@AlexanderGruber Did you get to "The Fix ♫♪"?

@PeterTamaroff which time was that?

36:00

i'll skip forward

When he presents Richard Hawley.

It is interesting how seemingly complex questions have easy solutions-

@PeterTamaroff that is pretty nice.

@PeterTamaroff also, that is pretty nice.

@AlexanderGruber Do you listen to music when you study/work?

@PeterTamaroff i am rarely not listening to music.

Hello everyone

@EduardoAlanDávalosPeña Everyone has left. Would you want to leave a message?

Hi @Eduardo.

@EduardoAlanDávalosPeña Here he is

what are the dicyclic groups about?

Hi @AlexanderGruber can someone help me to prove a proposition?

what proposition?

@EduardoAlanDávalosPeña i can try.

@EduardoAlanDávalosPeña Sure. What is it?

@caveman they're semidirect products that work kind of like dihedral groups.

this sounds interesting

It is about logic, is simple: prove that a proper initial segment of a sentence is not a sentence.

they're a pretty good class of nonabelian groups to know. did you see them on that one post i made?

@EduardoAlanDávalosPeña whew, i'm not a logician, i don't know anything about that. maybe somebody else can help you. if not, i'll try to help you, but you'll have to give the definitions of all that from the ground up.

I think this can be done with induction, but I don´t know how to start

@EduardoAlanDávalosPeña You can just post it on main.

this is so nice en.wikipedia.org/wiki/Binary_polyhedral_group

@AlexanderGruber, I think I did

@AlexanderGruber, oh no I didn't, this is great!!

@caveman i am still trying to figure out the answer to that one guy's comment. the "make your own" section makes it really tough

oh that's an evil question!

i don't even know if i can answer it, lol.

what is a double or triple cover?

@EduardoAlanDávalosPeña, can be used as.____?

@EduardoAlanDávalosPeña, oh it's because you use prefix notation

that no initial segment is another one

@EduardoAlanDávalosPeña, one way to prove it is to have a counter function travelling along the string that gets to zero only when it reaches the end

@EduardoAlanDávalosPeña You're interested in logic?

@PeterTamaroff we can say yes more or less, I intersted in logic more for foundational stuff

@EduardoAlanDávalosPeña Foundational?

@caveman a double (triple) cover of $G$ is a central extension of $C_2$ ($C_3$) by $G$.

@PeterTamaroff for make rigorous arguments say in topology or group theory

clearly it doesn´t need that but i feel incomplete with out learn logic

thanks

@EduardoAlanDávalosPeña You needn't worry about logic now, really. It is something pretty specific. You'll never see someone trying to prove some theorem in topology using first order logic.

@EduardoAlanDávalosPeña, e.g. "not and x or y z" -> "1 not 1 and 2 x 1 or 2 y 1 z 0" the number is how many things you need left

lemma: never 0 except right at the end, corollary: no proper initial segment is a sentence

@caveman it may be that the extension should be strict, but i'm not sure.

@caveman what do you mean whith a counter function in your first comment

@EduardoAlanDávalosPeña, take your sentence and add n-1 for each n-arity operator you see as you walk through the string

variables are countable as 0 arity

Hola

@EduardoAlanDávalosPeña, maybe you should post is as a question I will write a more clear answer

hi @TobiasKildetoft!

@EduardoAlanDávalosPeña Sure, you'll use $\neg$; $\forall$; $\exists$, $\exists !$; $\vee$; $\wedge$, but that's about it.

@caveman hi

@Caveman thanks a lot! I will post it

so are the non-abelian p^3 groups always dihedral or dicyclic?

that can't be right

@Haytham hola ¿De donde eres?

@caveman no, they can also be heisenburg groups, quaternions

@EduardoAlanDávalosPeña soy americano hehehe

and i think i'm forgetting one

@caveman what does dicyclic mean again?

@Haytham ¿Porque aprendiste español?

@TobiasKildetoft, I think it's the group genrated by the quaternions $a = e^{i\pi/n}, x = j. $

oh right there's this other $\Gamma$-group of it

but there is also presentation $$\langle a,x \mid a^{2n}=e, x^2 = a^n, xax^{-1} = a^{-1} \rangle $$

there are two non-abelian groups of order $p^3$ for any prime p

if $p = 2$ these are the dihedral and the quaternion group

@EduardoAlanDávalosPeña para saber otra lengua

if $p$ is odd, then one has an element of order $p^2$ and the other has all elements of order $p$

@Haytham es una buena lengua jejeje mi querido castellano

@TobiasKildetoft, I did manage to get that for p=2, but my proof is very bare hands

and doesn't seem to generalize for p>2

then you've got the heisenberg group ($3\times 3$ unitriangular matrix group over $\mathbb{F}_p$) and this other weird one which is a $\Gamma$ group, consists of matrices $\left(\begin{array}{cc}pa+1&b\\0&1\end{array}\right)$ where $a,b\in \mathbb{F}_{p^2}$

@EduardoAlanDávalosPeña diferente!

@Haytham ¿A que te refieres? :D

@EduardoAlanDávalosPeña diferente de lo inglés

@Haytham Diferente del Inglés.

@caveman you make those without any of that dark magic though. math.stackexchange.com/a/279102/12952

@Haytham si es ciertamente diferente

@PeterTamaroff tu si eres español o hispanoamericano

?

@EduardoAlanDávalosPeña Argentina.

@PeterTamaroff es un lindo país!

@EduardoAlanDávalosPeña Supongo que si.

I only need to show $[a,b] \in Z$ to see that it generates the center.. but why is it ?

@caveman why do you only need to show that or why is it in there?

why is it in there

let $g \in G$ then I think that $a,b,z$ generate $G$, so need to show that $ababzbzba[a,b] = [a,b]ababzbzba$

@caveman if the group is not abelian, the derived subgroup must be equal to the center

@PeterTamaroff verdaderamente piensas que no es debido o importante preocuparse por el formalismo de la lógica.

?

(since the order of the group is $p^3$)

because $G'\leqslant Z(G)$ b/c $G/Z(G)$ has order $p^2$.

that feels like something I'd never come up with :(

@caveman how are you today?

it is also good to know that any normal subgroup of a p-group intersects non-trivially with the center

@EduardoAlanDávalosPeña Sólo necesitas lo basico y casi obvio, por ejemplo lo que dice "Provable identities" bajo "Deducitive systems".

Es decir, estar familiarizado con la lógica basica.

@caveman workin' with $p$-groups all the time makes that sort of thing easier to come up with

Pero luego dedicarse a el estudio realmente formal de la logica es una materia en si.

@amWhy Amy.

that's really cool

I heard this guy classified the p-groups up to something like p^10

it must get pretty wild

@PeterTamaroff Hi, Pedro!

@caveman that doesn't sound right

given that there are about 50 billion groups of order $2^{10}$

@amWhy I'm trying to explain to @EduardoAlanDávalosPeña he needn't study hard logic to make rigorous statements in mathematics.

it says on wikipedia:

> the sheer number of families of such groups grows so quickly that further classifications along these lines are judged difficult for the human mind to comprehend

"[to] make rigorous arguments say in topology or group theory, clearly it doesn´t need that but i feel incomplete with out learn logic"

@amWhy

@PeterTamaroff, I disagree with that - you need to master to logic to properly understand proofs etc.

@caveman not formal logic

@caveman I didn't say that. I meant formal / hard logic.

@PeterTamaroff Well, it helps, a lot! I can't say, because in my case, I studied both. So I don't know how well I'd do in math if I didn't also study logic. I find logic indispensable to really understanding the logic of proofs, at least it has helped me.

@amWhy Yes, of course. What I meant is stuff like this

@PeterTamaroff do you see? I´m like amWhy, maybe others don´t worry, but I do.

At the very least, one should have a semester of symbolic/formal logic as an undergrad.

@EduardoAlanDávalosPeña, did you ask your question?

@amWhy OK, but Eduardo is 15.

He has a lot of time to get into symbolic/formal logic.

@caveman oops not yet

@PeterTamaroff I think symbolic, formal logic, first-order (propositional logic, predicate logic, and quantifiers) is crucial, not necessarily advanced math logic. I really think students in high school/secondary school should encounter such basics.

@amWhy I agree, the basics.

Say, this thing you're dealing with now @amWhy

@PeterTamaroff Exactly: but also predicates, and quantifiers...

@amWhy Still agreeing. =)

@PeterTamaroff If you like the answer you can upvote it ;-) (double wink!)

How can anyone understand a math proof without knowing how to work with, e.g. $\forall x...$ and $\exists x...$ etc.?

@PeterTamaroff But I see you're agreeing with that...;-)

@amWhy What's your age?

@PeterTamaroff That's a secret! ;-)

Buuuuuuuu!

Near $30$.

she's 16

@caveman Sure...

@caveman In two months! ;-)

haha

@PeterTamaroff you don't ask a woman's age!

@Haytham That's sexist!

@PeterTamaroff that's polite

bye

bye

@caveman oh wow. i had no idea he did a whole bunch of those expository papers. math.uconn.edu/~kconrad/blurbs

Cool

i really like this one on schur-zassenhaus

LAWL

@PeterTamaroff Wow!!!

@PeterTamaroff do you think the guy who accepted understood?

@Haytham I guess not. But it is damn funny.

@PeterTamaroff It's hilarious

so cool!

Well, I gotta go! Bye!

@caveman here is my question math.stackexchange.com/questions/329911/…

There are someone here....

?

Need help because maplesoft help pages are offline. How does one animate a power series?

what does that even mean, first of all?

@anon are you talking to me? I have to give a presentation in class and I wanted to animate the convergance of a power series, but I don't know how get that done in maple. I can graph the power series for only one n, but I wanted it to look like the gif on the wikipedia page for power series.

so, you want to animate convergence of the power series, specifically (which is a phrase that makes sense), and you can graph a partial sum of the power series. are you not able to view maplesoft.com/support/help/Maple/view.aspx?path=plots/animate?

@anon, yes I can't not load the page, it times out. Do you know of another source where I can find this information? Recently got maple 12, and I am learning how to use it and I thought it would make my presentation better to have this, plus, I learn how to use Maple.

@anon I didn't know where else to ask.

Are there are any set theorists around?

@MaoYiyi try it with a different browser, it's been up for me this entire time

@anon, Ok, I will try that. didn't know that mattered.

@MaoYiyi obviously the problem is not with the site but with something closer to your end

@anon You're correct, it works with ie but not chrome. odd.

@AlexanderGruber Do you know anything about the free profinite group?

is that the profinite completion of a free group?

yea

@anon I haven't studied it much and I needed it to have a property which I decided it had. Namely that the profinite free group on $\kappa$ generators contains the profinite free group on $\mu$ generators where $\mu,\kappa$ are infinite (actually uncountable) cardinals such that $\mu \leq \kappa$.

intuitively, the free profinite group $\widehat{F_\mu}$ for infinite $\mu$ should be the projective limit of all finite groups (with all possible surjective quotient maps), it's just a matter of cardinality to consider to distinguish them, so that seems to make sense

so for $\kappa\ge\mu$, the inverse system underlying $\widehat{F_\kappa}$ should have the exact same objects and morphisms as that underlying $\widehat{F_\mu}$, but with a whole lotta copy/paste business going on, so the latter should be able to sit inside the former

that's actually really cool that the free profinite group should be the projective limit of all finite groups

Okay, yeah. I couldn't think of an object off the top of my head where that property wouldn't hold.

That is neat. It makes sense since profinite groups are groups that are projective limits of their finite subgroups.

(finite quotients)

right

ah, that's an interesting galois theory question

Well, I think in general Galois theory won't be too useful

unless I can reduce to the case $k(t)$ somehow.

Otherwise in a lot of situations I think the absolute Galois group is poorly understood.

Hello

I have a question.

I want to prove that the intersection of any family of closed in the Zariski topology is closed

but

but what?

isn't that the definition of topology? arbitrary unions of open sets are open hence arbitrary intersections of closed sets are closed

nevermind

I read something wrong

I read that soemone wrote

that $V(I)\cap V(J)=V(I\cup J)$

but that is wrong

the correct statement should be $V(I)\cap V(J)=V(I+J)$

@KanyeWest no, $V(I)\cap V(J)=V(I\cup J)$ is correct, and $V(I)\cap V(J)=V(I+J)$ doesn't even make any sense (how do you add two sets of multivariable polynomials together? you don't - you take their union).

@DominicMichaelis right, but if one wants to define a putative topology one needs to check it satisfies the axioms, i.e. closed sets are closed under arbitrary intersection etc.

ah ok

so the question is more that if the topology is really a topology

yes

He added some step here he isn't serious or?

@JSchlather Nope! Too much "pro" for my taste.

my father wake me

@JasperLoy Does "godly powers" include access to email addresses?

I believe mods do have access to email addresses. However the rep-born demigods that are not official mods do not.

mh i hardly can sleep after getting waken

@jasper you can always ask for my email :D

so i will do some pancakes :D

where some is about for 8 persons pancakes

cause there is only something which is better than pancakes

more pancakes

Good, I won't have to delete my account then ;-)

or at least, set up a dummy email address.

@JasperLoy it never is!

@DominicMichaelis why did you father wake you up at this ungodly hour? ;-) (Big Ban Theory Imitation.)

do you know lindsey stirling

By the way, I don't think 7 in the morning is ungodly

@DominicMichaelis no, why?

I know now, thanks to google.

Lindsey Stirling

@DominicMichaelis nice.

Interesting...

i can't one of them

...takes practice.

I can dance alright I think. I know people can play guitar while dancing, never seen people dancing with violin.

3 kilo pancakes :d

Who is going to eat all that? ;-)

well someone has to do that job D:

:D :D

i think i am going to sacrifice myself for that one :)

3 kgs? I am not sure you could sacrifice if you wanted to, unless you are damn hungry.

Do you like waffles too?

...or crepes.

yeah

i have a sister and a firend is coming later

they will help if i am in need

That is cool I think then.

Do pancakes last long? (As in over the complete day?)

We have a some pancakes here that can last the complete day without going bad, but others, even though they do not go bad, become very hard and not pleasant to eat.

our pancakes are a bid like crepes

Yes, we have some pancakes made like crepes too. I see.

and at the moment i only make the batter

I like crepes :-D

and waffles.

@κρανίοπεριπολία I am not sure if you would get good Indian crepes at your place, but you schould try out "Masala Dosa" if you find it somewhere. It is one of my favorite dishes, or as @JasperLoy might call it Tosai.

@JayeshBadwaik I'll look for it, thanks :)

@JasperLoy You are killing me...

@JasperLoy Tit for tat, then ;-)

$$\frac{\pi^2}{3}-\frac{5}{2}=\sum_{\text{perfect powers p}}\frac{1}{p+1}=\frac{1}{2^2+1}+\frac{1}{2^3+1}+\frac{1}{3^2+1}+\frac{1}{2^4+1}‌​+\frac{1}{5^2+1}... $$

$\Huge\text{(}$removed$\Huge\text{)}$

Don Quijote here is still trying to disprove that if $X$ is compact and $f: X \to Y$ continuous then $f$ is uniformly continuous.

lol :D

maybe he should read the definition of compactness

@JasperLoy That's what I told him.

@JasperLoy to disprove ? !

But he won't listen.

@κρανίοπεριπολία $$\sum_{p\leq x}(-1)^{[\frac{x}{p}]}\sim (1-2\ln(2))\frac{x}{\ln(x)}$$

@DominicMichaelis I think he should sit down and try to first understand the theorem and thereafter its proof.

@JasperLoy Does it mean I get to ignore his comments?

@MattN. Yes! You should! He is not even the OP!

May be he just decide to troll some users to have some fun, who knows!

@JayeshBadwaik But that's no reason to ignore him I think.

jasper just that i make no mistakes you are jacob black right ?

I guess I'll let him think about it for a few days.

@MattN. Yes, that is not the only reason, but after all the comments it can be. If he was the OP on other hand, you would be obligated to answer all his doubts.

@JayeshBadwaik Sorry, I missed your joke :(

@JayeshBadwaik Haha. But I think this one is honestly trying to find a counterexample.

@JasperLoy McDonald's atleast in here generally makes everything shitty.

@κρανίοπεριπολία No, problem, I will write it again sometime.

@MattN. Yeah, let him work on it for a few days.

@JayeshBadwaik I guess I don't make this distinction.

I'll see you all later! : )

@MattN. okay. :-)

later.

later

see you later

@jasper great

charlie wrote me on facebook

or just a pit

a bottomless pit

...and you are at the bottom.

$\Huge\text{(}$removed$\Huge\text{)}$

Anyone know how I can center the "removed"?

why not making something like $\bigl(\text{removed}\bigr)$

why not just remove the comment

or $\Biggl( \text{removed} \Biggr)$

or $\Huge\text{(}$$\frac{removed}{}$$\Huge\text{)}$

@DominicMichaelis Thanks :)

those guys on tex.SE are so freaking good

$$\Biggl( \text{removed} \Biggr)$$

$$\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)$$

@DominicMichaelis yeah, the talk is that the best texperts hang out on Tex.SE but the best mathematicians hang out on MO, not MSE.

yeah i guess

everytime i read a question i always think wtf?!

and when i read the answers i think wtf?!

yeah pretty much like MO

$$\Biggl( \text{removed} \Biggr)$$ $$\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)$$ $$\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)$$ $$\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)\Biggl( \text{removed} \Biggr)$$

Ethan's triangle @Ethan

lolol

$$(a^4-2ab^3)^3+(a^3b+b^4)^3+(2a^3b-b^4)^3=(a^4+ab^3)^3 , \text{ Viete}$$

$$\sum_{n=1}^\infty\frac{\sigma_a(n)\sigma_b(n)}{n^s}=\frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}$$

$$\frac{\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}=\sum_{n=1}^\infty\sum_{k=1}^n\frac{\gcd(n,k)^s\zeta(s-b,\frac{k}{n})}{n^{2s-a-b}}$$

what is a perfect power ?

any integer power

2^2

3^3

5^7

4, 8, 9, 16, 25, 27, 32, 36, 49, 64

heres another

$\frac{\pi^2}{3}-\frac{5}{2}=\sum_{\text{perfect powers p}}\frac{1}{p+1}$

$$\frac{-\sin(2\pi x)}{\pi}=\sum_{n=1}^\infty \frac{\mu(n)}{n}{ { \{n x}\} }$$

$$\text{#number of perfect powers} \leq x = \sum_{n=2}^\infty\frac{\ln(x)^n}{\zeta(n)n!} + O(\ln(x))$$

I messed up, the last equality should be subtracted from x

batter finished now baking :D

what are you making

pancakes

what time is it in germany

8:51 a.m.

its 12;51 here

;?

india ?

oh no 12:51 is somehwere in russia isn't it ? about moskow

$$\sum_{n=0}^\infty \frac{2^n}{x^{2^n}+1}=\frac{1}{x-1}$$

$$\frac{1}{x-1}-\sum_{n=1}^\infty \frac{v_2(n)}{x^n}=\sum_{n=0}^\infty \frac{1}{x^{2^n}+1}$$

@DominicMichaelis he means 00h51 probably.

which would be properly in LA

yes

...or approximately any line of longitude through LA.

like portland, oregon?

Yep.

oh ok

wtf ? that is a weird comment math.stackexchange.com/questions/330115/…

Hi! What does mean notation $\mathbb{Z}_{2^{\infty}}$?

i guess it is not unique

but i think the p adic interpreation doesn't make sense

category theory belongs to algebra or ?

theory

;-)

sry i don't get it

There is nothing to "get", I'm just foolin' around with English.

it's sort of a hobby :-D

I picked-up from math...

ah you interpret it as a logical or ?

Logically.

oh now i get it :D

Hey

Hi*

hi

I am trying to find what the Jacobson Radical is for F[X]/(f)

where f is a non-constant polynomial

I was trying a few examples, and I think it is something along the lines of break f into irreducibles

f=g_1...g_k

and then this irreducibles would create maximal ideals in F[X]/(f)?

i.e. (g_i) would be a maximal ideal.

and then take the intersection of those?

is that correct?

$\Huge\text{Happy $\pi$ Day}$