Mathematics - 2018-04-24 (page 1 of 2)

Montel's theorem says that if you have a family $\mathcal{F}$ of holomorphic functions on some domain $\Omega$ which are uniformly bounded on compact sets, then the family is normal.

The proof of that basically amounts to Cauchy integral formula

Eric Silva

cauchy estimates 2 stronk

0celo7

1:02 AM

Cauchy estimates are harmonic estimates in disguise

Daminark

You use it to get equicontinuity, then you use Arzela-Ascoli to get that on a given compact set, you have uniformly convergent subsequences. To get a single subsequence that works for all compact sets you just do a diagonalization argument by choosing countably many increasing compact sets such that any other compact set is contained in one of them

Ted Shifrin

Yup, applications of Cauchy like that are things I used to love to put on qualifying exams.

Or getting control on derivatives in a family ...

Daminark

Now once you know that, and a lemma which says that the uniform limit of analytic injections is either constant or an injection, you can prove RMT. The idea of that proof is this. Let $\Omega \subsetneq \mathbb{C}$ be simply connected, then choose some $z_0 \in \mathbb{C}\setminus \Omega$. Then you can define $\log(z-z_0) = g(z)$

Eric Silva

@0celo7 they're not even disguised

Ted Shifrin

Well, they're disguised to some of us who don't live all the time in PDE land.

0celo7

1:04 AM

@EricSilva Most people probably see complex analysis before Evans Ch. 2

Ted Shifrin

That would include me, since I've never read Evans.

hi @MikeM

Daminark

Fix $w\in \Omega$. You know that you can't find a sequence $z_n$ such that $g(z_n) \to g(w) + 2\pi i$, since you can exponentiate the relation to get $z_n\to w$, so $g(z_n) \to g(w)$, which is shit. So you can find a $\delta > 0$ such that $g(\Omega) \cap B(g(w) + 2\pi i, \delta) = \emptyset$.

Eric Silva

i mean if youve seen both then the first thing people usually like to do is say "oh if youve seen complex this is that"

Ted Shifrin

I mean ... I know some PDE theory, but I've never looked at Evans.

Mike Miller

I agree with Eric; just because one language comes first doesn't mean the latter isn't illuminating

I do not like Evans tbh

Daminark

1:07 AM

This lets you define $F(z) = \frac{1}{g(z) - g(w) - 2\pi i}$. It's injective, holomorphic, and bounded, so if you just translate and rescale, you get a conformal equivalence between $\Omega$ and a subset of $B(0,1)$ containing the origin. So we assume wlog that we're in that case

Ted Shifrin

I liked a few books when I was teaching some of that stuff, but they're long gone ... I never owned Evans.

Eric Silva

@MikeMiller the thing about evans is that if you have developed some of the intuition already it's very useful to look shit up

Ted Shifrin

Oh, useful like Federer? :D

Eric Silva

Evans is more transparent than that at least

0celo7

I've always felt Evans was the place to learn, GT/Morrey/Hormander was the place to look up

Ted Shifrin

1:10 AM

Damning with faint praise, I'd say.

0celo7

The stuff on Sobolev spaces in Chapter 5 is done wonderfully

Daminark

Now you let $\mathcal{F}$ be the set of holomorphic injections $\Omega \to B(0,1)$ fixing $0$. This family is obviously uniformly bounded, so it's normal by Montel. Let $s = \sup_{f\in \mathcal{F}} |f'(0)|$. Then $s \ge 1$ since we have the inclusion, and $s <\infty$ by Cauchy integral formula. Choose a maximizing sequence and pass to a convergent subsequence. The limit $f$ is non-constant since $|f_n'(0)| \to |f'(0)|$ and the $|f_n'(0)|$ are bounded below

Ted Shifrin

I actually liked Trèves's introductory book. The advanced ones were ridiculous.

Eric Silva

i think Evans is better to learn than the others but it's still a useful reference

Mike Miller

Sure

0celo7

1:12 AM

Actually talking about traces is good, GT commit the crime of not mentioning them

Eric Silva

Taylor's books are good from what ive seen of them i think

Daminark

So $f\in \mathcal{F}$ achieves the max (it maps into $B(0,1)$ by continuity + max modulus). Finally, if it isn't surjective, you can create another function with larger derivative at 0 by playing with Mobius transforms. Makes sense @Akiva?

Ted Shifrin

I don't know Taylor's books, either. I was never as book-ish a person as lots of people here.

Demonark, is DogAteMy actually paying attention?

Akiva Weinberger

@Daminark Kinda? That's a lot of theory I don't know. But it's a good road map. Thanks!

Daminark

Apparently he is

:P

Ted Shifrin

1:14 AM

Well, thank goodness DogAteMy doesn't know all of 1st year graduate material before he starts college :P

Eric Silva

now prove the big RMT for him

0celo7

There's a big RMT?

Akiva Weinberger

\huge{RMT}

Ted Shifrin

LOL

I think it's little/big Picard?

Eric Silva

the beltrami equation solving stuff

Ted Shifrin

1:15 AM

hmm

Eric Silva

quasiconformal maps

etc etc

Ted Shifrin

oh, yeah, generalized RMT

Daminark

Lol I don't even know that. My undergrad complex prof never even did Montel or RMT, I just learned that because Marianna's like "Yeah you should know this, a lemma will be on the midterm" and I was like oh okay

Ted Shifrin

You're doing OK, Demonark.

Eric Silva

charlie didnt do it in my class whenever i took that either

Daminark

1:16 AM

It's a good theorem though, 10/10

Eric Silva

he did PMT tho

PNT*

i want pearl milk tea, freudian slip

Daminark

Lol Marianna wants to do that actually, and also talk about why Riemann hypothesis can tell you all that it does about number theory

Eric Silva

charlie presented the most opaque proof possible of pnt because he had no time to prove it

Daminark

From office hours: "If I can make sure that nobody leaving this class submits a paper proving Riemann hypothesis by manipulating $\sum_{n} n^{-s}$ for $Re(s) = \frac{1}{2}$, I'm happy"

Ted Shifrin

That sentence made no sense.

Eric Silva

1:18 AM

so he had to prove the version with all the intuition squeezed out of it

maximum cleverness

Ted Shifrin

Oh.

Eric Silva

minimum understandability

Ted Shifrin

Probably better to state a few lemmas without proof and do it intelligently.

Eric Silva

i dont disagree

Ted Shifrin

I knew you wouldn't :)

Eric Silva

1:20 AM

@Daminark is it that you cant prove it this way

0celo7

what are some big theorems that were open forever and had surprising solutions?

surprisingly simple

Daminark

Well, that'd be manipulating a divergent sum. She said that when she was in England and edited for a journal, she got papers almost every day trying to prove RH like that. Some people even proved and disproved it in the same paper

0celo7

@Daminark that's what I do when I'm running out of time on a test. Write everything I can think of, positive and negative, and submit

Daminark

A+ strat

0celo7

@Daminark I remember doing double degenerate perturbation theory in QMII. My final answer was 3 variations for the grader to choose from

maybe the same thing works for research papers...

Daminark

1:26 AM

What'd the grader do?

Eric Silva

@Daminark oh yes i see, Re(s) = 1/2 indeed

0celo7

@Daminark the first answer was right, full credit

don't think he got to the other crap

I just did the problem multiple ways and got different answers each time

Daminark

Lol, good thing the first answer was the correct one

Semiclassical

1:43 AM

@0celo7 ugh, degenerate perturbation theory

Prototank

hiya

I've got a question. I'm trying to show that a Hypergraph $H$ is $2-$colorable. I know that the edges of $H$ $e_i$ satisfy $\sum_{i=1}^m(1/2)^{-|e_i|}<1/2$.

0celo7

@Semiclassical we should do some degenerate perturbation theory for fun

to see if we remember that stuff

Semiclassical

lol

Prototank

I think I have an argument, it is short and goes like this. Color $V(H)$ uniformly at random. Then
$$\mathbb{P}(\text{some edge is monochromatic})\leq \sum_{i=1}^m \mathbb{P}(e_i \text{is monochromatic})=\sum_{i=1}^m(1/2)^{-|e_i|}<1/2$$

Semiclassical

try doing degenerate perturbation theory for the periodic eigenvalues of $-y''(x)+(\alpha \cos x)y(x)=\lambda y(x)$

Prototank

1:52 AM

since the probability is less than 1, there are examples where no edges are monochromatic

Semiclassical

it's...not fun

0celo7

lol unless I can write it as a 2x2 or 3x3 matrix problem, screw that

I had a 6x6 on HW once, was like wtf

Semiclassical

yeah, this one is worse

it's first order degenerate perturbation theory for the levels $\lambda=\pm 1$ (in powers of $\alpha$)

actually I guess it's $\pm (2\pi)^2$ but eh details

but the next degenerate pair requires second-order degenerate pert theory

and so on and so forth to absolute misery

0celo7

is there a recursive formula

it seems symmetric enough

Semiclassical

lemme find the results, they're on the dlmf

the version they use for Mathieu's equation is $w''+(a-2q\cos 2z)w=0$, for reference (28.2.1)

0celo7

2:03 AM

oh right this is a Mathieu equation

Semiclassical

yeah

because of the factor of two, you need to consider both periodic and antiperiodic boundary conditions to get the spectrum I referenced

first are labelled as $a_n(q)$, second as $b_n(q)$

they give expansions for small $q$ here: dlmf.nist.gov/28.6

so for instance if n=6 then the two series only disagree at terms of order q^6

with the generic statement being dlmf.nist.gov/28.6#E15

(in physics, you can interpret that difference as the level splitting of the original cosine problem, and you can obtain that using WKB if you have sufficient patience...which I had to, a few years ago)

Akiva Weinberger

> My Dad’s a physician. He told me he thinks psychologists should be banned from having children because every one of them he knew used their kids to test a pet theory on child rearing, and usually ended up fucking them up.

- A Reddit comment I just found

Semiclassical

they're not the only ones, I think.

0celo7

@Semiclassical you want to ban people from having children?

Akiva Weinberger

Not the only ones who should be banned from having children, or not the only ones who would test a pet theory on child rearing?

Semiclassical

2:11 AM

the latter

Mike Miller

The former is also true

nitsua60

@AkivaWeinberger That sounds to me like every parent I know, including me and my spouse. It's either try your pet theory or do it with no theory, best I can tell.

Mike Miller

Seems like a false dichotomy

Akiva Weinberger

It was on this thread for context

nitsua60

@MikeMiller Could be--what's the excluded middle you're seeing but I'm missing?

Daminark

2:19 AM

Using someone else's pet theory that was confirmed to work

Leaky Nun

Just use a pet instead of a pet theory

0celo7

@LeakyNun have the dog raise the kid?

Leaky Nun

right

Akiva Weinberger

Unfortunately I could not train the dog to be a chess prodigy

nitsua60

@Daminark I still think of those at pet theories. My spouse and I are adherents of Bruno Bettelheim's parenting theories/strategies, for instance. But "I'll use this, because I think based on reading a few things and having been parented once and having never parented before that it'll go well" is still, in my book, a pet theory. I.e. my pet theory is "Bettelheim seems like a good idea."

@LeakyNun That... I did not consider. Third way: identified.

=)

Mike Miller

2:23 AM

"If you're inclined to test pet theories on children, don't have children"

Daminark

But that can be made into its own pet theory! "My pet theory about raising kids is nahhhhh"

@AkivaWeinberger you had one job

Akiva Weinberger

He's only a 1400 elo or so

Daminark

2:36 AM

Smh

tatan

Is there a way one can easily check the differentiability of a function? For eg. f(x)=$1+(x-3)^{2/3}$ is not differentiable at x=3. How do I find it out rather simply (graphically maybe but how?) without having to analytically by finding the limits and seeing and it takes a lot of time.

Semiclassical

in general, $x^p$ is differentiable at $x=0$ when $p\geq 1$

(getting it to x=3 is just a change of variables)

simple way to understand that is to note that $\frac{x^p-0}{x}=x^{p-1}$ which goes to zero as $x\to 0$ if $p> 1$, stays at 1 if $p=1$, and diverges if $0<p<1$

tatan

Can you provide an example plz?

Semiclassical

2:52 AM

You already have one.

And I gave you the ingredients to figure out any other that you want, at least as far as exponents go

tatan

oh I forgot. Thanks for making me realise that ;-)

Semiclassical

the thing to note is that, when it comes to the basic examples, there's some obvious facts at play

but if you've got something complicated, it's not always going to be easy to see where the function is differentiable regardless

user537566

@mercio You're right. It needs solving $x^3-3x-18=0$. If rational root theorem isn't allowed etc, then it's a mess. I can't imagine that though. I learned a method of solving this by working sort of backwards from a journal once, but it's basically cheating (it's basically requires knowing the form the answer is gonna be, and it even looks prophetic when read).

Semiclassical

thank goodness for rational root theorem, really

user537566

@Semiclassical Very nice! I see what you did there! xD

Never mind, I'm sleep deprived, and finding puns/humour everywhere.

tatan

2:56 AM

@Semiclassical I did not understand the implication of this statement (the one to which I have replied here)

Semiclassical

i'm just writing out the difference quotient for $x^p$ at $x=0$

tatan

What does it really imply?

Semiclassical

it implies that the limit required for the derivative at $x=0$ to exist only does so when $p\geq 1$

tatan

Oh! Thanks ... I got it

tatan

3:09 AM

Is the converse of Rolle's Theorem always true?

nitsua60

Vieta's formulas

In mathematics, Vieta's formulas are formulas that relate the coefficients of a polynomial to sums and products of its roots. Named after François Viète (more commonly referred to by the Latinised form of his name, Franciscus Vieta), the formulas are used specifically in algebra. == Basic formulas == Any general polynomial of degree n P ( x ) = a n x n + a n ...

should help

uhoh

3:31 AM

1

Q: Do all rigid bodies of radius $r$ have at least one stable orbit with perihelion $p$ such that $2r >p > r$?

In reading about the TESS mission, I was ultimately led to this NASA commentary from 2006, discussing how lumpy Earth's moon is. One of the points made is that there exist only 4 orbital inclinations at which a (close) lunar orbit is stable, and that they all simply avoid mass concentrations. F...

I've voted to close...

But perhaps someone can help explain?

user537566

@nitsua60 I can't see it. Does it solve the polynomial?

nitsua60

Scipione del Ferro

Scipione del Ferro (6 February 1465 – 5 November 1526) was an Italian mathematician who first discovered a method to solve the depressed cubic equation. == Life == Scipione del Ferro was born in Bologna, in northern Italy, to Floriano and Filippa Ferro. His father, Floriano, worked in the paper industry, which owed its existence to the invention of the press in the 1450s and which probably allowed Scipione to access various works during early stages of his life. He married and had a daughter, who was named Filippa after his mother. He likely studied at the University of Bologna, where he ...

or this

@user537566 yeah, historically there is a lot known about solving cubics that we don't much teach these days

Nicholas Roberts

Condsider these two sets: Homomorphisms from the fundamental group of a manifold to the real numbers and the set of homomorphisms from the abelianization of the funamental group of a manifold to the real numbers. Are these sets isomorphic?

The manifold is connected and smooth.

This is bothering me. Im trying to prove the first de rham cohomology group is isomorphic to sets of homs btwn fundamental group of the manifold to real numbers and im stuck on this step....

nitsua60

Cubic function

In algebra, a cubic function is a function of the form f ( x ) = a x 3 + b x 2 + c x + d {\displaystyle f(x)=ax^{3}+bx^{2}+cx+d} in which a is nonzero. Setting f(x) = 0 produces a cubic equation of the form a x 3 + b ...

@user537566 that has the Cardano-Tartaglia methd, too

user537566

@nitsua60 That's very nice. Reading through the wikipedia page on cubic equation is a blast! It seems majority of methods for solving cubic equations have Italian names (Cardano, Tartaglia, del Ferro, Lagrange, etc).

Nicholas Roberts

3:39 AM

Let $Ab(\pi_1(M)$ denote the abelianization of the fundamental group of $M$. We are given that $Ab(\pi_1(M) \cong H_1(M)$, the first homology group

By the de Rham theorem we know that $H^1_{DR}(M) \cong H^1(M ; \mathbb{R})$, the first singular cohomology group.

nitsua60

@user537566 that's 'cause they couldn't/didn't read Arabic =)

(Khayyam,of course, was there before any of the Italians. As usual.)

gotta run. enjoy!

Nicholas Roberts

And then by the universal coefficent theorem $ H^1(M ; \mathbb{R}) \cong Hom(H_1(M), \mathbb{R})$.

And then we are given that $H_1(M) \cong Ab(\pi_1(M)$

Thus, $H^1_{DR}(M) \cong Hom(Ab(\pi_1(M)) , \mathbb{R})$

0celo7

@NicholasRoberts is this going somewhere

Nicholas Roberts

Now i need to show that this is in fact isomorpic to the set of homomorphisms between $\pi_1(M)$ to $\mathbb{R}$$

Any ideas?

Ya, i need that abelianization to go away

user537566

@nitsua60 That's hilarious! Language barrier forced them to discover/rediscover some good maths!

0celo7

3:48 AM

@NicholasRoberts $Hom(G_1,G_2)\cong Hom(Ab(G_1),G_2)$ if $G_2$ is abelian

user537566

I've just checked that one of my books has a method of solving $\displaystyle \sqrt[n]{a+\varphi(x)}+\sqrt[n]{b-\varphi(x)} = c$ via symmetric polynomials that looks suspiciously like Cardano's method. It says let $\displaystyle u= \sqrt[n]{a+\varphi(x)}$, $ v=\sqrt[n]{b-\varphi(x)}$. Then solves $u+v = c$, $u^3+v^3=a+b$ via symmetric polynomials.

Nicholas Roberts

Ah, ok. Is this from the unversal property of the abelianization?

I know given any hom from G to H (H abelian) we can get a unique hom from Ab to H

0celo7

Probably, but easy to check directly.

Yes, exactly.

Nicholas Roberts

Ok, so then every hom from G to H corresponds to a hom in Ab to H. but how do you know every Hom from Ab to H is hit?

Like there could be some members in there that do not correspond

0celo7

So you have $f\in Hom(Ab(G_1),G_2)$...take $[g]\in G_1$ and show that $\tilde f(g)=f([g])$ is a well defined thing and constant on the class of $g$

Nicholas Roberts

3:51 AM

Im not being clear lol but hopefully you see what i mean

0celo7

and that gives you the hom $G_1\to G_2$

roughly because any extra bits you get get canceled when you use the homomorphism property and then the abelianness of $G_2$

Nicholas Roberts

Ah I see. and this will show the two sets are in set-theoretic bijection??

0celo7

@NicholasRoberts yeah

Nicholas Roberts

THanks dude

0celo7

np

Leaky Nun

4:07 AM

@NicholasRoberts abelianization is the left adjoint of the forgetful functor from Ab to Grp

@NicholasRoberts they are in natural isomorphism

0celo7

4:20 AM

@LeakyNun **** ***

Daminark

@0celo7 I have never heard of a group operation denoted in quite that way

0celo7

@Daminark star morphism

Daminark

:O

Secret

5:40 AM

[Random]

1+1+1+1+... < w

1+1+1+...+1 = x

Secret

5:51 AM

Interfinite attempt 1:

1+1+1+1+...+1=a

a+1=1+a=a

a+a+a+...+a=b

b+1=1+b=b+a=a+b

n1=a=1n

ma=am=b

$1<_n a <_m b$

(m+1)a=ma+1a=b+a=b

0

1

11

111

1111...

Let (m1,m2,...) be (4,4,4,...)

0,1,11,111,a

Silent

6:08 AM

@0celo7, Artin says: 'A polynomial with coefficients in a ring $R$ is a finite linear combination of powers of the variable: $f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ where coefficients $a_i$ are elements of $R$.' Here, what is $x$, ie what is domain? real? complex?

Secret

a=a1=a11=...

a,aa,aaa,b

w=lim(1,11,111,...)

a=1111

hmm...

Tobias Kildetoft

@Silent There is no domain, as these are not functions

the $x$ is just a formal variable that helps you keep track of what each coefficient represents and helps you remember how to add and multiply these things

Silent

@TobiasKildetoft thank you very much

Vrouvrou

7:06 AM

hello, i want to study the convergence of $(v_n=\frac{1}{n})$ in the topology endowed with $\{(a,b], a<b, a,b\in\mathbb{R}\}$, i say let $l\in\mathbb{R}$ , if $l\leq0$ then $l$ is not a limit because $\frac1n\notin(l-\varepsilon,l]$

if $l=1$ there exists $\varepsilon=\frac12$ such that there is only $v_1\in (\frac12,1]$ so $1$ is not a limit

how to do the case $0<l<1$ and $l>1$

@AlessandroCodenotti hello, have you an idea please ?

Tobias Kildetoft

@Vrouvrou Why did you take the interval $(l-\epsilon,l]$ rather than $(l-\epsilon,l+\epsilon]$?

Actually, why did you take an epsilon at all?

Vrouvrou

because $(l-\varepsilon,l]$ is open in this topology and it is small then $(l-\epsilon,l+\epsilon]$

Tobias Kildetoft

ok, and why the epsilon at all?

Vrouvrou

to work with a small neighborhood

i can say $(a,l]$ a<l

nothing change with a instead of $l-\varepsilone$

Tobias Kildetoft

What special property of $l=1$ did you use? Why can you not do all the $l > 0$ cases like that?

Vrouvrou

7:22 AM

i tryed but i don't find how to explain that why we can't find $n_0$ such that $\forall n\geq n_0, $\frac1n\in (a,l]$

Alex D

Hi all it's my first time in the chat and I was wondering if I could ask a question in here instead

Tobias Kildetoft

@Vrouvrou Just pick any $a$ such that $0 < a < l$

then from some point onwards, the sequence will be smaller than $a$

@AlexD Ask away

Vrouvrou

@TobiasKildetoft i don't understand

Tobias Kildetoft

@Vrouvrou You are trying to pick an open set which contains $l$ and which does not contain the tail of the sequence, right?

Vrouvrou

yes

Tobias Kildetoft

7:27 AM

So given $l > 0$ pick $a$ as I said and take $(a,l]$

Alex D

I was working on a problem from my intro to analysis class. I need to find the limit of $x^x$ as $x$ approaches 0. From calculus I know its 1 using l'hopital's rule, but we havent covered that in this course. The hint is that I choose a sequence converging to zero such that the limit of the sequence $(f(x))$ is easy to determine

Tobias Kildetoft

the argument is basically the same as for why this sequence does not converge to something larger than $0$ is the usual topology

Alex D

My teacher couldn't give us any further hints as we tried solving it in class

Vrouvrou

@AlexD can we write $x^x$ as exp ?

@TobiasKildetoft i will try to write this , but i need something exact

Tobias Kildetoft

@Vrouvrou That was exact

Alex D

7:33 AM

@Vrouvrou You mean as an exponential sequence?

Tobias Kildetoft

@AlexD What do you get if you take log twice?

Actually never mind, that probably requires you to already know that taking log once makes it go to $0$, rather than to something negative

and that is basically the issue anyway

Abhinav

I am kinda stuck in a question.The question is: five balls are to be placed in 3 boxes

Vrouvrou

@TobiasKildetoft how i prove that for any $0<a<l$ card\{n\in\mathbb{N}^*, \frac1n\in(a,l]\}<+\infty$

Abhinav

Each can hold all the five balls.In how many ways can we place the balls so that no balls remain empty if balls and boxes are different

Tobias Kildetoft

@Vrouvrou By noting that there is some $m$ such that $1/m < a$.

Vrouvrou

7:39 AM

ohhh Archimed?

Tobias Kildetoft

right

Vrouvrou

there exist 1! how to prove that there is an ifinity ?

Tobias Kildetoft

@Vrouvrou Because the sequence is decreasing

Balarka Sen

@Abhinav "Balls and boxes are different" That means the balls are distinguishable, and the boxes are distinguishable?

Abhinav

yes

Balarka Sen

7:43 AM

So, eg, if the balls are numbered and the boxes are denoted by carets, 1|2|345 and 2|1|345 are different arrangements

Abhinav

I think so

I proceeded to do by diving them into groups of{3,2,1}and {2,2,1}

But I didn't get my answer after calculating this

150

Balarka Sen

Sorry, uh, that was a wrong count. Hm.

Abhinav

Each can hold all five balls....I forgot to add "all"

Balarka Sen

Wait, but you said boxes can't be left empty!

If no box is left empty, I want to start by placing three balls in three boxes. There are 3! ways to do it. Then place the rest two in, which you can do in 3x3=9 ways. The answer should be 9*6 = 54, in that case.

Tobias Kildetoft

@BalarkaSen You also need to choose which balls to place from the start

Balarka Sen

7:49 AM

Crap you're right.

Vrouvrou

@TobiasKildetoft Archimed say that $\forall \alpha,x\in\mathbb{R}_+,\exists m\in \mathbb{N}, 0\leq \alpha\leq n x$ i applied it for $\alpha=1$ and $x=a$ then i must appy it for what ? to get the infinity ?

Tobias Kildetoft

But some final configurations may be identical despite having different starting configurations

Balarka Sen

Yeah.

Tobias Kildetoft

@Vrouvrou Did you see what I wrote about the sequence being decreasing?

Abhinav

@BalarkaSen I think they said that for confusion

Balarka Sen

7:50 AM

@TobiasKildetoft Eg, 1|2|3 and put 4 and 5 in the first and 4|2|3 and then 1 and 5 in the first

Vrouvrou

@TobiasKildetoft yes how we use that it is decreasing ?

Balarka Sen

That's annoying.

Tobias Kildetoft

@Vrouvrou What you are trying to show follows directly from that, since then all natural numbers larger than the $m$ you found will also work

Vrouvrou

@TobiasKildetoft what about the $m$ that i found ican be large!

Tobias Kildetoft

so what?

Vrouvrou

7:54 AM

i can't see why there is an ifinity $\frac1n$ from the fact that there exist $m$ such that $0<\frac1m\leq a$

Abhinav

Can anybody point out what am I missing

Tobias Kildetoft

@Vrouvrou I think you need to take a step back and revisit the proof that this sequence converges to $0$ and not to anything else in the usual topology. Because the things that are causing confusion are really things that ought to feel natural before you start considering other topologies.

Balarka Sen

@Abhinav Ok, I think inclusion-exclusion does the trick. There are 3^5 ways to put the balls in the boxes, but we have to subtract the number of arrangements which leaves one box empty. That's possible in 3*2^5 ways (select a box which would be empty, and you put the other crap in the rest of the two boxes), but you're overcounting: subtract the number of arrangement which leaves two boxes empty. That's 3C2*1^5.

So it's 3^5 - 3 * 2^5 + 3C2 * 1^5, which seems to be 150

Tobias Kildetoft

@BalarkaSen Right, we are counting surjective maps from a set with 5 elements to one with 3 elements

Balarka Sen

I suppose, yeah

Tobias Kildetoft

8:01 AM

and counting surjective maps is always a pain

Balarka Sen

Yeah

Abhinav

@BalarkaSen oKAY

@BalarkaSen Also can you tell me the difference between (3^5)and (5^3)here

Balarka Sen

The balls go the boxes. There are 3 ways a ball can go in a box, and there are 5 balls, so 3 x 3 x 3 x 3 x 3 = 3^5 ways to do this. You can't use the same argument for boxes, because a box can "take in" multiple balls.

Each ball goes to a unique box, but the converse (each box contains a unique ball) is false.

So the 5^3 thing doesn't work.

Vrouvrou

9:31 AM

hello

@TobiasKildetoft thank you for your help

I have $w_n=\sin(n\pi/4)\sin(n\pi/2)$ i want to find the adherent value of $w_n$

$w_0=w_2=w_4=...=w_{2k}=0$

so 0 is an adherent value

but

$w_1=w_7=w_9=\sqrt{2}/2$

$w_3=w_5=w_{11}=w_{13}=-\sqrt{2}/2$

how i can summerise this ?

Tobias Kildetoft

What is an adherent value?

Vrouvrou

the limit of subsequences

we are in the usual topology

Tobias Kildetoft

So the set of values of the sequence is finite and hence the only convergent subsequences are those whose value becomes constant from some point.

So the adherent values are precisely those values that the sequence takes infinitely many times (and it takes all of its values infinitely many times)

Vrouvrou

yes

Tobias Kildetoft

So you now have all of the adherent values of the sequence.

Vrouvrou

9:46 AM

$l$ is an adherent value iff $\forall V\in \mathcal{V}_l, card\{n\in\mathbb{N}, x_n\in V\}$

@TobiasKildetoft i want to write the subsequences as $w_{2k}$

i want to find all the subsequences that takes $\sqrt{2}/2$

until now i found $w_1=w_7=w_9=w_{15}=w_{17}$

Tobias Kildetoft

Why do you want to find all of them?