Mathematics - 2013-04-01 (page 1 of 5)

12:34 AM

Why is mathematics so developed?

I mean You need a lot of time to get to research level mathematics

user19161

@JorgeFernández Because it happened over thousands of years.

I also have a theory that mathematicians are in general extremely good at what they do

user19161

@JorgeFernández There are many bad ones who can still earn a living.

yet abel died because he couldn't get cash

user19161

Times have changed.

user19161

12:44 AM

In the past they eat bread, now they eat pizza.

perelman used to eat cheese and bread

user19161

He is crazy.

lol

hes very good at math though

user19161

Yes. It takes a crazy man to solve a millennium problem. The non-crazy ones can only win medals.

there are some cool medals though

user19161

12:53 AM

Yes, I have some chocolate medals in the fridge, lol.

lies

do you really?

user19161

No, just kidding. But I have eaten them before.

me too, but the chocolate isn't that good in those medals

Talking of unusual mathematicians, did anyone else see the rumors about Grothendieck

user19161

1:11 AM

@AlexJBest Is he still alive?

@AlexJBest Nope. What happened?

@JasperLoy Yes, and apparently someone claiming to be him phoned the IHES and asked for a book to be posted to him. They normally don't post books, but asked the director who agreed. I hope it was a mathematical book and he's still doing maths!

BTW, games of thrones is premiering worldwide right now

math.columbia.edu/~woit/wordpress

Whoa, link?

user19161

@AlexJBest Ah, maybe I am the one who phoned, lol.

1:14 AM

@JasperLoy Maybe, but he must have convinced the director, so it seems relatively likely it was him.

user19161

@AlexJBest Some people are good at lying. Various people on this site have told me many lies.

Damn it! I got only one upvote today...

@JasperLoy True, nevertheless I think it reasonably plausible, or maybe it just a hope that he's been working on something amazing all these years.

user19161

@AlexJBest He is probably working on his mental health.

@JasperLoy The last I heard he was writing a huge book about evil :/

user19161

1:18 AM

@AlexJBest Haha, well this world is full of stupidity and evil.

user19161

@PeterTamaroff It seems you have a few now.

user19161

@PeterTamaroff Good night Pedro! Happy April Fool!

1:37 AM

@JasperLoy When is April Fools'?

Oh, tomorrow.

But it is tomorrow already where you are.

2:05 AM

@PeterTamaroff HMM

2:23 AM

27^27=443 undecillion O.o

anorton

2:55 AM

I have 3K! I can now place close/open votes! :)

congrats :)

anorton

thanks

and... now to bed. DiffEq test tomorrow.

'night all.

3:25 AM

Hah! math.stackexchange.com/questions/347807/…

@MarianoSuárez-Alvarez Hey.

hello ;-)

I drop by for a peek and get called an asinine pedant

I should call it a day

I love the adjective

@MarianoSuárez-Alvarez hahaha let me see

@MarianoSuárez-Alvarez didn't know that word.

@MarianoSuárez-Alvarez I was going to ask you something but now I've forgotten!

would anyone help me with normal series groups please

ask the question

3:30 AM

I can't get my definitions straight

asking for help on a statement that you have not given is not something that is often met with success

who would venture a «I will!» not knowing what you are going to ask?

only a fool

A series for a group $G$ is one of these $1 = G_0 \le G_1\le G_2 \le \ldots \le G_n = G.$

and you do not want a fool's help!

and a normal series is when $1 = G_0 \unlhd G_1\unlhd G_2 \unlhd \ldots \unlhd G_n = G.$

@MarianoSuárez-Alvarez Mariano, now I remember.

3:32 AM

And the factors of a normal series are $G_{i}/G_{i-1}$

but what about $G_i \unlhd G$?

does that ever have to hold

and a composition series is a normal series that can't be refined

I will make it short, since I have to wake up early tomorrow (I agreed to give a tennis class at 10 am, I am an asinine)

not now, though

I am in the middle of leaving :-)

@MarianoSuárez-Alvarez, te rason I ask is because I didnt want to type it al out to be ignored

@caveman no, normality is not transitive (however, add characteristic into the mix...)

I know that

3:35 AM

seems Yoda got into mathematics

@caveman It is a matter of convention, as far as I can tell.

"uniformly converges.... hmhmmm"

what is?

@caveman well, at least from me you are never ever going to get a positive answer that way

@anon I know a very nice exercise that shows that

3:35 AM

@caveman Deciding whether the term normal series means each $G_i$ is normal in $G$.

In the years I have been online in math fora, i have never ever answered «yes, I know about subjct X» ;-)

I have answered, though, many many questions!

@MarianoSuárez-Alvarez OK.

@user1, I amcouldn't get it straight, I think he gave that when lecturing but then later it couldn't be assumed and stuff I get confused

@caveman If he uses the term subnormal series as well, then you can probably assume they are all normal subgroups of $G$ when he uses normal series.

I should write out all the theorems I have about series and then figure out the definition from the proofs

Will Jagy

3:42 AM

@MarianoSuárez-Alvarez, were you satisfied with the octonions answer? I couldn't tell.

amWhy

@caveman Hey, caveman! how goes it? Good to see you!

hi

good to see you too

@WillJagy, I knew the content of the answer. I tend to think the answer is there is no answer: Marty sort of agreed in his last comment.

@anon all cyclic groups are abelian because addition is comunative right

Will Jagy

@MarianoSuárez-Alvarez, alrighty then. Marty works near enough to me, and is evidently close with Pete Clark.. There is this question here that i think, in dimensions 5,6,7, comes from integral octonions but I never finished it: math.stackexchange.com/questions/261204/… although I showed it false in dimension 9, true in 2,3,4,8.

3:52 AM

@Ethan Yes.

$a^ma^n=a^{m+n}=a^{n+m}=a^na^m$

@Ethan Woah! No.

no

but they all have order

@Ethan What makes you think that?

they all have finite order?

wait let me check my definition of cyclic

@Ethan There is but one cyclic group of each order.

(Up to isomoprhism.)

oh, lol, nvm I got my definitions messed up

3:54 AM

The only (UTI) cyclic group of infinite order is $\Bbb Z$.

Will Jagy

night.

@PeterTamaroff UTI?

Good night Will!

@anon up to...

@amWhy whats up

3:55 AM

oh. to most it means urinary tract infection.

@anon HAHAHHAHAHAHA

LOL (for real)

amWhy

@caveman Not much, lots down... How about you?

Arkamis

That's why we pee after establishing isomorphisms.

I'm trying to get back to studying, took a few days off

amWhy

@caveman We ALL need time off! Good for you.

3:57 AM

What's this "CHAT WITH AN EXPERT" thing that keeps popping up?

"connecting you to an expert..."

oh my oh my ohmy

Arkamis

@PeterTamaroff Terence Tao is taking questions

Sorry, no expert responded. Awwww... bummer.

if anyone wants to discuss series

@caveman Analysis or Group Theory?

Ahh, I should go to sleep.

group theory

3:58 AM

@caveman Don't know about those.

cant find a pencil sharpner anywhere

Hi @caveman How are you?

hi

@caveman Use a cutter, be a man!

thats my only option

4:08 AM

...or switch to a pen.

NEVER

Not even temporarily?

(to get the job done)

No! Work in your head rather than using a pen.

You should be constantly working in your head, no?

Arkamis

I constantly work in other people's heads. I'm a brain doctor.

4:13 AM

:-)

4:31 AM

Asinine pedant *and* petty tyrant...
I am on a roll tonight.

@AndrewSalmon are you interested in series in group theory

Andrew Salmon

@caveman series?

Sure

chat.stackexchange.com/rooms/8154/series-of-groups

I think these are important

in group theory

Andrew Salmon

I have not studied much group theory. But I'm looking at it now.

I saw it first in galois theory

where solvable groups are related to solvability by radicals

I don't see how $[G_i,G] \le G_{i-1}$ implies $G_{i-1} \unlhd G$

where $G_i,G_{i-1} \le G$

i found one way to do it but don't know how to write it nicely, what do you think?

we want to show that for all g, $g G_{i-1} = G_{i-1} g$

that's the same as showing $g^{-1} G_{i-1} g \le G_{i-1}$

we can just prove that for all $h^{-1} \in G_{i-1}$, $g^{-1} h^{-1} g \in G_{i-1}$

which is the same as showing $[g,h] \in G_{i-1}h = G_{i-1}$

which is immediate from the assumption since $G_{i-1} \le G_i$. $\square$

but how can I make that quicker and better

@MarianoSuárez-Alvarez, can you help

pls

I don't even understand why it's $[G_i,G] \le G_{i-1}$ rather than $[G_i,G] \le G_{i}$

oh wait I got it

what's the relation between nilpotent and solvable?

nilpotent groups are solvable

why do conjugacy classes have sizes powers of p?

in a p-group

is it because they're conjugate to a subgroup? how do we know that?

5:16 AM

they are not conjugate to subgroups

they generally do not contain 1

@caveman It is an application of the orbit-stabilizer theorem, when considering the action by conjugation of the p-group.

@user1, ah! the size of a conjugacy class would be equal to $|G|/|P_g|$ then!

thanks

so how do I show a P group has nontrivial center? if I assume |Z| = 1 for contradiction, I have |G| = |Z| + sum over conjugacy classes, I gather that there must be kp-1 size 1 conjugacy classes

Again, what is the size of a nontrivial orbit?

Then, reduce mod p.

a power of p

I don't see what you mean

1 is a power of p

oh wait

if a conjugacy class had size 1 it would be in the center

Yes, all nontrivial orbits have sizes divisible by p. :)

5:22 AM

great

now I can show p-groups nilpotent by induction

How do I go from $$\frac{P_i/Z}{P_{i-1}/Z} \le Z\left(\frac{P/Z}{P_{i-1}/Z} \right)$$ to $$P_i/P_{i-1} \le Z\left(P/P_{i-1} \right)$$ though?

mathworld.wolfram.com/FourthGroupIsomorphismTheorem.html doesn't say anything about center

user19161

6:22 AM

@anon Are you going to change your name and pic today like last year?

I'm pretty bogged down with catching up on work I didn't do over spring break. not sure I can fit in anything fun.

user19161

I just got the Functions Badge.

user19161

@anon Ah, must be too trivial for you.

user0 sounds nice.

user19161

I now have 6 red badges, yay!

user19161

6:26 AM

@user1 Yes, I prefer naturals to start from 0.

user19161

The word annoying is so annoying.

user19161

Ah, I see that the backstabber has no Sportsmanship Badge, lol.

user19161

@κρανίοπεριπολία Did you chat with the bot?

user19161

@PeterTamaroff Didn't you read meta?

@JasperLoy No, have you?

user19161

6:40 AM

@κρανίοπεριπολία Yes, it was fun!

user19161

The great anon is busy catching up, let's not ping him, shh...

@JasperLoy They should send some bots into chat.

user19161

@κρανίοπεριπολία We are the bots!

:-)

do bots initiate conversation or always respond? someone should make two of them talk to each other.

sos440

6:54 AM

italic bold link

... asynchronous javascript formatting test complete.

I don't understand this en.wikipedia.org/wiki/Upper_central_series#Upper_central_series

How does this definition work? $Z_0(G) = G$, $Z_{i+1}(G)$ is unique such that $$\frac{Z_{i+1}(G)}{Z_i(G)} = Z\left( \frac{G}{Z_i(G)} \right)$$

a subgroup $Z(G/Z_i)$ (the center) of $G/Z_i$ by lattice theorem corresponds to some $Z_{i+1}\le G$ s.t. $Z_{i+1}/Z_i=Z(G/Z_i)$

@anon can you look at my previous question please chat.stackexchange.com/transcript/message/8759507#8759507

think of it as an ascending onion. the core layer is the group's center, and each successive layer is the union of the cosets in the center of G mod the previous layer

I see the onion but Idon't know what the theorem you use is

My lattice isomorphism doesn't have anything about centers

Would it just be that if N is normal in A and B with A = Z(B) then A/N = Z(B/N)

7:09 AM

the lattice theorem applies to all subgroups of a quotient, and the center of the quotient G/Z_i is in fact a subgroup of quotient.

I don't understand

which part?

oh wait I think I see it

it's just that if X is a subgroup of Z(Y), then X/N is a subgroup of Z(Y)/N = Z(Y/N)

is that right?

yes

thanks a lot!

I'm not sure how that's related. You're talking about things inside Z(Y), but the point of this sequence is you get ever-larger subgroups building out from the center.

what? :( I thought that just resolved everything

7:12 AM

oh, you're talking about the earlier question

I feel lost

at any rate, to show finite p-groups are nilpotent by induction, use the following two lemmas: (a) if G/Z(G) is nilpotent, then G is nilpotent, and (b) finite p-groups have nontrivial centers.

I.e. assuming all p-groups of order <p^n are nilpotent, and |G|=p^n, then |G/Z(G)|<p^n so G/Z(G) is nilpotent so G is nilpotent.

hardly any need to mess with iterated quotients and centers and mumbo jumbo

wait a second this looks better than the proof I get lectured

I cannot think of any other proof by induction.

the proof I have doesn't use the lemma "G/Z(G) is nilpotent so G is nilpotent", instead I take the abelian central series for P/Z(P) and then do this chat.stackexchange.com/transcript/message/8759507#8759507 (which I still don't know how to do) to get a central series for P

thisis so weird

ok that is really nice thanks

now ill try to understand the Z_i again

7:21 AM

@caveman I am unable to distinguish this from a proof of (a).

oh...

$Z_0(G) = 1$

that makes more sense

7:54 AM

@WhitAngl Have you found out anything about 5pm yet?

9:02 AM

Where's Old John been?

9:39 AM

@pen Hanging around under an assumed name :)

@anon I once read a transcription of a "conversation" between two bots ... it was pretty dull, and full of fairly irrelevant responses., but quite funny

I once heard an audio recording of two phone sex hotline girls connected to each other. It was hilarious when they figured it out at the end.

2

bot v bot

@CтарыйДжон How did you even figure I was talking about you? :-P

@anon I can imagine :)

@pen I break cover sometimes, and reveal who I really am :)

The part when the expert says "O... kay..." and "Truth." is pretty annoying

9:58 AM

Must go buy food - later

Good morning!

@κρανίοπεριπολία knock knock knock

@pen hi young man

How are you @anon ?

@Charlie Hey! :-)

forcibly awake

@pen wassup?

@anon oh!

10:27 AM

@anon When younger me and my friends would have group skype chats, and call two freephone numbers at once then listen to them argue about them pranking each other, that was pretty hilarious.

too easy

huhu

when i reread my answer here i guess this would count as a april fool too

10:42 AM

@Charlie Who's there?

bananinos :D

bananinos who?

@DominicMichaelis Not sure if you are serious or it shall be an april fool

ahaha

I don't know if i can take my own answer serious at all :)

I think it as serious as it can get when you're in the context of geometric intuition.

10:52 AM

@κρανίοπεριπολία Chucky

i think when you have improper integrals geometric intuition should be avoided

@κρανίοπεριπολία if you make any prank, mr. Patrol..... you will see my darkside...

@GitGud i mean from the very first intuition one may think that every non trivial series diverges

@DominicMichaelis I suppose that's natural, at least to me there was no way I could have conceived an infinite sum of positive numbers to be anything less than infinity.

Even after seeing it happen (given the usual definitions, of course), I felt that something wasn't right, that maybe the definitions as they stood didn't really mirror what one would expect.

same here, although something like 1/3 can be written as an infinite sum of it's digits

11:00 AM

Only after some training and thinking about Zeno's paradox my intuition got good.

I frown upon anything related to decimal representation of real numbers. I've never seen it treated rigorously.

i don't know sometimes though digit representations helped me, for example at the cantor set and devils staircase, even though it really took me some time

Even though I have seen it treated rigorously, I do not think that matters. It gives intuition for why an infinite sum of positive values can converge.

@DominicMichaelis Of course it helps, I just really dislike using it in a reasoning which is supposed to be rigorous.

@user1 It matters to me because I'm not able to translate a mathematical proof using infinite representation of a real number into a formal proof.

@GitGud It is extremely easy to bound a number in its decimal expansion.

mh why would someone this question?

11:05 AM

Both below (truncation) and above (increment a digit and truncate).

@user1 Are we addressing the same issue? I have no problems with series converging.

@GitGud The point I was making: "It gives intuition for why an infinite sum of positive values can converge."

@user1 Ah, OK.

@DominicMichaelis Why would someone what?

oh downvote

I really dislike getting downvotes without comments

@GitGud But I am curious, what kind of proofs do you have in mind, which give you difficulties "translating" from their current form involving decimal expansions?

11:13 AM

Sometimes it's just misclicks. Often times when I see a weird downvote I ask why and within the next minute it's changed to an upvote.

@user1 Cantor's diagonal argument. The problem is not the proof itself, but rather what meaning to give to the expression $(0.a_1a_2a_3\ldots)$.

I simply don't know how would I read that if I was devoided of decimal representation intuition.

I gotta go in 5 minutes. I'm getting dressed. Just letting you know so we don't get into anything too deep.

you can go to $2^\mathbb{N}$ from the decimal expansion i guess

@GitGud OK. I think I agree with you on this point.

See you later.

see you

hi

11:28 AM

hi

@Charlie Is that a threat or a promise?

:-)

11:45 AM

Don't know if I enjoy waking up early, or it is just the coffee...

:-)

@κρανίοπεριπολία That is most definitively not for Mat, sorry.

I don't even see him on chat,.

11:59 AM

@anon Morning.

ugh

@Expert Oh?

also, refresh your tab

@Expert Let me see.

oh, you manually typed anon, nevermind

12:01 PM

@Expert HAHAHAH

What the fuck?

Check meta.

@Expert Will people be knocking on your interwebs asking for official help now?

@Expert This?

@Expert Man, no expert responded to me yet.

open up a tab on an mse question and leave it be for awhile

My internets is working like horseshit this morning.

O... kay...

12:06 PM

@Expert No expert yet.

And you're trying to contribute to this?

Of course. There are a pool of selected users the system connects everyone else to, and these are the "experts," but mostly it's just bots. Also I changed my gravatar and name to be more realistic.

"Sorry, no expert responded."

Bleh.

Really? hmm.

I caught myself an "expert" yeee haaw

@Expert The bots are afraid of me.

12:11 PM

what do you want me to ask?

You: are you female

Expert: Would it make a difference?

You: yes

Expert: Noted.

You: what did you note?

Expert: The answer is pretty complicated.

I destroyed my expert with 2 sentences :(

When you put cleverbot and expertbot in a room together they sure want to end the conversation asap.

@DominicMichaelis what were the sentences?

Don't know them verbatim but it should be something like "What are you thinking about?"

got the answer "What would you make with the information?"

I said "I guess selling it"

and got the answer "Why would you guess selling it?"

You: can i swear at you?

Expert: I may can you swear at me if you can me swear at me.

12:21 PM

wtf?

You: wtf?

Expert: I don't understand.

than you are not a expert. q.e.d. puta

@DominicMichaelis "puta" means whore in spanish!

@anon This.

oh sry

i just know te quiro puta from rammstein :D

12:36 PM

@DominicMichaelis "Te quiero, puta" means "I love you, whore."

oh :D lol look at WA wolframalpha.com/input/?i=int+1%2F%281%2B1%2Fx%5E5%29

Man, all this contour stuff in complex analysis (like choosing it, deforming it, making it circle up stuff) feels like complete wizardry to me!

@DominicMichaelis How awful! My eyes! Comic Sans!

yeah complex analysis is like magic

@PeterTamaroff I feel exactly the same way

about compelx analysis that is

A lot of the little set theory that I know feels that way.

12:49 PM

just a lil bounded and then constant and i can show everything D :

@DominicMichaelis well, at least I somewhat understand the proof of Liouville's theorem. Picard is still a complete mystery to me

the large one that if $z_0$ is an essential pole of $f$ for every $\varepsilon>0$ $f(U_{\varepsilon}(z_0)\setminus \{z_0\} ) \supseteq\mathbb{C}\setminus\{ x\}$

by Picard I mean "the image of an entire function is either $\mathbb{C}$ or $\mathbb{C}\setminus \{x\}$ for some $x$"

now i got it :D

it just seems to mysterious that it is possible to be missing one point but no more

12:57 PM

@Tobias isn't that a consequence of lioville ?