@Charlie k

I'm on gmail...

ohh

Good morning.

@JayeshBadwaik I'll upload the full version to you, it has covers, soundtracks, etc.

@Charlie Hello. =)

@GustavoBandeira Bom dia!

@Charlie Tá bem?

@GustavoBandeira tô. e tu?

@PeterSheldrick Wolfram's rival.

@Charlie Tambão.

@GustavoBandeira que bão

@Charlie O que faz?

@GustavoBandeira revendo uns troço pruma prova

@Charlie Tu escreve blog ou alguma coisa?

@GustavoBandeira não, já tive vontade de fazer um, mas tava com preguiça....

@Charlie XD

Faz, mas faz em inglês.

@GustavoBandeira tá

@Charlie Marília, moça de boa família.

@GustavoBandeira Quê?

Zuando. É que rima.

ah

não deixa de ser verdade :P

A Eduarda tava me falando uma coisa.

@GustavoBandeira o quê?

Lá perto da casa dela tem uma tia que tem problema mental, né? Aí ela casou com um cara lá, a família abandonou e ela sai fazendo filho ad infinitum com o cara. As crianças são mal cuidadas, vivem fedendo e com piolho.

O pai - pra resolver o problema dos piolhos - raspou a cabeça delas com a faca, elas tão com a cabeça toda cortada.

@GustavoBandeira Geez.....

Foda. =/

@GustavoBandeira vish...que horror....

Ontem eu tava comentando um barato com o professor de piano: Já percebeu que boa parte do povo que tem filho num aparenta ter 1/90 do que precisa ter pra criar o moleque?

@GustavoBandeira hoje em dia é assim..nego põe filho no mundo e num tá nem aí....

É. Ter filho é um ato tipo peido - cê faz inocentemente e já era.

hahaha

Porra, se eu fosse ter filho, tinha que ter tempo pra colocar o moleque pra estudar piano e matemática, observar/auxiliar ele aprendendo.

sim, claro

Tem uma coisa muito escrota em educação de criança, o povo pensa que só colocar na escola já é suficiente.

@GustavoBandeira exatamente

Eu vejo a escola como um ambiente que muito mais atrapalha do que ajuda, quero colocar meus filhos com professor particular.

Fazer encontros do moleque com gente fodona.

Ia ser umagracinha.

@GustavoBandeira que engraçadinho!

Porque?

@GustavoBandeira fofinho,ué

=)

@JayeshBadwaik Uploading - It's gonna take some time: ~1,5GB.

@csss Hello!

@GustavoBandeira My brother says that if i have a daughter she will look like this

@Charlie Cute. Haha

@JayeshBadwaik Hi

Have a group theory exam in 4 hours...in no mood for it

@csss Don't worry, be happy.

I'll be happy when it's done alright hehe

@csss i have an exam IN 20 MIN :)

End of semester exam?

@Charlie This is very introspective.

@Charlie Are you on a notebook?

I am looking for the non normal subgroups of S3. Say X = {x y z} and consider S(X). Why are the subgroups <(xy)>, <(xz)> and <(yz)> not normal? I read because 'they are conjugate in S3'...conjugate with what exactly?

@GustavoBandeira no

@csss yup

@csss Have you tried conjugating these subgroups to check whether they are normal? Note: to say that two subgroups A and B in G are conjugate is to say that there is a g in G such that gAg^-1=B. The act of conjugating a subgroup A is picking a g in G and applying the conjugation map x|->gxg^-1 to the subgroup A, which will give you the conjugate subgroup gAg^-1.

@AlexJBest Hello!

Anon always appear from the shadows to answer math questions.

@GustavoBandeira exactly!

Let me go, bye bye!

@Charlie Cya.

@anon Does that mean I have to conjugate the element, say (xy), with every other element for a question like this? i.e. there is no quicker way of doing it?

@csss If you were to show that <(xy)> were normal (it isn't) by brute force, you would have to conjugate it by every element (or at least a generating set, but that is perhaps too sophisticated at this point). To show it is not normal, you only need one g in G such that g<(xy)>g^-1=/=<(xy)>.

It is quite easy to pick such a g. Say, g=(xz) or g=(yz).

Also note that g<a>g^-1=<gag^-1> always (do you see why?), which (especially when we restrict ourselves to elements of order 2, where the cyclic subgroups are comprised of only the identity element and the element itself), justifies why we only need to conjugate a=(xy) by g in order to see that <(xy)> is not normal.

In general, we do not bother using brute force (i.e. "check every possible conjugation) to prove that a subgroup is normal. Almost always, we use tricks at hand.

@peoplepower Could you elaborate on your earlier description of the cycle type of the permutation $x\mapsto x^3$ on $\Bbb Z/p\Bbb Z$ (when $p\equiv2\bmod3$)?

(Which I have promptly forgot the statement of.)

@anon I might have to review the transcript.

I remember leaving out a few steps anyway. We're trying to prove that the permutation is even iff $p\equiv -1\pmod{4}$ right?

The original question was why the permutation is even iff p=3(4), but you went further and described the cycle type in terms of $2\varphi(d)$'s for divisors $d\mid?$, if I recall.

A generalized question would be the cycle type of $R\to R:x\mapsto rx$ for $r\in R^\times$ a unit and $R$ a finite ring.

(To see why this is a generalization, we recall that (Z/pZ)^x is cyclic hence iso to Z/(p-1)Z as a group, and the cube map viewed in the latter is x->3x, which can be viewed as multiplication by a unit in the ring Z/(p-1)Z.)

Our permutation is an automorphism of the multiplicative group of our field.

Right.

hrm, what's the smallest noncommutative finite (unital, associative) ring, out of curiosity?

Some matrix ring.

sure, though a subring of a full matrix ring in order to get minimality no doubt

characteristic either 2,3 or 4 probably

Ideals correspond, and the entries will necessarily be from a field.

So we cannot mod out an ideal.

Supose you have the set of natural numbers, what's the cartesian product of N and {0}? {1,{0}} or {1,0}? I believe it's the first choice.

Nx{0}={(1,0),(2,0),(3,0),...} (or it might include (0,0) if you have 0 in your N)

Thanks.

First time I see disjoint union, what's it useful for?

for writing unions and forcing things to be disjoint, duh

(-:

Yes, I've noticed that but why would I want to do this?

$\Bbb F_3\langle x,y\rangle /(x^2,y^2,xy+yx)$ is probably pretty small

3^4=81 elements, maybe not that small

@Mark Hello!

@GustavoBandeira well, what area of math do you want an application for?

@anon That's the problem, have nothing specific. I'm just reading the book you've recommended me: ALGEBRA, Chapter 0.

But relax, I'll just keep reading - I understand the concept now.

direct limits, then. coproducts in Set.

$\Bbb F_2\langle x,y\rangle/(x^2,y^2,xy-x,yx-y)$ has cardinality 8 I think.

That's got to be the smallest.

$x=xy=xyx=x^2=0$.

drats

:)

perhaps /(x^2-x,y^2-y,xy-x,yx-y)

it's commutative if we can obtain x=y from the relations

not quite sure whether or not it's possible

Yeah, I misread your relations.

In the meantime, I think that the cycle type of a "Cartesian product" of two permutations (viewed as a permutation of a product of two given sets) is given by ordering the pairwise lcms of the two cycle types' entries and rearranging to be nonincreasing.

That seems to be a noncommutative ring with four elements. If it is noncommutative, there no beating it.

four elements?

right, four

0,1,x,y

wait, no: 0,1,x,y,1+x,1+y,x+y,1+x+y. that's 8.

Right.

Still seems impossible to beat.

There are two operations on the book, which I've seen no description on them:

come to think of it, that's a nontrivial multiplicative structure on the Klein-four group

@anon I have trouble understanding the algorithm you are explaining here.

algorithm?

It reads like an algorithm.

@GustavoBandeira Depends on context. In the most general context, they are product and coproduct in an arbitrary category. In Set, they are cartesian product and disjoint union.

@anon Oh, I thought it was a different operation, because he was denoting cartesian products with a X. Got it now.

@peoplepower Say $\sigma\in\mathrm{Sym}(X)$ and $\rho\in\mathrm{Sym}(Y)$ have respective cycle types $(\lambda_1,\cdots,\lambda_n)$ and $(\mu_1,\cdots,\mu_m)$. Then I think the cycle type of $X\times Y\to X\times Y:(x,y)\mapsto (\sigma(x),\rho(y))$ is given by taking the multiset $\{\mathrm{lcm}(\lambda_i,\mu_j):1\le i\le n,1\le j\le m\}$ and ordering it into a nonincreasing sequence.

seriously, why make a \gcd command but not a \lcm one?

:mind=boggle:

@GustavoBandeira The $\times$ operation is generally taken to be binary, or n-ary for finite n, whereas $\prod$ extends it to any indexed family of objects in some category.

@ano Yes.

Actually it is pretty clear, because we can write $\sigma\times\rho = (\sigma\times1_Y)(1_X\times\rho)$.

the latter is some kind of disjoint decomposition

(decomposition into mutually central elements; then just look at the orbits of the cyclic groups the generate, etc)

@anon There's a LaTeX website where you draw some symbol and it tries to suggest what command is needed to type it, I don't remeber what site is this, do you know it?

Found detexify.kirelabs.org/classify.html

@GustavoBandeira lemonparty is another great site

It is much easier to remember what sorts of things to google than to remember things.

@N3buchadnezzar Oh yes, that's the one with the girl in the tub innit?.

And some sort of goat.

@anon No, I think it is the one with the old people in it.

$\rm\color{Gray}{sarcasm}$

@anon You are probably thinking about tubgirl.

you failed the bme sarcasm olympics

keyword = allusions

@anon Thats the one with the cup in it right, and chocolate?

right, the food contest involving lemons, cups, spinning meat, and colored waffles.

@anon Sounds like a delicious breakfast, just needs some broken jars.

my reaction to all of the above upon my first viewings

well, actually lemon and spin are understandable and waffle is n/a

I just laugh at these things, there at much worse things out there.

Guro, snuff, and people abusing notation.

Oh, you mean like $\{\emptyset\}\cup\{\{\emptyset\}\}+4=6$

@anon Cheers for the answer on normal subgroups anon...just saw it now, was afk for an hour

Hmm. Assuming ZF, both ZF+C and ZF+~C are consistent. Are they ω-consistent? (Which I understand to mean all of the existential and universal quantifications to be true as we interpret them.)

I am looking at some notes here for group theory and I think they are wrong. It defines the dihedral group D_n as = $\{a, b | b^n = 1, a^2 = 1, aba^{-1} = a^{-1}\}$. Shouldn't it be b^{-1} at the end

?

correct

it should be b^-1 not a^-1 at the end

Yes. In fact, if we assumed the third relation as given then $1=a^2ba^{-1}=ba^{-1}$ implying $a=b$.

Grrr.

I've been having a headache all day and now I was typing up an answer when suddenly this...

Not sure cause posting the same answer 5 minutes after someone else has posted theirs already looks idiotic.

The only thing that looks even more idiotic than posting a duplicate answer is taking someone's comment one minute after they post it and post it as your answer.

e.g.

I agree but you can tell them to make their comment into an answer, like e.g.

: )

Any 10k or diamond mods here who could give some hint about why this question was deleted? math.stackexchange.com/questions/116497/…

It feels a bit annoying to have it just disappear, since I put some effort into the answer.

(I wouldn't even have noticed, if someone hadn't apparently suggested an edit to my answer just before it was deleted. Apparently the deletion didn't make the notification disappear.)

...or I could just post on meta, but I figured I'd try here first.

@IlmariKaronen I don't know. That's odd.

@MattN. Sire

@PeterTamaroff Aye?

I just got a down vote on this one. What's wrong with it?

@MattN. Let me see.

@MattN. Hey.

@PeterTamaroff What's up?

@MattN. I'm doing some exercises on uniform convergence

@PeterTamaroff Aye?

@MattN. I'm having some dificulties with two of them

THe first one is the following.

Suppose that $\{f_n\}$ is a sqnce of differentiaable functions that converge pointwise to some $f$

over $[a,b]$

Suppose that $\{f_n^\prime\}$ converges uniformly to some continuous $g$.

The theorem I have proved that $f$ is differetiable and $f'=g$

Now I have to prove that $f_n\to f$ uniformly.

What's the meaning of the relation ~?

Got it.

"Similar"

@PeterTamaroff Not sure but $f_n \to f$ uniformly means that $\sup_{x \in [a,b]} |f(x) - f_n(x)| \to 0$ for $n$ to $\infty$. This means $\max_{x \in [a,b]} |f(x) - f_n(x)| \to 0$ because $[a,b]$ is compact. Then if you find the extremal point $x_0$ of $f -f_n$ and show that $|f(x_0) -f_n(x_0)| \to 0$ then you're done.

@GustavoBandeira It depends

@PeterTamaroff On what?

So I suspect you differentiate to find the extremal points...

...for which you probably need your previous result.

@MattN. That is interesting. I was thinking about using some Lipschizity.

@GustavoBandeira Well, any equivalence relation, or just any relation is usually denoted by $\sim$

So you need to see how the relation was defineed.

@PeterTamaroff Well if they are differentiable then they are Lipschitz, yes. Because $f'$ is bounded because the domain is compact.

@MattN. Indeed.

But what do you do if $f_n$ are all Lipschitz?

@MattN. I have to think.

This is an honest question that is, I don't know the answer.

@PeterTamaroff What's the difference of ~and =? In my book there are three properties, the first one is reflexivity: $(\forall a \in S)a \sim a$

@GustavoBandeira For matrices, we do denote $A\sim B$ and say $A$ and $B$ are similar if there exist an invertible $C$ with $A=CBC^{-1}$

@PeterTamaroff Did you have a look?

@GustavoBandeira Well, $=$ is a type of equivalence relation. But not the other way around. In fact, the properties of equivalence relation indeed mimic those of usual equality.

@MattN. I'm trying. =)

: )

@PeterTamaroff not the other way around?

@GustavoBandeira Nope.

@PeterTamaroff What you mean with this?

@GustavoBandeira That what I say is the way it is.

@MattN. I think I got it!

@PeterTamaroff $\lvert f(x_0+\delta) - f(x_0) \rvert < f'(x_0)\lvert \delta \rvert$, and then taking the maximum of derivative, you have got a bound.

@PeterTamaroff Tell me.

@PeterTamaroff You mean a=b but b$\neq$a?

@GustavoBandeira no, he meant, an equality relation is an equivalence relation

but every equivalence relation is not the equality relation

@JayeshBadwaik I dunno what's this other way around.

@MattN. Given any $\epsilon>0$, uniform convergence implies that for each $x$ in the interval $|f'(x)-f_n^\prime(x)|<\epsilon$.

@GustavoBandeira refer to my corrected statement

This means that $|f(x)-f_n(x)|$ is "almost constant" in every nbhd of $x\in [a,b]$ with large enough $N$

But pointwise convergence should imply that $f(x)-f_n(x)$ must go to zero.-

I'm tempted to use compactness of $[a,b]$ to get a finite cover and work with that

@MattN. It does make sense, right?

@GustavoBandeira I'm saying that if you look at usual equality, then defining $a\sim b$ if $a=b$ produces an eqv. relation

And basically, reflexivity, symmetricality and transivity are defining properties of equalirty. they are the nice properties equality has

Yep, I'll search about the difference between equality and equivalence.

@GustavoBandeira Just from logic: two statements can be different but cna be equivalent!

@PeterTamaroff Not sure I understand. I thought you wanted to show "Lipschitz" =>"uniform convergence" , not <=. Also: do you mean if $f$ is Lipschitz then $f$ is almost linear at every $x$?

@PeterTamaroff I'm not sure I'm still with you.

@MattN. No, no. I mean that since we can make $g'=f'-f_n^{\prime}$ as small as possible for each $x$, then the slope of $g=f-f_n$ must approach zero. Moreover, pointwise convergence must imply the difference approaches zero

I think I still don't follow. But I am cooking at the moment so I'm a bit distracted.

@MattN. Oh, man. Let me try and put it formally.

Apologies : )

@MattN. I this will be better: since both $f$ and $f_n$ are continuous, $f_n-f$ attains it maxima at some $x_n$ for each $n$. Let's say $\sup\limits_{[a,b]} |f-f_n|=f(x_n)$

For this point $f'=f_n^\prime$

FUUU

My train of thought had to make an emergency stop

@PeterTamaroff I don't think this is true. Pick continuous functions that attain the maximum at $a$ or $b$ (such that $f_n' \neq 0$).

(sorry still cooking)

@MattN. Oh, right. Darn. But you can make the differnc eof derivaives as small as you want!!!

@PeterTamaroff This is saying the derivatives are continuous?

@MattN. cause of uniform convergence

@anon

Man

wut

If I have two matrices that represent a two automorhisms over some finite dimensional vector space

say $A$ and $B$

and I know the eigenvalues of them

how can I know the eigenvalues of $AB$ or $BA$?

do you think the eigenvalues of AB (or BA, which has the same) are determined by those of A and B?

pretty sure BA has the same. I think we need to modify the proof of det(I+AB)=det(I+BA).

@anon Well, now that you say... no?

@anon Yes.

@anon I mean, if you can diagonalize them, you can just multiply them, and you'll get another diagonal matrisx

well, if you can simultaneously diagonalize them...

@anon yes, but that is too much to asl

@anon SO in general the evalues of AB needn't depend on thoseof A and B

@sunflower Is P(A) not representable as hom(2,A)?

hmm

somebody help me to solve this equation \sqrt [ 3 ]{ 3+x } -\sqrt [ 3 ]{ 3-x } \quad =\quad \sqrt [ 6 ]{ 9-x^{ 2 } }

$\sqrt [ 3 ]{ 3+x } -\sqrt [ 3 ]{ 3-x } \quad =\quad \sqrt [ 6 ]{ 9-x^{ 2 } } $

@sunflower I tried already this but I didn't notice anything helpful?

I guess it isn't representable by hom(2,A); that would be the contravariant case.

err, right. so when A->P(A) is covariant how can it make sense to say it is naturally iso to a contravariant functor?

A->P(A) is being considered covariant. A->hom(2,A) is contravariant. right?

derp I am all over the place today

:salmonfacepalm:

@anon You have 21?

I'm a member for 9 months! The baby is born! Yay!

@sunflower how did u do that?

@robjohn I guess I've lost some rep - for unkown reason: I had like 1700 rep, now I have 1200, is there something wrong?

meh. math.stackexchange.com/questions/252425/… anyone have the SLIGHTEST idea? Its rushing down to thread hell again

@CBenni five of the comments are because of you! anyway, let me look at it.

well yes, 5 comments were written by me, I dont like double/triple posts, but not being able to edit after 5 minutes was the reason I did. I could delete the old ones if you like.

@GustavoBandeira 500? let me see if I can see anything

someone help to solve this equation $\sqrt [ 3 ]{ 3+x } -\sqrt [ 3 ]{ 3-x } \quad =\quad \sqrt [ 6 ]{ 9-x^{ 2 } } $

@GustavoBandeira I don't see anything strange. The last recalc in your history was on Nov 9 and adding the rep changes since then, I get 1201

@pourjour let $u=\sqrt[6]{3+x}$ and $v=\sqrt[6]{3-x}$. then u^2-v^2=uv.

@pourjour from that you can compute the possible values of $u/v$

@pourjour raise that to the sixth power to get the ratio $\frac{3+x}{3-x}$ and solve

@robjohn did u notice that the left hand set is compounded of radical of 3th order and the right hand set is radical of 6th order?

@pourjour and does that not match what I said?

@robjohn the reality I didn't understand where did u get this relation u^2-v^2=uv.

@pourjour look at the substitutions

@robjohn Rooooob

Could you look at the messages of mine after this?

@pourjour I get $x=\frac6{\sqrt5}$

@robjohn can u please show me the full steps?

@robjohn can u please show me the full steps?

@robjohn can u please show me the full steps?

@pourjour have a bit of patience. Look at the steps I showed you (which are pretty much all there is except for arithmetic)

@pourjour I am looking into something for Peter that he asked first

Hi all

@PeterTamaroff I have a method of showing that $f_n$ converge uniformly. What have you tried?

@robjohn Did you read what I wrote?

@robjohn Maybe you can hint me,.

@PeterTamaroff consider the sup of the $f_n'$ (they are uniformly bounded, since they converge uniformly)

@pour assalaamu alaykum - are you still struggling with the equation?

@pourjour what do you get for $u/v$?

@robjohn What does uniformly bounded mean?

@PeterTamaroff You can find an $M$ so the $|f_n'|\le M$

@robjohn Oh, OK.

@robjohn Did you read what I wrote above?

@PeterTamaroff where?

@PeterTamaroff that is sort of right. Can you make it rigorous?

@robjohn I'm thinking that for a given $x$, in each nbhd of $x$ we can be sure that $g$ is "locally" constant... Yes, I'm trying.

@robjohn Now, wouldn't getting a finite cover help here?

SO that I get some nbhds and take some mins and BAM!

@PeterTamaroff Say I give you an $\epsilon>0$, can you find an $N$ so that for $n\ge N$, we have $|f_n(x)-f(x)|<\epsilon$ for $x\in[a,b]$?

@robjohn For $x$ in a certain nbhd yes... I think.

@robjohn (You mean "for all $x$" there?)

@PeterTamaroff Say for $x\in[0,1]$

@PeterTamaroff I don't think we can do this for all $x\in\mathbb{R}$.

@robjohn Right,

@robjohn I meant $x\in[a,b]$

@robjohn =P

DIdn't see the edit,.

OK

I'm trying to get one.

@PeterTamaroff I just made it. One of the advantages to being a mod :-)

@robjohn You sneaky mod!

@PeterTamaroff I cited it so that you would notice it.

@robjohn Wait.

What I'm claiming is that $|f(x)-f_n(x)|<C_n\pm \epsilon$

@PeterTamaroff Divide $[a,b]$ into segments no longer than $\frac{\epsilon}{M}$ long

@robjohn Where $M$ is $\sup_{n\in \Bbb N}f_n^\prime(x)$?

@PeterTamaroff yes that is the uniform boundedness of the derivatives that I was mentioning before

@robjohn But then... should this sup range through the $x$ and the $n$?

@PeterTamaroff that means all $x,n$

@PeterTamaroff that is what uniform boundedness means

@robjohn Because of uniform convergence $|f_n^\prime(x)-f^\prime(x)|<1$

@PeterTamaroff yes

For some $N$, $n>N$

OK

But why do you choose this $M$?

@robjohn Why?

@PeterTamaroff I don't choose it, it exists

@robjohn I mean, what are you up to?

Really

@PeterTamaroff So, you've divided $[a,b]$ into segments no longer than $\epsilon/M$, let's call the inter-segment points $\{x_k\}$ where the intervals are $[x_{k-1},x_k]$

@robjohn Very well then