Mathematics - 2014-02-03 (page 4 of 6)

Frank Science

10:00

Did you try some concrete finite groups $G$, such as $G=C_n$ or $G=S_3$, etc?

Mike

@anon things go fine if you're considering, say, a six-dimensional vector space over $\Bbb F_3$

Balarka Sen

@PedroTamaroff Then stop chatting useless and do some real math. (At least thats what I do)

Mike

(for other related counting problems)

Pedro Tamaroff

@BalarkaSen Chatting useless? If you say so.

@anon Composition series are going to get me.

anon

@Mike the structure of the group of units should be sensitive to the family of irreps, which is in turn sensitive to |G|'s invertibility

Mike

10:01

alright, granted

anon

also,

21 mins ago, by anon

I figure $K[G]^\times$ will be $K[G]$ minus a set number of left/right ideals (principally generated by left/right nonunits), so in particular $K[G]$ minus a bunch of proper $K$-subspaces, which can be partitioned according to dimension. as long as tensoring up from ${\Bbb F}_p$ to ${\Bbb F}_{p^n}$ doesn't get too wild it shouldn't destabilize this family of subspaces, which would mean the size is in fact polynomial in $q=p^n$.

Mike

so for convenience, we pick $|G| = 1$ :)

anon

(:

Mike

what would you expect $\lim_{q\rightarrow 1}$ is

anon

:O

Pedro Tamaroff

10:03

@Mike Damn it, Mike.

Mike

@Pedro Does that limit bother you?

Pedro Tamaroff

@BalarkaSen I have real math problem for you.

@Mike You're always pushing things.

Frank Science

Eh, now that you have mentioned limits.

Balarka Sen

@PedroTamaroff Fire away, I'll do after I have done with my eating.

Frank Science

I wonder whether we can take limits on physical objects.

Pedro Tamaroff

10:04

@Mike I couldn't click the link.

Again!

Mike

anon

gb2/b/

Karl Kronenfeld

you have to wait a minute before removing to allow us non-moderators a chance

Mike

@Karl my link was busted

Frank Science

For example, could we take limits to a volume distribution of charge?

Karl Kronenfeld

10:05

@Mike ah

Mike

truly a great loss for all of mankind

Frank Science

To obtain surface charge, or point charge.

Or taking limits to get dipoles, etc.

Mike

don't ruin this for me

Karl Kronenfeld

See, @Mike, that's how long you have to wait before removing a message

Mike

: - (

Frank Science

10:07

Or something like $\lim_{\epsilon\to0^+}\delta_\epsilon(x)=\delta(x)$, etc.

Pedro Tamaroff

@BalarkaSen Suppose that $t_1,t_2,\ldots$ is a sequence for which there exists another sequence of positive numbers $\varepsilon_1,\varepsilon_2,\ldots$ converging to $0$ for which $t_{n+1}>t_n-\varepsilon_n$. Then $\{t_1,t_2,\ldots\}$ is dense in $[a,b]$ where $a=\liminf\limits_{n\to\infty} t_n,b=\limsup\limits_{n\to\infty} t_n$.

Mike

btw the response to : - (

should always be "why the long face"

Frank Science

It's the Dirac delta function.

Balarka Sen

That's not what I call real math, @Pedro.

Karl Kronenfeld

@Mike I was waiting for a second long face before asking.

Pedro Tamaroff

10:09

@BalarkaSen Now you're just being a fool.

Balarka Sen

@PedroTamaroff Am I?

Pedro Tamaroff

So I will desist from talking to you.

Balarka Sen

I have much more works than talking here in chat. I will just go then.

Pedro Tamaroff

@Mike

Mike

music

Pedro Tamaroff

10:13

Joining.

Mike

anyone else is free to as well

Balarka Sen

Is it me or the icosahedral polynomials pops up in the modular equation for j too?

Mike

@Pedro Not gonna join the DJ waitlist?

Pedro Tamaroff

Done.

Thought I had done so.

@Mike Would you like to think about a proof of my problem above? It is from Polya and Szego.

A profesor in my university gave me a really cute proof.

It's a proof by contradiction, still nice proof.

I'd like to see a direct proof, though. =)

Mike

@Pedro I can't promise I'll look hard (doing reading for a test) but I'll look.

Pedro Tamaroff

10:28

The point is that if an elt $\ell$ is not a limit point, there must be a $\varepsilon_0>0$ such that there are finitely many $t_n$ inside $(\ell-\varepsilon_0,\ell+\varepsilon_0)$, hence for a smaller $\varepsilon_0'$, we have an nbhd of $\ell$ which has no elt of $t_n$ inside.

Mike

ugh@elt

Pedro Tamaroff

Since we're inside the limsup and liminf, we have infinitely many points to the left and right of this interval.

Mike

Coulda easily skipped the 'finitely many' step, but go on

Pedro Tamaroff

In particular, we can find $N$ large enough so that $\varepsilon_N<2\varepsilon_0'$, and such that $t_N$ is to the left of the interval and $t_{N+1}$ to the right, so $2\varepsilon_0'<t_N-t_{N+1}<\varepsilon_N$ which is impossible.

Mike

Cute, I like it

Pedro Tamaroff

10:33

Yeah, it is very nice.

A direct proof would probably be very messy, but still doable.

Though in this case the indirect proof is quite likeable.

@Mike This one is a crazy problem.

I haven't solved it yet.

Consider a cube $[-n,n]^3$.

Where $n$ is a positive integer.

Mike

I'll try.

Pedro Tamaroff

Consider another integer $s$.

Mike

One more consideration and my brain will go bust.

Pedro Tamaroff

Then the number of lattice points inside the cube, for which $|x+y+z|\leqslant s$ is...

Mike

Three.

@Pedro That was my last cocaine song.

Pedro Tamaroff

10:37

$$=\frac{1}{2\pi}\int_{-\pi}^{\pi} \left(\dfrac{\sin\frac{2n+1}2 t}{\sin \frac t2}\right)^3\dfrac{\sin\frac{2s+1}2 t}{\sin \frac t2}dt$$

I shit you not, Mike.

Mike

Didn't know Chris's Sis was here ;)

Pedro Tamaroff

That's the number of lattice points.

Mike

Seems awfyk.

awful

Pedro Tamaroff

@Mike Well, let's try to prove the integral is a positive integers first. =D

Mike

No.

Pedro Tamaroff

10:39

@anon You?

anon

me

Mike

how do you feel about music @anon

anon

most feels

Mike

invite remains open

we're out of cocaine songs, tho

Karl Kronenfeld

@PedroTamaroff Difficult decision: (1) Do a typical messy differential geometry calculation or (2) Figure out whether that thing is an integer

Pedro Tamaroff

10:43

@KarlKronenfeld I am reading the solution by Pólya himself.

Wanna see it?

Mike

clearly one should prove that it's an integer

err

should prove it's the number of lattice points

and thus, it is an integer

Karl Kronenfeld

@PedroTamaroff If it's not a burden to type, sure.

Pedro Tamaroff

Mike

nope

Pedro Tamaroff

@Mike ??

Mike

10:45

what is that the #30 of?

Pedro Tamaroff

It's in the book "Problems and Theorems in Analysis Vol. 1".

Karl Kronenfeld

heh

Pedro Tamaroff

It is the 30th problem in the first part of the book.

ERMAGHERD. I'm in love with that singer @Mike.

skullpatrol

Hey @KarlKronenfeld wazzup ... Did you watch any of the game?

Mike

Previously, or just now?

Pedro Tamaroff

10:48

The woman, right now.

Mike

She's good

my singercrushes are sarah blasko and kate earl

Karl Kronenfeld

@skullpatrol hey, no I didn't

Pedro Tamaroff

@Mike What about Norah Jones?

Mike

@Karl You didn't miss much

skullpatrol

oh, true dat^

Mike

10:50

@Pedro not so much

Pedro Tamaroff

@Mike corinne bailey rae?

Mike

bro keep your playlist fresh

Pedro Tamaroff

Sawry.

Mike

I'll play a Sarah Blasko song for you

Pedro Tamaroff

Gut.

Mike

11:00

@Pedro bro, throw some new songs on

skullpatrol

May I join your community?

Mike

feel free

skullpatrol

how?

Mike

starred link on the right

skullpatrol

i did, what is the name?

Mike

11:01

you'll need a twitter or somesuch, but you can change your username before entering if you're worried about privacy

oh, if you've registered, just click the link again maybe? that links to a specific room

Pedro Tamaroff

@Mike Electricity went off.

But I had thrown in two songs.

Mike

@Pedro Damn.

Pedro Tamaroff

DUnno what happened.

skullpatrol

Hot Chip - Night & Day current dj: mike_m

Mike

that's the one

you can play songs of your own with stuff in the bottom left

skullpatrol

11:03

\o/

Mike

the botom left button lets you pick a playlist, and once you've got a few songs on there, add the DJ waitlist

skullpatrol

thanks :D

Mike

np

Balarka Sen

11:24

@robjohn @Chris'sis $$\sum_{n=0}^{\infty} \binom{2n}{n} H_{n} \ x^{n} = \frac{2}{\sqrt{1-4x}} \log \left( \frac{1+ \sqrt{1-4x}}{2 \sqrt{1-4x}} \right)$$

Integrating w.r.t. $x$ and subbing $x = 4^{-1}$ does the trick.

Reference

Studentmath

11:41

@KarlKronenfeld Interesting. The only thing that bothers with me with that is the fact that every subset of Q in the way of [a,b) Intersection Q is also dense, and therefore won't be equal to the ordinal of w, unless you figure out a way to get it to be so. Note that [0,infinity) Intersection with Q is obviously not alike to N in that sense.. But I guess I am missing something you did manage to prove?

Karl Kronenfeld

@Studentmath I am not sure if I understand what you are saying. All I was able to prove is that we can find a set in $\mathbb Q$ (actually in $[0,\infty)\cap\mathbb Q$) of the same order-type as any ordinal smaller than $\omega^\omega$. I did not get any results determining the order-type of the whole $\mathbb Q$ or $[0,\infty)\cap\mathbb Q$.

Studentmath

How did you get to that proof?

(Not that I doubt you did, I am just interested)

Karl Kronenfeld

@Studentmath For $a,b\in\mathbb Q$, let $\phi:[0,\infty)\cap\mathbb Q\to[a,b)\cap\mathbb Q$ be an order-isomorphism (e.g. the inverse of $x\mapsto\frac{x-a}{b-x}$). Then, you can use $\phi$ to copy sets of order-type $\alpha$ into the sets $[n,n+1)\cap\mathbb Q$. The union of these sets will have order-type $\alpha\cdot\omega$.

Pedro Tamaroff

@KarlKronenfeld I'll try to prove Jordan Hölder now.

Studentmath

Oh, I get it now

Question is how (and if) we can get it from there to being certain we can get to w^w too

Karl Kronenfeld

11:55

That's a good question. I haven't resolved it.

It would require a different idea than what I have here.

Studentmath

If, at all, it is possible..

Balarka Sen

@Mike You there?

Karl Kronenfeld

Right. I still don't think we could use an idea like this to reverse engineer a given set of order-type $\omega^\omega$ for a contradiction.

Balarka Sen

Or perhaps @anon?

anon

?

Balarka Sen

12:00

Has there been any research work on whether or not quintics can be solved in $\Bbb E$?

The best I could get is a 'no' by Schanuel.

Studentmath

I am certain it's not possible actually and the upper limit is obviously w^2, but. Yes.

It's problematic, or well, beyond my abilities. Speaking of which, heading back to my pre-test studies (test on the 10th)

Pedro Tamaroff

@anon Why should one think of a composition series as a "factorization" of a given group?

anon

think of the composition series of C(n)

(namely, the multiset of subquotients)

Studentmath

Another question - do you guys ever sleep?

anon

@BalarkaSen dunno

Balarka Sen

12:02

@Studentmath (It's 5:30 PM)

Pedro Tamaroff

@Studentmath Hehe, today was a particular day. I usually sleep.

Karl Kronenfeld

@Studentmath I'll gladly hear any specific objections (I agree this is counter-intuitive, and I will see if I can construct any meaningful examples above $\omega^2$).

Pedro Tamaroff

I skipped sleep today.

Balarka Sen

@anon You are familiar with $\Bbb E$, right?

anon

@BalarkaSen euclidean or constructibles or something like that?

Pedro Tamaroff

12:03

@anon That would be an ascending chain of divisors of $n$?

anon

@PedroTamaroff yes, and the multiset of subquotients?

Pedro Tamaroff

@anon What do you call subquotients?

Balarka Sen

@anon It's the smallest subfield of $\Bbb C$ closed under $\exp$ and $\log$ (with appropriate branches, of course)

anon

@BalarkaSen didn't know that

@PedroTamaroff the composition factors, whatever

Pedro Tamaroff

@anon Oh, that was my guess. Well, still cyclic subgroups but must be of prime order yes?

Since simple.

anon

12:05

the composition factors will be cyclic of prime order yes

Pedro Tamaroff

Well, I guess C(n) is an easy one.

=)

Balarka Sen

@Mike?

anon

the multiset of composition factors will be C(p) for every prime divisor of n, with multiplicity equal to the exponent of p in n's prime factorization. so is it clear now why the composition factors are like a factorization of a group?

Dunno

Hi, can someone help me with notation? Is $2^{\mathbb{N}}$ the same as $\{0,1\}^{\mathbb{N}}?

Pedro Tamaroff

@anon Oh...! So if $n=p^2q^2$ say, the composition series would be $1\lhd C(q)\lhd C(qp)\lhd C(qp^2)\lhd C(p^2q^2)$?

anon

12:07

a composition series

Karl Kronenfeld

@Dunno unless you're doing category theory, that'd be my guess.

Pedro Tamaroff

@anon Right.

@anon Well, I guess you convinced me a little. =)

Dunno

@KarlKronenfeld No, it's just elementary set theory stuff. Thanks

Mike

@BalarkaSen ?

@Karl what would that denote in category theory?

Balarka Sen

8 mins ago, by Balarka Sen

Has there been any research work on whether or not quintics can be solved in $\Bbb E$?

anon

12:09

@Dunno yes 2^X is another name for the power set of X. notice |P(X)|=2^|X| is true in terms of cardinal numbers. furthermore Y^X is the set of functions X->Y, and satisfies |Y^X|=|Y|^|X|. (specifying a subset is the same as specifying a function X->{in,out}, or X->2)

robjohn

@BalarkaSen Integrating with respect to $x$ puts an $n+1$ into the denominator, not an $n$.

Balarka Sen

Schanuel gives something.

Mike

no clue

Karl Kronenfeld

@Mike Category of functors from discrete category $\mathbb N$ into the category with a single nonidentity arrow, namely $a\to b$.

Mike

@Karl Gotcha

Balarka Sen

12:10

@robjohn Well, how about cancelling out a $x$ or somewhat? To give $n-1$ as the exponent?

I am sure a tweak would do.

@robjohn Have you seen my recent TNT/Galois theory Q?

2 mins ago, by Balarka Sen

8 mins ago, by Balarka Sen

Has there been any research work on whether or not quintics can be solved in $\Bbb E$?

robjohn

@BalarkaSen Something is wrong with your formula. At $x=0$, the left side is $1$ and the right side is $0$

skullpatrol

$1 \neq 0$

Balarka Sen

@robjohn Assuming $0^0 = 1$?

Karl Kronenfeld

@skullpatrol Oh! Damn, I had used $1=0$ in my proof of RH. I guess I should retract my paper eh.

4

Mike

@Karl No need, just fill the gap and prove $1=0$

skullpatrol

12:16

divide by 0

Mike

$\frac 10 = \infty = \frac 00$

we're saved

Pedro Tamaroff

@Mike Yay Mike!

Karl Kronenfeld

Yay thanks, @Mike

Now, my proof works.

Pedro Tamaroff

Mike > Jesus

Mike

I'm gonna stay miles away from that one

Pedro Tamaroff

12:18

@Mike Staying in topic, how did that Religion course go?

Mike

@Pedro It's gotten better but it's still awful.

Balarka Sen

A crank in a forum said "I gve 30000$ of teh prize money t anyone hlps me to publishh my rh proof."

Pedro Tamaroff

@DanielFischer Ahoy.

robjohn

@BalarkaSen Take the limit. In any case, we almost always use $0^0=1$.

Balarka Sen

@robjohn Limit of what?

Daniel Fischer

12:21

@PedroTamaroff Ahoy too.

robjohn

@BalarkaSen The left side. That is where I was assuming your comment about $0^0=1$ came from

Pedro Tamaroff

I posed this problem today

2 hours ago, by Pedro Tamaroff

@BalarkaSen Suppose that $t_1,t_2,\ldots$ is a sequence for which there exists another sequence of positive numbers $\varepsilon_1,\varepsilon_2,\ldots$ converging to $0$ for which $t_{n+1}>t_n-\varepsilon_n$. Then $\{t_1,t_2,\ldots\}$ is dense in $[a,b]$ where $a=\liminf\limits_{n\to\infty} t_n,b=\limsup\limits_{n\to\infty} t_n$.

Shortly afterwards I provided a proof, if anyone is curious.

Balarka Sen

@robjohn Okay. Ignore that discontinuous ungrateful little beast.

Do things for $x \neq 0$

Wait.

Isn't $H_0 = 0$?

robjohn

@BalarkaSen It is not discontinuous.

Balarka Sen

Let's assume that $H_0 = 0$, dual to the paper.

Both sides are now equal at the point.

robjohn

12:25

@BalarkaSen Ah. I missed the $H_n$. okay it's fine

@BalarkaSen Did you look at Corollary $7$ of that paper?

Balarka Sen

@robjohn Not really. I got that paper from S&Co., referring the methods to that.

Nice result.

Mike

@Pedro You're missing out on "Bearforce One" right now

robjohn

@BalarkaSen Corollary 7 and the data presented in the proof give the sum as $\frac{\pi^2}{3}$

since $x=\frac14\implies y=0$

Balarka Sen

Hmmphm.

robjohn

@BalarkaSen Since the $n=0$ term is $0$, sum from $n=1$ :-)

Then it is easier to see that you can divide by $x$ before integrating

Chris's sis

12:52

Greetings

skullpatrol

Greetings @Chris'ssis

Chris's sis

@robjohn are you around? I think I've found an amazing relation.

robjohn

@Chris'ssis yes

Chris's sis

@robjohn OK

@robjohn sent it

Chris's sis

13:13

@robjohn may I delete those? I have to go for 1 h or 2.

robjohn

13:26

@Chris'ssis I'll delete after I read them. Sorry, I was away.

Chris's sis

@robjohn OK

I have to go now.

Later.

skullpatrol

Bia!

robjohn

@Chris'ssis Cya

Dave