anyone feel like taking on a basic linear algebra question on AI: ai.stackexchange.com/q/3734/1671

Abcd

1

Q: Determining the value of trigonometric expression

If x and y are real numbers such that $\dfrac{\sin x}{\sin y}= \dfrac{1}{3}$ and $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$, then how do I find the value of $\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}$ My attempts: Using the given condition and double angle formula, $\dfrac{\sin2x}{\sin2y} = \...

@LeakyNun Please see my question^

Leaky Nun

15:31

@Abcd answered :)

@Abcd not me

Abcd

@LeakyNun yes.

AlwaysNeedHelp

15:53

0

Q: Generating a finite simple group via rational conjugacy classes.

I'm currently new to GAP and computer algebra systems and was wondering whether anyone knew of the easiest method to achieve the following. Suppose I have a finite simple group - let us take $M_{11}$ (a Mathieu Group) and let $C_1, C_2, C_3$ be a triple of rational conjugacy classes (for the def...

abstract-algebra group-theory representation-theory computer-algebra-systems gap

Jasper Loy

@AkivaWeinberger I think Wikibooks is really a bad idea. It is not good for so many people to work collectively on one book. Books need a lot of uniformity to pass.

@aminliverpool I don't think science should be free for all, but I do think it should be affordable, and many books and journals are indeed absurdly expensive.

Secret

16:11

Alice: balls have zero to me to me to me to me to me to me to me to me to

Bob: you i everything else

Alice: balls have a ball to me to me to me to me to me to me to me to me

Bob: i . . . . . .. . . . . .

Alice: balls have zero to me to me to me to me to me to me to me to me to

Bob: you i i i i i everything else . . . . . . . . . . . .

Alice: balls have 0 to me to me to me to me to me to me to me to me to

frequency analysis is going to fail badly for this one...

Semiclassical

16:30

...

Tamay Besiroglu

Ooh yea that was the AI that 'invented' a new language right?

It seems more like it just failed at english, rather than having invented a new language

Could someone answer my question?

1

Q: Probability of having the biggest draw

I'm reading a probability theory text book and came across a puzzle Suppose N people draw gold nuggets from an urn. Person $i$ takes $k_i$ gold nuggets. The weight of the gold nuggets are given by the cumulative distribution function $F^{k_i}(x)$, where $F$ has the support $[0,\bar{x}]$. The fun...

probability probability-theory probability-distributions order-statistics

Secret

in English Language & Usage, 13 mins ago, by Jakub

@Secret Maybe it's organically born exploits. Alice adds "to me" a few times after saying "balls have zero (value?)' because Bobo the robo perceives it as strong assurance.

ongoing discussion with english guys in the english room

Secret

16:48

I am starting to wonder whether that cumulative distribbutive function is monotonically increasing, and that $j\neq i$ might be the first N draws except the ith draw of the gold which does not attain largest value

but to simplify into 2) either means it might be related to the details of F, or it might be some sort of property ofprobability is being used when you integrate the N-1 trials

But all of this is just wild guess, I don't know enough measure theory to work out this probability

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$Mathematics$ Mathematics