Mathematics - 2013-03-27 (page 1 of 4)

In particular, if this happens $a\prec b$ for any pair $(a,b)\in \hat a\times \hat b$, so in some sense we produce disjoint sets of upper and lower bounds.

caveman

@PeterTamaroff, I've posted an answer that I do like

Everything else on hat page is far too strange and complicated

Peter Tamaroff

:8696930 Right.

Alexander Gruber

2:15 AM

ah yes i see. i think maybe i have seen this in physics in approximation theory.

Peter Tamaroff

@AlexanderGruber Yes, indeed. We produce a (partial) order on $A/\sim$.

caveman

@PeterTamaroff, hello

Peter Tamaroff

@caveman Aha?

caveman

were you just begging for upvotes or did you want to discus that question

I posted an answer for you

amWhy

@caveman Yeah...I saw the comment... not called for...

Peter Tamaroff

2:16 AM

@caveman I wanted to know if there was something that made my answer unupvotable.

caveman

@amWhy, thanks amy

@PeterTamaroff, it's needlessly complicated and strange

Peter Tamaroff

@caveman Really?

caveman

I have posted the true answer

Peter Tamaroff

Whatever suits you...

amWhy

@PeterTamaroff which question? I'm not getting any upvotes tonight, either... :-(

caveman

2:17 AM

@PeterTamaroff did you see it

Peter Tamaroff

@caveman Yeah.

caveman

@PeterTamaroff Do you understand the comment here math.stackexchange.com/questions/342168/…

Peter Tamaroff

@caveman Yep. I plead guilty. I had upvoted but then changed my mind so I edited to unvote =D

Scumbag me!

caveman

that's really annoying, you shouldn't do that

I've seen you make these pointless trivial edits before

Peter Tamaroff

@caveman Not really. I edited one of your answers cause it was lacking punctuation, for example.

amWhy

2:19 AM

upvote for you each... ;-)

Peter Tamaroff

So I actually made a pointless answer pointfull.

Alexander Gruber

what's wrong with pointless trivial edits? that's basically half of what i do here.

amWhy

@caveman Did you do that to one of my posts?

caveman

@amWhy, I don't think so

amWhy

oops, I meant @Peter

Peter Tamaroff

2:20 AM

I cannot believe people keep using "proof" and "prove" interchangeably.

caveman

the worst is derive instead of differentiate

I can't stand that

Peter Tamaroff

@caveman Well, but that is because in spanish we say "Derivar".

caveman

mm but that only excuses the spainish

Peter Tamaroff

That is, "Differentiate a function" is "Derivar una función".

@caveman Sure.

@amWhy Where?

@amWhy Do what? Minor edit to unvote? Nah, it's the first time I do it.

amWhy

Occasionally I've lost an upvote...

@PeterTamaroff good to know..I upvoted your answer, and caveman's too...

Alexander Gruber

2:23 AM

@amWhy is it from old posts? sometimes that happens to me when things get deleted

amWhy

@AlexanderGruber Ohhh, that might explain it. Yes, it's usually from old posts. Not often, but bothersome. As long as it's not personal, I guess I can live with it.

Alexander Gruber

hey, has anybody ever seen additive notation used for direct sums/products of abelian groups?

user19161

@PeterTamaroff I see that you just got 20k, congrats!

Alexander Gruber

like i'd write $(\mathbb{Z}_2)^3$ or $\mathbb{Z}_2^3$ instead of $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$

but maybe i should be writing $3\cdot\mathbb{Z}_2$ or something

caveman

that's crazy

user19161

2:30 AM

@julien I see that you now have 15k, congrats!

Peter Tamaroff

@AlexanderGruber Hmm, curious.

caveman

@amWhy how was your day

Charlie

@jasper :D

user19161

Hmm, maybe I should delete my account soon.

amWhy

@caveman so-so, again. @Jasper :^D

Charlie

2:35 AM

@JasperLoy NOOOOO

user19161

Why is everyone smiling at me today?

caveman

$$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{itx} \, dt$$

I dont get this

This isn't valid for Lebesgue integral is it?

Jayesh Badwaik

@JasperLoy YES!!!!

user19161

@JayeshBadwaik I said maybe, so I was serious.

user19161

I have become rather unhappy about a number of things on this site.

amWhy

2:38 AM

@JasperLoy :-?

@JasperLoy Like what?

user19161

@amWhy That is my secret.

Jayesh Badwaik

@JasperLoy Ahh. Sorry then.

user19161

By the way guys, I just found the best lecture notes on complex analysis online.

amWhy

@JasperLoy send me an email...Have you read my email, tonight? ;-)

Alexander Gruber

@JasperLoy is it that rep-crack? are you losing control of your life?

user19161

2:40 AM

Check out the notes by Michael E Taylor and Wilhelm Schlag on complex analysis, they cover riemann surfaces as well.

user19161

@AlexanderGruber I lost control of it the day I was born. That was a very sad day...

user19161

@PeterTamaroff I totally love that picture Pedro.

amWhy

@AlexanderGruber who or what is "that rep-crack"

Peter Tamaroff

@JasperLoy Share, share!

user19161

@PeterTamaroff I will provide the link here just for you, hang on...

user19161

2:45 AM

math.uchicago.edu/~schlag/complex.pdf

user19161

math.unc.edu/Faculty/met/cpx.pdf

Peter Tamaroff

@JasperLoy Thanks

I'm leaving now.

Bye byes, peoples of the math.

user19161

@PeterTamaroff Good night!

Alexander Gruber

@amWhy grade-A uncut columbian reputation, carnal as the red in the up arrow.

caveman

lol

hey @Ethan

Ethan

2:47 AM

hi

caveman

@Ethan, look at my two answers math.stackexchange.com/questions/342241

user19161

Guys, I just discovered an orange square on this site.

user19161

mottese

41 4

user19161

The coloured squares are taking over MSE...

Alexander Gruber

@PeterTamaroff adios holmes

caveman

2:48 AM

@Ethan, also how to prove the ramanujan sum is always an integer?

amWhy

@AlexanderGruber got it...hahahaha!

Ethan

If, $c_q(n)= \sum_{a=1}_{(a,q)=1}^{q}e^{2\pi i\frac{a}{q}n}$

@caveman then $c_q(n/d)$ counts the roots of unity where a,q have a common factor of d

caveman

I don't understand

Ethan

one sec

I messed up it should be q/d

1 sec

κρανίοπεριπολία

1 second later...

user67848

2:58 AM

Oh god it was quite an ordeal for me to figure out how to get the network to recognize that I have enough rep to chat.

caveman

@user67848, that is so annoying

julien

@JasperLoy @JasperLoy Thanks. This website is addictive...

Ethan

@caveman

caveman

hey

Ethan

Using the special case $c_q(1)=\mu(q)$ shows the sum of the primitive qth roots of unity is equal to $\mu(q)$, also if you use a similar argument there you can get sums of the form $$\phi(n)=\sum_{k=1}^n\gcd(k,n)e^{2\pi i k/n}$$ Where $\phi(n)$ is eulers totient function

caveman

3:14 AM

Very nice!!!!

@ the two sheets

that even proves its multiplcative

user67848

lol. I got a badge for upvoting.

caveman

yeah you get lots of silly badges when you join

user67848

Cool. Does one get privileges from having lots of badges?

caveman

no

awllower

3:41 AM

hello!

caveman

hey!

amWhy

hello!

Alexander Gruber

4:10 AM

@user67848 feature request please :)

user67848

@AlexanderGruber Hm.. What privileges have yet to be implemented?

Alexander Gruber

@user67848 gosh, i don't know. maybe if you have a lot of badges you should be allowed to make up some new ones.

user67848

@AlexanderGruber Yes, I would form the "badge ordinals": i.e. badge badges, badge badge badges, etc.

awllower

@AlexanderGruber IIRC, one day, when some people are discussing group theory, you appeared, saying: my third eye tells me that someone here is discussing group theory, right?

And I am interested in knowing how did you know!
Thanks for any explanations. :D

Alexander Gruber

@awllower clairvoyance badge

awllower

4:26 AM

Oh!
Forgive my ignorance: I knew not about this badge before.
:)

Alexander Gruber

@awllower if you did, you'd have it. :)

awllower

:D

I see.

user67848

@AlexanderGruber The clairvoyance badge must be visible only to other owners of it.

What does it mean for an algebraic set to be defined by two polynomials f,g? I.e. is it the zero set of the ideal generated by f and g or is it the union of the zero set of f and the zero set of g?

Alexander Gruber

@user67848 @user67848 exactly as one would expect

@user67848 what's the context?

user67848

@AlexanderGruber A problem in Hartshorne's algebraic geometry defines $Y$ to be the algebraic set in $\mathbb A^3$ defined by $x^2-yz$ and $xz-x$.

Alexander Gruber

4:42 AM

@user67848 yeah it's just the zeroes shared by both polynomials

@user67848 so you need $x,y,z$ so that $x^2-yz=xz-x=0$

user67848

@AlexanderGruber Ok, thanks. :)

caveman

5:05 AM

I like tis one a lot math.stackexchange.com/questions/342496/…

amWhy

I'm not getting any (or many) upvotes tonight! Why do I let that get to me!?

It's midnight, my time, and I hate going to bed without much in the way of rep...I've become a night-owl as a result...and worthless in the early a.m.!

@caveman What time is it where you are?

caveman

5a m

Ethan

im bored

amWhy

@caveman Did you stay up all night?

caveman

yeah

Ethan

5:11 AM

caveman how old are you

caveman

23

amWhy

@caveman You must be in Europe, yes? East Coast US is an hour ahead of me...so

user67848

@caveman The sum of the digits in your age is the time where you are. Coincidence?

caveman

yes

@user67848 lol

Ethan

yes

its also a prime

amWhy

5:13 AM

Hah, the sum of the digits in my age is the time where I am :-)

So is your age, Ethan!

Ethan

yes

my birthdays in october

@amWhy why do you say that?

amWhy

@Ethan I just meant that age is temporary ;-)

Ethan

nvm

amwhy how old are you

Ben

Can I bet 39?

Ethan

what kind of music do you guys listen to

Ben

5:26 AM

I like mathrock and older punk. You?

Ethan

can I listen to some of it?

Ben

I like this song youtube.com/watch?v=N0sck9To_VA

Here's an old punk song I've been listening to recently: youtube.com/watch?v=Q2ICpPhRE2M

amWhy

@Ben No! hardly!

user67848

28

Ben

@amWhy Isn't it a little past 12 in Wisconsin? I couldn't think of another suitable age whose digits sum to 12.

amWhy

5:33 AM

@Ben I was teasing...thinking of midnight as 00:00 ;-)

:8699009 there you go! I'm a prodigy!

Ben

@amWhy Are you a grad student at Wisc?

amWhy

@Ben yes...ABD...on a bit of a sabbatical!

Ben

@amWhy May I ask at what campus?

amWhy

@Ben Madison

Ben

Oh cool, amazing school I hear.

amWhy

5:38 AM

@Ben Yeah, it is. It's an amazing city, too!

Alexander Gruber

@amWhy you're 70?

amWhy

@AlexanderGruber 71, to be exact ;-)

Ethan

lol no

you are not

amWhy

@Ethan you're right, I'm not 70 or 71...not anywhere near there!!

Ethan

are you really?

amWhy

5:46 AM

Am I really what? 70? no...71...no...just teasing.

Ethan

lol

caveman

I'm stumped math.stackexchange.com/questions/342518/…

Ethan

its a no

when he says evaluate

in closed form

what does he have in mind?

caveman

maybe it's a hypergeometric function

Alexander Gruber

@amWhy are you 8?

:)

amWhy

5:57 AM

@AlexanderGruber How'd you figure that out? Darn, now it's not a secret!

26? 35? 44? 53? 62?

10(-2)?

48?

@AlexanderGruber Okay, so let's just say I skipped a few grades in school... ;-)

It's now 1:00 a.m. here, so I guess now I'm 100.

Alexander Gruber

interesting

amWhy

6:03 AM

I'm probably going to need to think about getting some sleep...I'm not getting anywhere on main...I'll just have to go to sleep feeling defeated :-(

Alexander Gruber

need questions?

amWhy

Questions, upvotes, questions, did I say upvotes? ;-) I spent too much time tonight on more trivial questions...and too much time for OP's who put in little effort.

Alexander Gruber

hmmm.

caveman

I didnt realize it until after

Gustavo Bandeira

@caveman Yo! Where's the Ph.D?

amWhy

6:10 AM

@caveman Yes, that happens.

Alexander Gruber

if a group has a normal subgroup $X$ which is nilpotent so that each sylow subgroup of $G/X$ is cyclic or generalized quaternion, why would this prove that the fitting length of the group is bounded by $4$? @amWhy @caveman

caveman

amWhy

@caveman hahahahaha!!

Gustavo Bandeira

I want to be a rich idiot.

caveman

I'm looking up fitting length now

Gustavo Bandeira

6:11 AM

@amWhy Are you antropomorphic?

Alexander Gruber

oh: group is solvable also

caveman

I think that's too difficult fo r me

Alexander Gruber

i think it has to be a frobenius complement or something but i'm blankin'.

hm

amWhy

@GustavoBandeira Not sure what you're asking...I looked up anthropomorphic, but also saw the connotation for a similar word in Spanish that has to do with psychology?

Alexander Gruber

so equivalently, every sylow subgroup of a group quaternion or cyclic should imply fitting length bounded by 3

Gustavo Bandeira

6:16 AM

Basically, it's human form.

Alexander Gruber

something has to be metacyclic

Gustavo Bandeira

But yes, it's related to something in psychology. If you're not looking for physical form.

Woody does not have a human form, but it shares some behaviors.

Antropomorphic woody - it could be.

amWhy

Okay...got you...that's kind of what I thought. Yes, of course, I am of human form...Yes, I understand I think. Attributing human thought/feeling/behaviors to non-humans

Gustavo Bandeira

What have you learned?

amWhy

@GustavoBandeira Is it because of my "smiley" question mark: yes indeed, antropomorphic "questioning smiley"

@GustavoBandeira What have I learned in my life? Oh, my, do we have a few years to talk ;-)

Gustavo Bandeira

6:20 AM

If you could generalize, what would you say?

amWhy

@GustavoBandeira Wow...that's a tough one. I'm pretty sleepy right now. I have learned humility. That's says a lot, in general.

Gustavo Bandeira

I'm sleepy too

But I have to get a few things done.

Jayesh Badwaik

@GustavoBandeira I would say that I cannot generalize in general.

amWhy

@AlexanderGruber Sorry @Alexander: I got sidetracked, and truth is, my brain is pretty worn out. I need to eat, then sleep. Sometimes I resent being embodied as a human being, subject to the need for food, and sleep, etc.

Alexander Gruber

@amWhy no problem. let me know if you figure it out.

(this is a genuine question, not a puzzle. should pop out of the definitions though, guy states it without proof.)

luckily i am not a human being so i will remain not sleeping here

amWhy

6:25 AM

@AlexanderGruber Oh, my! Have we a deity among us?!

Alexander Gruber

@amWhy possibly :p

amWhy

@AlexanderGruber ;^)

Jayesh Badwaik

finite p-groups are solvable. a cyclic or a generalized quaternion has a unique p-group, so that is one unit in the fitting length of 4.

G -> X -> Q -> P

G is the original group, X is the normal subgroup

P is the finite p-group

Q is the cyclic or generalized quaternion

I am missing something here but the idea seems okay, let me see.

Alexander Gruber

what about the action between the Sylow subgroups?

Jayesh Badwaik

6:47 AM

what about it? I am not comfortable with actions. I was thinking of using canonical homomorphism to some effect.

Alexander Gruber

well, our $X$ is going to be either $F(G)$ or a subgroup of $F(G)$, so if we have $\ell_F(G/x)=3$ then we have $\ell_F(G)=4$ in the former case and $\ell_F(G)=3$ in the latter

so really we just need to prove that $\ell_F(\overline{G})=3$ whenever $\overline{G}$ has only cyclic or gen. quart. sylow subgroups.

when we take $F(\overline{G})$, say it's divisible by $q$ - i don't see why the condition implies that $\overline{G}/F(\overline{G})$ shouldn't have a Fitting subgroup divisible by $q$ also (i.e. it may not take the entire Sylow $q$-subgroup)

(i can certainly prove this is true for $\ell_F(G)\leq 5$)

Jayesh Badwaik

okay

Alexander Gruber

yeesh - i'll think about it tomorrow, i am sleepy

thanks man

Jayesh Badwaik

7:06 AM

:-)

night

Dominic Michaelis

ah finally the 1990 of xkcd worked for me :D

(the first tower crashed)

Jayesh Badwaik

??

Dominic Michaelis

sry 1190

this

i don't know how that tower gets destroyed but it does some time

caveman

7:24 AM

> var r = Math.round(Math.random() * 55);
n("reloading in", r + 5, "seconds"), setTimeout(function () {
t("reloading"), setTimeout(function () {
location.reload()
}, 5e3)
}, r * 1e3)

http://c4.xkcd.com/stream/comic/time?method=EventSource

://imgs.xkcd.com/comics/time/9c8b0e5c90dd5b110ad625410c2e84cc4c676065dff14540e326f04e0685d958.png

Dominic Michaelis

thats cheating :D

caveman

im trying to cheat but its not working

Jayesh Badwaik

I downloaded the image js

caveman

looks like I have to send something to the event source

Jayesh Badwaik

it seems there is a randomizer

out of every 7 times

the tower will break once may be

Dominic Michaelis

7:28 AM

i saw the destroyed one

i heared a story about david from tex.se. he answered a question, not by looking in the documentation, but directly going to the source code ...

Jayesh Badwaik

sometimes that's the best method

Dominic Michaelis

i write numerics today -.-

Jayesh Badwaik

http://primis.org/time/output.gif

Transcript for

$Mathematics$ Mathematics