Mathematics - 2017-09-16 (page 3 of 9)

@Daminark About the book, I read it in the library, I tought my professor wasn't teaching in the subjects the same way (both intensity and focus) that the book did

so I got afraind in studying using Hoffman and taking bad grades

which sadly I got even tought I didn't use it

Daminark

Yeah don't use that book with the aim of supplementing your class if it doesn't match

Balarka Sen

@Semiclassical what about them raviolies

Daminark

I recommended it less for that reason and more because I had fun with it

Balarka Sen

ravioli ravioli where is your omelettoli

Daminark

So I'd say to read it if you have spare time, or after this class is over, for a more theoretical treatment, since that subject is just beautiful

@BalarkaSen kek

Leaky Nun

4:27 AM

$$\int_0^1 \dfrac{\mathrm dx}{\sqrt{-\ln(x)}}$$

Xam

Hi @Daminark

Daminark

How's it going?

user2860452

@Daminark yeah... I believe in you

this subject is beautiful, I love math, I don't know why the chemistry isn't working

Xam

Well, I'm searching about an alternative proof of Akizuki's theorem: an commutative artinian ring is notherian.

Balarka Sen

I have always found that fact interesting. I don't know a proof though.

Xam

4:39 AM

@BalarkaSen are you referring to Akizuki's theorem??

Balarka Sen

yes

Semiclassical

@LeakyNun Whence this problem?

user2860452

Balarka, how's reasearching something?

I never did that before

Leaky Nun

@Semiclassical some discussion group

Semiclassical

hm

Balarka Sen

4:40 AM

Uh, researching what?

user2860452

" I'm searching about an alternative proof of Akizuki's theorem"

"I have always found that fact interesting"

this

Xam

@BalarkaSen the classical proof is based on length of modules, but I'm looking for a free module approach.

Semiclassical

I find it hard to be excited by integrals which are only hard because someone has done a substitution to make it look different.

Balarka Sen

@user2860452 I am not the one searching for a proof :) I just think it's an interesting result.

@Xam I see.

Xam

@user2860452 doesn't searching and researching have different meanings?

Daminark

4:42 AM

@user2860452 fwiw, my first analysis pset was basically "row reduce 20 matrices" and they made me wanna stab someone, so if you've just been toying with numbers every time, I think it's just that part of math that doesn't click

Xam

@BalarkaSen yes, indeed. But I don't like much the proof xd

Semiclassical

@Daminark ugh

"I am not a computer, I am a human being!"

Balarka Sen

Ok, I got to run now. See everyone in about 3 hours.

Semiclassical

("But didn't computers used to refer to actual human beings who did calculations?" "Quiet, you.")

Daminark

I kid you not it was like, a few problems that were fun, one of them involved proving that row echelon form was unique, that sorta thing

4 non-linalg problems that were from the set theory chapter of our book that our prof wanted us to read

user2860452

4:45 AM

@BalarkaSen Taking care of health, congratulations! :D

Daminark

And then 20 something problems that were like "row reduce this matrix" or "solve this system of linear equations using the augmentation thing"

user2860452

math community, I have to sleep, it's 1:47 AM in Brazil.

have a good day

Daminark

See you @user2860452!

user2860452

#australia

Daminark

5:50 AM

@Fargle yo

Fargle

oh nice

yo

I didn't even mean to enter the chat; I just still had it open when I closed my laptop

but here I am

Daminark

Lmao, welcome back!

How goes life?

Leaky Nun

6:02 AM

@TobiasKildetoft regarding $\operatorname{Aut}(\Bbb Z/n\Bbb Z) \cong (\Bbb Z/n\Bbb Z)^*$: why exclude $n=0$?

$\Bbb Z/0\Bbb Z = \Bbb Z/\{0\} \cong \Bbb Z$

The units of $\Bbb Z$ are $1$ and $-1$

so it's still correct for $n=0$...

Also, I presume that the LHS is talking about the automorphisms of a group while the RHS is the units of a ring?

Tobias Kildetoft

@LeakyNun Yeah (though the course the exercises were made for introduces the notation $(\mathbb{Z}/n\mathbb{Z})^*$ as a group before introducing rings).

Leaky Nun

@TobiasKildetoft then what does $*$ mean?

(also did you see my earlier comment?)

16 hours ago, by Leaky Nun

@TobiasKildetoft typo on exercise 3? (x,0) is not an element of the infinite dihedral group you constructed

Tobias Kildetoft

6:18 AM

The star indicates those elements that are coprime to $n$

Leaky Nun

@TobiasKildetoft :O

I've never seen that notation :P

you haven't even shown that it is a group then

Tobias Kildetoft

it is inspired by the fact that those are precisely the units of the ring

Leaky Nun

do I need to show that it is a group?

Tobias Kildetoft

The book the course uses introduces the notation even before groups

If you haven't seen that done then yes. The exercises takes that as given

Leaky Nun

:O

@TobiasKildetoft well of course I know how to prove that it's a group, I'm just asking you if I need to do so

Tobias Kildetoft

6:20 AM

@LeakyNun You don't need to do anything :) But also no, not in the context of these exercises

And you are right, it is a typo in exercise 3. I changed the notation in the first part at some point without updating the second part

Leaky Nun

I see

@TobiasKildetoft I quite enjoy doing your exercises :P

Abcd

$$L_1 \equiv 5x-y-4 = 0$$
Any point A on this line is $$(5t-4, t)$$ Please verify if you can.

It's opposite $$(t, 5t-4)$$

Tobias Kildetoft

@LeakyNun Good to hear

Leaky Nun

@TobiasKildetoft what do you use if $\varphi$ is taken?

Tobias Kildetoft

use for what?

Leaky Nun

6:24 AM

mapping

isomorphism

@Abcd it's $(t,5t-4)$

Tobias Kildetoft

typically $\psi$

Leaky Nun

@TobiasKildetoft oh, right

Tobias Kildetoft

or I start using indices

Abcd

@LeakyNun Realised that and edited but how do I identify without plugging in and checking?

Tobias Kildetoft

Funny how certain letters tend to be used for certain types of objects

Leaky Nun

6:25 AM

@Abcd let $x=t$. find $y$.

Abcd

@LeakyNun Why not the other way out? Let y= t?

Leaky Nun

@TobiasKildetoft certainly. $f,g,h$ are functions, $g$ is a group element, $h$ is an element of a normal subgroup, etc.

@Abcd you can also do that.

Relevant joke (paraphrased):

> let $N$ be an integer. Wait, $N$ is too big; let $k$ be an integer.

Abcd

@LeakyNun So $((t+4)/5, t)$ is also correct?

Leaky Nun

@Abcd certainly.

Abcd

Okay :)

Leaky Nun

6:29 AM

@TobiasKildetoft wait, I need to use a third isomorphism

gg

Abcd

$$\Re$$

How do I make $\theta$ straight instead of italic using math jax?

Leaky Nun

> The MathJax web fonts only include lower-case Greek letters in italic form, so there are no upright versions available. You can use something like \unicode[Times]{x3B1} to obtain the alpha symbol from the Times font (assuming the user has that installed), which will be an upright version.

github.com/mathjax/mathjax-docs/wiki/…

Tobias Kildetoft

@LeakyNun There, Abcd gave you a new letter to use for it, $\theta$

Leaky Nun

@TobiasKildetoft it sounds strange, but I'll use it

Abcd

$\unicode[Times]{x3B1}$

Leaky Nun

6:31 AM

$\unicode[Times]{x3B8}$ \unicode[Times]{x3B8}

Abcd

@LeakyNun very small

Leaky Nun

@Abcd don't complain about everything

Abcd

xD

$$\unicode[Times]{x3B8}$$

@LeakyNun It's barely visible, not complaining, speaking the truth

$\eth$

$\mathfrak{F}$

Leaky Nun

$\mathfrak{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$

$\mathscr{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$

$\mathbb{ABCDEFGHIJKLMNOPQRSTUVWXYZ}$

Abcd

$$\mathbb{CHEMISTRY} \mathbb{LEAKYNUN}$$

How to get spaces between words in mathjax?

Like if I want to write $How are you?$ with spaces between the words.

Leaky Nun

6:38 AM

~

\quad

@TobiasKildetoft What is $\operatorname{Aut}(\langle \Bbb Q,+\rangle)$ isomorphic to?

Abcd

$$\mathbb{ABCD} \quad is \quad \mathbb{OFFLINE}. \quad \mathbb{BYE}$$

Tobias Kildetoft

@LeakyNun $\mathbb{Q}^*$

This is because $\mathbb{Q}$ is uniquely divisible

Leaky Nun

@TobiasKildetoft :O

But isn't $\Bbb Q^*$ equal to $\Bbb Q\setminus\{0\}$?

Tobias Kildetoft

yeah

Leaky Nun

@TobiasKildetoft hmm?

oh, I get it now

Tobias Kildetoft

6:41 AM

This is just because homomorphisms of uniquely divisible groups are really $\mathbb{Q}$-linear maps

Leaky Nun

now what is $\operatorname{Aut}(\langle \Bbb R,+\rangle)$ isomorphic to?

Tobias Kildetoft

something astronomical

Leaky Nun

all hell breaks loose

@TobiasKildetoft I think it's isomorphic to $S_\Bbb R$ given choice

or $S_{\Bbb R/\Bbb Q}$ with no choice

amirite?

Tobias Kildetoft

@LeakyNun No, it has more structure than that

Leaky Nun

@TobiasKildetoft how so?

Tobias Kildetoft

6:43 AM

Those automorphisms are really $\mathbb{Q}$-linear, so it is the general linear group on an uncountably dimensional vector space

Leaky Nun

:O

you're right, we aren't just permuting the basis

right, it's $GL(|\Bbb R/\Bbb Q|,\Bbb Q)$

@AlessandroCodenotti hi :D

Alessandro Codenotti

Hi @Leaky and everyone

Leaky Nun

@AlessandroCodenotti amirite?

$$\operatorname{Aut}(\langle \Bbb R,+\rangle) \cong GL(|\Bbb R/\Bbb Q|,\Bbb Q)$$

Alessandro Codenotti

Can't answer those questions before my morning coffee!

Leaky Nun

@TobiasKildetoft Should I post this as a self-question?

Tobias Kildetoft

6:46 AM

Maybe I should add an exercise about the automorphisms of the rationals

@LeakyNun I have a feeling it might well have already been asked and answered

Leaky Nun

@TobiasKildetoft I can't find it

Alessandro Codenotti

$\text{Aut}(\Bbb R)$ as a field is much nicer

Daminark

Hey friends

Leaky Nun

Do you need choice to prove that $\langle \Bbb R,+ \rangle \cong \langle \Bbb C,+ \rangle$? I have a proof with choice.

@AlessandroCodenotti and much boring-er

Hello darkness my old friend

Tobias Kildetoft

@LeakyNun I think so, or at least some restricted version of choice

Leaky Nun

6:48 AM

@TobiasKildetoft should I ask that on main? :D

Tobias Kildetoft

Sure, if you don't find it then go ahead and ask. Asaf might have a good idea of how much choice is needed, as he has done a lot of stuff with vector spaces in the absence of choice

Leaky Nun

who is Asaf?

What is $GL(|\Bbb R/\Bbb Q|,\Bbb Q)$ isomorphic to?

Tobias Kildetoft

Asaf Karaglia. He does not use the chat as he doesn't like it

I don't think there is any "nicer" description of that group

Leaky Nun

11

A: Group of automorphisms of $(\mathbb{R}, +)$

No, it's much, much larger. Assuming the axiom of choice, $\mathbb{R}$ has a basis as a $\mathbb{Q}$-vector space, so it's isomorphic to an (uncountable) direct sum $\mathbb{R} \cong \bigoplus_I \mathbb{Q}$. Its automorphism group is $\text{GL}_I(\mathbb{Q})$, which is very big, and in particular...

You're right

and it has been asked

Leaky Nun

7:03 AM

0

Q: Proving $\langle \Bbb R,+ \rangle \cong \langle \Bbb C,+ \rangle$ without choice

In this question, we have proved that $\langle \Bbb R,+ \rangle \cong \langle \Bbb C,+ \rangle$: Pick your favourite Hamel basis $H=\{U_\alpha \mid \alpha \in I\}$ where $I$ is an indexing set. Then, $H \cup iH$ is a basis of $\Bbb C$ as a vector space over $\Bbb Q$. Pick your favourite bijecti...

@TobiasKildetoft :)

Alessandro Codenotti

If you google "pathological solutions Cauchy functional equation" you'll find more about constructing those weird automorphisms with choice

Leaky Nun

@AlessandroCodenotti can't I just think of it as $GL(|H|,\Bbb Q)$?

Alessandro Codenotti

What's H?

Leaky Nun

your favourite hamel basis

Alessandro Codenotti

Hm, I'm not sure what the notation $GL(n,\Bbb K)$ means for $n$ an infinite cardinal

Leaky Nun

7:06 AM

all invertible linear transformations :)

$A$ is invertible if there exists $B$ such that $AB=I=BA$

Leaky Nun

7:18 AM

how is it possible to have a gold badge in the axiom of choice...

@TobiasKildetoft the legendary Asaf appeared

> Ph.D. student in the Hebrew University of Jerusalem.

Set theorist to-be.

Felix.C

In currently getting a bit into digital data processing and radio transmission and I met the term of a chirp, being just a frequency starting at a lower bound and ramping up to a higher bound again and again in time (up-chirp), or vice versa (down chirp), it might also start in between but will end at the same position as the beginning after all (modulated chirp)...

It's obvious that if we add the frequencies of a up-chirp&down-chirp over time we have a constant function...modulation would just mean a offset..I'm just in my first university yet, so I'm not too much into fourier and so on, but any idea how one could represent such signals mathematically?

Leaky Nun

$2^A<3^A$ can be true without choice :O

Leaky Nun

7:45 AM

the world without choice is a weird one

the world with choice is also a weird one

Daminark

7:55 AM

@AlessandroCodenotti you can think of R^n as the set of maps from {1,...,n} into R

So I'd guess that in the case of an infinite cardinal it'd be the space of maps from a set of that cardinality into R

Leaky Nun

@Daminark I think it has to be a direct sum, not a product

Alessandro Codenotti

$\bigoplus_H\Bbb R$ is the object I was missing I think

Daminark

And invertible transformations on that space

Leaky Nun

@AlessandroCodenotti $\bigoplus_H \Bbb Q$?

Alessandro Codenotti

Doesn't matter the field, I wanted to know what $GL(\kappa,\Bbb K)$ means

Leaky Nun

7:58 AM

@AlessandroCodenotti right, it's $\bigoplus_\kappa \Bbb K$ then

not product :)

Daminark

Probably that I guess. Like thinking of GL(n,K) as GL(K^n), since they're isomorphic

Also isn't the direct sum defined as the Cartesian product?

Leaky Nun

@Daminark only if n is finite :)

direct sum is product with finite support

Daminark

Okay, so why are we preferring direct sum in the infinite case?

Alessandro Codenotti

Usually it's nicer. It's kinda like the product vs box topology

Leaky Nun

@Daminark because we're considering a vector space

Daminark

8:02 AM

Actually even in that case, like what are some of the things you get out of it

Leaky Nun

every element in a vector space is a (finite) linear combination of the basis elements

so the vector space is isomorphic to the direct sum, not product, of as many copies of the underlying field as the basis cardinality

Daminark

Oh so you want the (1,0,0...) and all to be a basis?

Leaky Nun

right

Alessandro Codenotti

Why can't the basis elements have infinitely many nonzero components? The space of all real functions doesn't look like a direct sum to me

Leaky Nun

Because the context is $\operatorname{Aut}(\langle \Bbb R,+\rangle) \cong GL(|\Bbb R/\Bbb Q|,\Bbb Q)$ @Daminark

@AlessandroCodenotti it's still a direct sum of its basis

but the basis of all real functions has cardinality $2^\mathfrak c$ (to be verified)

Daminark

8:06 AM

I mean if we're in infinite dimensions thinking about matrices is pretty shit anyway. Is it still true that if two vector spaces have the same cardinal dimension (including infinite cardinals), they must be isomorphic?

Leaky Nun

@Daminark yes, because you can just biject between the basis elements

(over the same field I presume)

Alessandro Codenotti

It looks like the direct product of $\Bbb R$ with itself uncountably many times to me

Leaky Nun

@AlessandroCodenotti yes, but $\mathfrak c$ isn't its basis cardinality

Daminark

Okay so in that case you can just think about transformations and use some direct product as many times as you need

Leaky Nun

Wait, isn't the space of all real functions just the dual space of $\Bbb R$?

yep, that justifies the doubly uncountability of the basis

Alessandro Codenotti

8:08 AM

@LeakyNun I don't know what you were arguig there then

Daminark

He wanted to have the e_i as basis vectors

Alessandro Codenotti

@LeakyNun What about non linear functions?

Leaky Nun

sorry, I was explaining something to someone else

I tried to think about both things at the same time, and I failed

@AlessandroCodenotti ignore that :P

@AlessandroCodenotti let $V$ be a vector space over field $F$. Let $B$ be a basis (assuming choice). Then, $V \cong \bigoplus_B F$.

If $|B| \ge \aleph_0$, then $V \ncong \prod_B F$.

I'm starting to think that the basis of all real functions is singly uncountable @AlessandroCodenotti

@Daminark so what's the question now?

@BalarkaSen hi

Balarka Sen

8:25 AM

h

Daminark

i

Leaky Nun

@AlessandroCodenotti wait, the dual space of $\Bbb R$ is a subspace of all real functions, so the basis of all real functions is doubly uncountable, just like the basis of the dual space of $\Bbb R$ is.

Balarka Sen

@Daminark youtube.com/watch?v=wOvsmAngDuM

Daminark

Peace out everyone!

And lol I'll check it out soon @Balarka

Balarka Sen

peace

Alessandro Codenotti

8:31 AM

Hmm I'm missing something. So you say that the space of all real functions is $\bigoplus_{2^{\mathfrak c}}\Bbb R$, which is sequences of real numbers of length $2^{\mathfrak c}$ with finitelt many nonzero elements, while I'd say it should be $\bigotimes_{\mathfrak c}\Bbb R$, sequences of real numbers with infinitely many nonzero elements

Leaky Nun

@AlessandroCodenotti and I would say that they are isomorphic.

FuzzyPixelz

Hello

Leaky Nun

@AlessandroCodenotti but the basis is doubly uncountable

FuzzyPixelz

So if someone happens to have studied chemistry in french, please give me this term in English:

Dosage

En chimie analytique, le dosage est l'action qui consiste à déterminer la quantité de matière, la fraction, ou la concentration d'une substance précise (l'analyte) présente dans une autre ou dans un mélange (la matrice). == Classification des méthodes de dosage == Il existe trois familles de dosages : méthodes physico-chimiques de dosage : méthodes d'électroanalyse ; méthodes chromatographiques ; méthodes physiques de dosage : méthodes spectrophotométriques : détermination de la quantité d'analyte par mesure de l'absorbance de la lumière à une longueur d'onde donnée par celui-ci (dans l'UV, le...

Leaky Nun

@AlessandroCodenotti and what we're saying basically amount to this:

> let $V$ be a vector space over field $F$. Let $B$ be a basis (assuming choice). Then, $V \cong \bigoplus_B F$.

> If $|B| \ge \aleph_0$, then $V \ncong \prod_B F$.

Abcd

8:33 AM

Is $$\sqrt{Mmcos^2\alpha+M+m}$$ equivalent to $$\sqrt{(M+m)(M+msin^2\alpha)}$$? Wait, I'll plug in some values and check.

Leaky Nun

@Abcd you just have to expand the latter to find out

Abcd

@LeakyNun No, I wish to reach the latter from the former. It's a part of a physics problem. I obtained the answer that is given as the first expression but the answer given is in the second expression form.

Leaky Nun

@Abcd ... to find out that it's wrong

Do we have a big list here or on mathoverflow of statements equivalent to AoC?

@Daminark @AlessandroCodenotti

Jasper

8:58 AM

@LeakyNun Wikipedia et al might have them, I don't know. I don't know why people look on SE for such things when there is Wikipedia, lol.

Leaky Nun

@Jasper I searched Wikipedia before asking that.

I'm dissatisfied with the brevity of the list provided therein.

they do have Tarski's theorem, lol

Jasper

@LeakyNun Wikipedia has quite a big list...

Leaky Nun

@Jasper hmm?

link please

Jasper

Axiom of choice

In mathematics, the axiom of choice, or AC, is an axiom of set theory equivalent to the statement that the Cartesian product of a collection of non-empty sets is non-empty. It states that for every indexed family ( S i ) i ∈ I {\displaystyle (S_{i})_{i\in I}} of nonempty sets there exists an indexed family ( x i ...

Leaky Nun

right, that page

Jasper

9:02 AM

Yes, so many, more than you need this lifetime.

Leaky Nun

9:23 AM

@MartinSleziak hi

Martin Sleziak

Looking at questions tagged both (big-list) and (axiom-of-choice) on Mathoverflow and on this site might be a reasonable place to look.

Leaky Nun

@MartinSleziak thanks

Martin Sleziak

But I think that all important ones are mentioned at Wikipedia.

If you're looking for more, try to get Rubin-Rubin, Howard-Rubin or Herrlich. (All of them are listed among references in the Wikipedia article and they are books devoted exlusively to Axiom of Choice.

Howard and Rubin's book is also accompanied by a website with a database. But it seems that the website is under reconstruction: What is current status of Consequences of the Axiom of Choice website?

Jasper

@LeakyNun @MartinSleziak is the most helpful person on Math SE.

Martin Sleziak

Don't talk non-sense.

3

Leaky Nun

9:40 AM

in Set theory, 43 secs ago, by Leaky Nun

Is there a model of PA in which Goodstein's theorem doesn't hold?

Martin Sleziak

Discussion about AC reminded me of this comment from the main site:

If I wanted to know erverything in greatest detail about some oscure conseqeunce of AC, I'd ping Asaf Karagila - probably much more complete than any dead database.. ;) — Hagen von Eitzen Jun 6 at 10:27

Leaky Nun

I've heard that Asaf doesn't use chat

Martin Sleziak

As you can see from this old post on meta, for some time he was the most active user of chat. But he avoids chat for some time. (I do not know the exact reasons why he decided to do so - but I can understand that chat can be time-consuming.)

Balarka Sen

i have heard rumors that he used to visit this place on a distant past

Leaky Nun

Aug 16 '11 at 11:08, by Feeds

Asaf Karagila has been automatically appointed as owner of this room.

Aug 16 '11 at 11:54, by Asaf Karagila

I thought about the mathematician, who tried to prove that all cardinalities are equal, but ended up misreading Bernstein's thesis and proving something most useful.

I've spent the last seven years working on choice related research. So that kind of makes sense, why not? — Asaf Karagila 2 hours ago

He spent seven years working on choice.

He joined SE seven years ago.

The above quotes are taken 6 years ago

It feels strange to be able to access a time where an expert hasn't become an expert.

Aug 16 '11 at 8:55, by Asaf Karagila

Interesting. About fifth of the traffic in this chatroom is me.

Jasper

9:54 AM

While Asaf is the expert on the axiom of choice, I think Waiting is the expert on limits, series, and integrals. =D

Quantitative analysis (chemistry)

In analytical chemistry, quantitative analysis is the determination of the absolute or relative abundance (often expressed as a concentration) of one, several or all particular substance(s) present in a sample. == Methods == Once the presence of certain substances in a sample is known, the study of their absolute or relative abundance can help in determining specific properties. Knowing the composition of a sample is very important, and several ways have been developed to make it possible, like gravimetric and volumetric analysis. Gravimetric analysis yields more accurate data about the composition...

I am not sure if this is the term, but it may help you. ^

Leaky Nun

@Jasper je l'aurais donne cette page si je n'ai decouvert que cette page a une version francaise

Perturbative

10:31 AM

Hey everyone!

Leaky Nun

@Perturbative hi

Perturbative

Quick question, if $X$ is a CW-Complex, the $n$-skeleton of $X$ is defined to be the subspace $X_n \subseteq X$ consisting of the union of all (both open and closed) cells of dimensions less than or equal to $n$ correct?

Balarka Sen

Right.

Perturbative

Ayt

Leaky Nun

Solve the diophantine equation $\dfrac{x}{y+z} + \dfrac{y}{x+z} + \dfrac{z}{x+y} = 4$

with $x,y,z \in \Bbb N \setminus \{0\}$

Balarka Sen

10:45 AM

Hm, that's the same as saying $1/(y + z) + 1/(x + z) + 1/(x + y) = 7/(x + y + z)$.

Faust

anyone online that understands graph theory?

Leaky Nun

@BalarkaSen Shouldn't it be $7/(x+y+z)$?

Balarka Sen

Ah, yes, thanks.

Leaky Nun

@Faust just ask. You know the rules.

Faust

i have a solution to a question that i cant understand for the life of me

Alessandro Codenotti

10:47 AM

@LeakyNun that seems one of those innocent looking problems with a 40 digits solution

Leaky Nun

@AlessandroCodenotti shh...!

Faust

in this problem https://math.stackexchange.com/questions/2428623/if-degree-of-each-vertex-of-a-graph-is-big-then-each-its-edge-belongs-to-k-4

$|V’|\ge\frac{n+2}3$ where does this come from?

Leaky Nun

Have you read my rant about basis earlier? @AlessandroCodenotti

Alessandro Codenotti

More or less, I was doing some uni work at the same time

Balarka Sen

This looks like a greedy expansion problem

Leaky Nun

10:48 AM

@BalarkaSen tell me more

(I don't actually know how to solve it)

@AlessandroCodenotti so $\displaystyle \mathscr F \cong \prod_{\mathfrak c} \Bbb R \cong \bigoplus_{2^\mathfrak c} \Bbb R$

Faust

ive spent over an hour working on this problem and i cant for the life of me understand where he came up with that lower bound from

Leaky Nun

@Faust what is $\delta$?

Faust

minimum degree of any vertex

im assuming hes getting it from assuming what the minimum degree of the two end points is but i see no logical connection. i actually found anther hint elsewhere and it appears to imply the same is true but i cant understand it

Jasper

Hi @Faust, lol.

Faust

bashes head against wall

sup jasper

Hint: Let $(u,v)$ be an edge, and let $N(u)$ and $N(v)$ be the neighbours of $u$ and $v$ amongst $V(G)\setminus \{u,v\}.$ Show that $$\left|N(u)\cap N(v)\right|\geq \frac{n+2}{3}\geq 2.$$

Balarka Sen

11:08 AM

@LeakyNun Yeah I don't know how to do this. I was going for AM-HM.

But that doesn't help here.

(3/7 < 2/3)

Leaky Nun

but what is greedy expension?

Balarka Sen

lmgtfy

It's a way to write k/n as an Egyptian fraction

mostly a heuristic though

AM-HM does say that if I take $4$ variables instead of $3$, there is no solution, so that's something I guess :p

Leaky Nun

but there's a solution

Balarka Sen

in the three variables case you mean

sure, i can believe that

Leaky Nun

29 mins ago, by Alessandro Codenotti

@LeakyNun that seems one of those innocent looking problems with a 40 digits solution

Balarka Sen

11:23 AM

but like if i have $\sum_{i} 1/(x_i + x_{i+1}) = n/\sum x_i$ with index on the first cyclic sum going over $m$ terms $i = 0, \cdots, m-1$ modulo $m$, then AM-HM poses the restriction that $m/n \leq 2/m$, or $m^2 \leq 2n$ if you prefer.

So that gives an immediate restriction on which diophantine equations of this form don't have solutions

maybe that's an obvious point to make, i dunno

it sounds like a good question to ask that for the particular case you wrote down

Perturbative

11:40 AM

There's this problem that I'm working on, "If $X$ is any CW-Complex, the topology of $X$ is coherent with the collection of subspaces $\{X_n | n \geq 0\}$"

Showing that the topology of $X$ is coherent with $\Phi = \{X_n \ | \ n \geq0\}$ means that a subset $U \subseteq X$ is open in $X$ iff $U \cap X_n$ is open in $X_n$ for each $X_n \in \Phi$

In the problem the forward implication is trivial, but I'm not sure how to do the reverse implication

Balarka Sen

The first thing to do would be to understand that $X$ is obtained from attaching higher dimensional cells to $X_n$, I think.

And that means, $X$ is a quotient space $X_n \bigsqcup_\alpha e_\alpha^k /\sim$ where $\sim$ is an equivalence relation keeping track of all the attaching maps in all dimensions ($k > n$)

What are the open sets of that quotient space?

Perturbative

Just give me a few mins to work it out the open sets

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