Mathematics - 2013-02-25 (page 5 of 5)

Git Gud

21:00

@Charlie Don't be mad.

Charlie

@Argon how were the things????

Argon

@Charlie Not fantastic. :/

Charlie

@Argon ow

user58512

I can't use the sylow theory or whatever to prove a group extension can't exist :(

Gustavo Bandeira

@Charlie Care. @robjohn is going to fire his lazors.

Charlie

21:03

@GustavoBandeira what in tarnation? we can't even swear anymore???

Arkamis

Gustavo just wanted to wake robjohn from his slumber by tagging him

Charlie

aah

user58512

How do I find a normal subgroup of a group generated by permutations?

pls

Arkamis

Any normal subgroup? or all of them?

user58512

any

Gustavo Bandeira

21:07

@Charlie Tu é muito gracinha. Nhóóów.

Arkamis

Is the order of the group even?

user58512

yes

robjohn

@GustavoBandeira Now I just have to find a target... let's see, who pinged me last?

user58512

haha

Peter Tamaroff

Lovely to wait for your teeth to get drillen. Fuuuu

Arkamis

21:07

Hm, I may not remember correctly since i haven't looked at group theory in over a year

but isn't any subgroup whose order is one half the order of the group normal?

Peter Tamaroff

Hmm... that didn't sound good.

robjohn

@PeterTamaroff You at the dentist?

user58512

that sounds plausible

Peter Tamaroff

Aye @rob

Arkamis

So... find an element of order |G|/2, and whabam, generator.

robjohn

21:08

@PeterTamaroff My sympathy.

Peter Tamaroff

@rob lawl thanks

Argon

@Charlie :(

Peter Tamaroff

@jacob i am not bold!

@argon hey

Arkamis

@robjohn Thanks for the answer on the complex analysis problem

I think I understand. I'm going to try to write up a solution where I do use that hint, though.

Argon

@PeterTamaroff What's up?

Peter Tamaroff

21:10

@argon deeeeeeentist

Argon

@PeterTamaroff Bleh. They cause trouble

Arkamis

@PeterTamaroff Are you at least going to be getting work done? Something to justify the good painkillers?

robjohn

@Arkamis The Schwarz reflection answer?

Arkamis

@robjohn Yeah

robjohn

@Arkamis it took a bit of manipulation, but it was not really what one would call deep.

user58512

21:12

I have a group of order 6*5*4, and I want to show no extension of it of size 7*6*5*4 can exist

how could I do it?

I wanted to get a normal 7-sylow subgroup but there might be 1,8,15 or 120 of them

(because then I could look at normalizer and maybe get a divisibility contradiction)

Peter Tamaroff

@arkamis oh, no. i dont get no pain killers. Maybe an eventual needle, but i dont think it is that serious

Arkamis

@robjohn I'm sure it wasn't. I was going down the wrong path trying to solve that.

robjohn

@Arkamis I did, too, at first. A different path, but still not the right one.

Cortizol

Can someone give me a hint for $y(y'+3)=ax^2+bx+c$. Thanks :)

Peter Tamaroff

@rob what are you guys working on?

anorton

21:15

How do you add a link in a comment (just wondering)

(aside from c/p it)

robjohn

@PeterTamaroff You mean Arkamis and me?

Peter Tamaroff

@cortizol model a polynomial solution and find approproqte coerficientsq

@rob aye

robjohn

@PeterTamaroff this question

@anorton you mean as I just did to Peter?

Arkamis

It is a homework problem, so robjohn's answer is excellent but too complete! So I am going to accept it, but also try a solution that uses the hint.

Cortizol

@PeterTamaroff Is that some method? Sorry for asking stupid question

Arkamis

21:19

@Cortizol Think about it like this

If $y$ is a polynomial in $x$, then so too is $dy/dx$

Cortizol

ok, i undestand

robjohn

@Arkamis Oops! I didn't notice the homework tag. Sorry!

Arkamis

two polynomial multiplied together give you another polynomial of higher degree

@robjohn ;)

Cortizol

ok, ok, but I think it not that easy

Arkamis

It is

Cortizol

21:20

mathematica can't solve it :(

anon

in characteristic zero, anyway

Arkamis

The degree of $yy'$ is at most 2. The degree of $y'$ is one less than the degree of $y$

user58512

anon can you help me with group theory?

anon

10 mins ago, by user58512

I have a group of order 6*5*4, and I want to show no extension of it of size 7*6*5*4 can exist

that problem?

Peter Tamaroff

@cortizol oh sorry my bad your solution isnt a polynomial. Try using an integrating factor maybe.

user58512

21:22

yeah

anon

@user58512 is it an arbitrary group of that order or does it come from somewhere?

user58512

its a group I have generated by permutations

Cortizol

@PeterTamaroff Thank you anyway :) This problem is from my professor and I think it's joke, who knows. Bad English, sorry...

Peter Tamaroff

Nah, it def has a solution

anon

@user58512 are you going to tell me more about it, or are you hoping that we can prove no extension exists using only its order?

or link me to old convo

user58512

21:25

I think it's possible only by order

that's the only way I've seen it shown a group cant exist

Cortizol

probably, but I put this equation in wolfram, and it couldn't find it
if we change ax^2+bx+c with x then wolfram find solution, but is pretty big...

user58512

@anon, it's sharply 4-transitive too

Arkamis

@Cortizol Try solving it for specific values of a,b,c

user58512

but that doesn't really imply anything, I think

Cortizol

@Arkamis for example, if a=0, b=1, c=0 it's pretty big solution wolframalpha.com/input/?i=y(y'%2B3)%3Dx&t=crmtb01 :(

i don't know how to start solving it, that is problem

user58512

21:28

oh wait a second I have a general theorem about this, I should run through the proof with my specific group

Peter Tamaroff

Lawl there are 3 kids in the wting room and the TV has Cheaters on.

Charlie

@PeterTamaroff funny!

I was at corner of discipline

But I was reading something else and couldn't think about what I did

user58512

I don't understand this: If G has a regular characteristic subgroup P then isn't N_G(P) = G?

in fact that should be true for any normal subgroup, isn't it?

Peter Tamaroff

corner of discipline? @charlie

user58512

I got it

Charlie

21:39

@PeterTamaroff I got suspended

skullpatrol

Hi @Charlie

Charlie

@skullpatrol Hi, Skull

Peter Tamaroff

@charlie when? from where?

Charlie

@PeterTamaroff I told gitgud to have coitus with himself, and it seemed agressive, offensive, inappropriate, whatever

Peter Tamaroff

hahaqhahahaha why, what did he say?

For you to flip him

Charlie

21:43

nothing important

Peter Tamaroff

Buuuuuu

Charlie

I'm not in a good day

skullpatrol

Who is gitgud?

Charlie

Then I say what i want to

skullpatrol

I will bring you his head!!!

Peter Tamaroff

21:45

Buuuuuuuuu

Arkamis

Charlie got internet stalked. And so she did her ninja thing.

Charlie

@skullpatrol here

@skullpatrol I appreciate :)

In school no boy played with me

Peter Tamaroff

@arkamis shit gon' get serious.... Stalking?

Arkamis

@PeterTamaroff Not really. That's just what I call it when people try to get other people's FB info, or whatever.

Peter Tamaroff

Ohhh, Ok.

Charlie

21:48

@skullpatrol that's what happens when you are not around

2

Peter Tamaroff

@charlie what happened?

skullpatrol

@Charlie :''(

Peter Tamaroff

Lol a little 2 yr old is trying to play dinosaurs with me...

user58512

I actually don't see at all how you can knot a sphere

Peter Tamaroff

@user58512 what do you mean?

skullpatrol

21:51

@Charlie OHHH Mr Kentucy Fried Chicken better not come here again...

Peter Tamaroff

its actually zermelo frankel continuum @skull

Charlie

@skullpatrol yeah

skullpatrol

@Charlie Here is the type of boyz you should of played with at school ;-)

Charlie

@skullpatrol hahahahaha

Arkamis

@Charlie This will make you feel better

i.chzbgr.com/maxW500/7082259968/h6C5FA6DF

Charlie

21:57

@Arkamis haha cats...

Arkamis

Cats make everything better.

Except sleeping. My cats do not make my sleep better >:(

Charlie

@Arkamis how many there are?

Arkamis

three

Charlie

oh! nice!

I will have dinner, bye guyss

skullpatrol

later

Charlie

22:00

@skullpatrol bye Skull

skullpatrol

@Charlie bye Charlie

Charlie

@skullpatrol :D

skullpatrol

@Charlie Do you mind if I call you Chucky?

Argon

@skullpatrol Doesn't this one girl always call Charlie Brown "Chuck?"

hhh

Conditional probability: Suppose $T=A \cap B$ (AND) where $P(A)=P(B)=0$. What is $P(T|\neg B)$? Is it 0 or 1?

skullpatrol

22:03

@Argon Yeah, Peppermint Patty?

Argon

@skullpatrol AH! Yes, that is her.

hhh

If you have an unit that works with 0 probability. What is its negation?

Amzoti

@robjohn: Has MSE ever considered writing an e-book using the collective knowledge of MSE? For example, "Problem Solving Strategies and Approaches".

skullpatrol

@Argon I want to use it because that was the nice name of the last Raiders' head coach that took us to the Super Bowl.

Argon

@skullpatrol You are such a skullpatroler

robjohn

22:06

@Amzoti Not that I know of. Perhaps some of the people who've been here longer might know of something.

skullpatrol

Chucky

Argon

This Chucky came to mind as well :)

Arkamis

@Amzoti I seem to remember such an idea being floated on Meta.

@skullpatrol You are a Raiders fan?

skullpatrol

@Arkamis Yep

Arkamis

I'm so sorry.

skullpatrol

22:09

@Arkamis Don't be

robjohn

@skullpatrol I know Brad Dourif who does the voice for Chucky :-)

Arkamis

You have Carson Palmer. You should accept my apologies.

skullpatrol

@robjohn Orly?

Argon

@robjohn HAHA THAT IS AWESOME

robjohn

@skullpatrol Yep, and he played Wormtongue in Lord of the Rings.

skullpatrol

22:10

@Arkamis He can't be worser than Jamarcus ;-(

Arkamis

True. You could do worse.

Mark Sanchez, Tim Tebow, Ryan Pickspatrick...

Theorem

why does $\int f\circ u (x)dx \le$ sup $f' \int u(x)dx$ hold ? any one help me if this holds ?

skullpatrol

@Arkamis Jamarcus took us through the worst loosing streak in NFL history!

Theorem

on some arbitrary region .

Arkamis

To be fair, you had Al Davis micromanaging from the upstairs suite for way, way too long.

skullpatrol

22:14

@Arkamis Big Al made a lot of dirty money in Vagas on us

Arkamis

Not sure why he's so beloved by the fan base

Same thing with Jerry Jones and the Cowboys.

Theorem

@robjohn : may i ask u for a look into my doubt ?

Arkamis

The Cowboys are mired in mediocrity, and it basically comes directly from the top.

skullpatrol

The orginators are always that way

robjohn

@Theorem which doubt?

Arkamis

22:15

Of course, not every team can have Bill Belichick...

Theorem

@robjohn : i got it i think :)

robjohn

@Theorem okay :-)

skullpatrol

@Arkamis Beli-cheat? ask the Jets

Arkamis

Whaaattttevvverrrrr

Theorem

it was a doubt regarding the inequality $\int f \circ g (x ) \le \sup f' \int g(x) dx$ but its just a substitution i realized .

@robjohn

Arkamis

22:18

The notion that any of the film could possibly be used in a game is comical. Not to mention their record post-spygate is better than pre-spygate.

skullpatrol

@Arkamis The GIANTS broke that team's spirit

Arkamis

As far as asking the Jets anything, all I have to say is thebiglead.fantasysportsven.netdna-cdn.com/wp-content/uploads/…

The Giants sold their soul to the devil. That's the only explanation.

skullpatrol

Al Davis was SATAN RIP

Arkamis

Eli Manning is one miracle catch and two inexplicable drops away from being Donovan McNabb

God I hate the Giants. I hate them so much.

skullpatrol

It has always been and always will be $\Huge\text{RAIDERS 4 LIFE}$

Arkamis

22:24

Can we talk about how the NHL's NE division has four teams within 4 points of first place? Of course, the Bruins have 4 fewer games played than all the others, but still.

robjohn

@Theorem are we to assume that $g$ vanishes at $\infty$ and $f(0)=0$?

Arkamis

Montreal, Toronto, what the hell are you doing here?

Argon

Go Leafs

skullpatrol

@Arkamis What 'bout dem Hawks???

Arkamis

The Hawks are ridiculous right now. Like, I don't even understand.

skullpatrol

22:26

During the lock-out they were practicing there asses off!!!

while everybody else sat on theirs

Arkamis

They're essentially 7 wins away from clinching a playoff spot, already

skullpatrol

Yep

Theorem

@robjohn : $f$ is just a $C^1$ function . and $u$ is a function such that $\int_\Omega g dx $ and $\int_\Omega Dg dx $ are $< \infty$ . this is all the facts that i have . what do u think ? I thought it was just substitution but it doesn't look like that .

robjohn

@Theorem Then it is false for $f=1$ since $f'=0$

pourjour

what is the set of points $M(z)$ that fulfill the relation $\dfrac{MA}{MB}=k / k\in \mathbb{R}^{+*}$

Theorem

22:32

@robjohn i am trying to prove that if $g \in C^1 (\Omega)$ and $\|f'\|_{L^\infty (R)} < \infty$ and $g$ satisfies the such that the $\int_{\Omega} f < \infty$ and $\int_{\Omega}Du < \infty$ then $f \circ g(x)$ also has the property of $g$ like above .

robjohn

@Theorem $\Omega=[0,1]$, $f=1$, $g=\text{any}$.

Theorem

@robjohn : yes , exactly , u are right . but my lecture notes are troubling me this .

robjohn

@Theorem perhaps there are some other assumptions

user19161

@skullpatrol That video scared me.

Theorem

@robjohn : let me tell u the precise statement .

skullpatrol

22:39

@JacobBlack Wild boyz?

Theorem

@robjohn Let $f \in C^1(\mathbb R)$ with $\|f'\|_{L^\infty (\mathbb R)}$ , $u \in H^1_1(\Omega)$ then $f\circ u \in H^1_1(\Omega)$ .

user19161

@skullpatrol Yes. I hate the devil.

skullpatrol

@JacobBlack I try not to hate.

Theorem

@robjohn : i gotta go . will look further with this problem .

:)

skullpatrol

later

robjohn

22:44

@Theorem That is quite different from the estimate you were proposing where the bound was on $\|f'\|_{L^\infty}$ and $\|u\|_{L^1}$

user19161

@skullpatrol I am not as noble as you.

Theorem

@robjohn : oh .

sorry

Argon

@JacobBlack How about me?

}:)

user19161

@Argon Same, I am only a banana.

skullpatrol

@Argon You are a gas

Argon

22:45

@JacobBlack Hmmm... I am a mere gas

@skullpatrol A noble one!

skullpatrol

Indeed

Theorem

@robjohn : but your contradiction applies even in this case isn't it ?

robjohn

@Theorem no, because $1\in H_1^1(\Omega)$

@Theorem but you don't have control of $f\circ u$ based on $u$

Ben W.

@skullpatrol Are you an SF Giants fan? Or A's?

skullpatrol

@BenW. Pirates

Yo ho ho

Ben W.

22:55

@skullpatrol Oh, I'm sorry to hear that.

:/

skullpatrol

@BenW. It's not about winning...it is the mystique of the image.

Ben W.

@skullpatrol The Pirates have a mystique to them?

We'll see how it goes this year. I wish them the best.

skullpatrol

@BenW. Thanks, who's your team?

Ben W.

Giants actually. So it's been a good few years.

skullpatrol

Nice.

Ben W.

22:58

I hope robjohn's not a Dodgers fan.

skullpatrol

He doesn't like sports

user43758

Can someone help me with this? math.stackexchange.com/questions/314371/…

user19161

@user43758 What help is needed?

user43758

I need to show that the dimension of the row space equals the rank of a linear application.
But the problem is I can't use the fact that the rank of the transpose of a matrix equals to rank of the matrix itself

@jacobblack

Ben W.

So are you just trying to show that the column rank and row rank are the same?

user19161

23:01

@user43758 I thought you needed help with the comment, lol.

user43758

@BenW. Yes if I can show that, that would be helpful

@JacobBlack :)

Ben W.

@user43758 If it helps, a proof that the column and row rank of a matrix is equal can be found as Thm 1.16 in Steven Roman's Advanced Linear Algebra.

user43758

@BenW. I am not sure that's what I need to use.. I don't like too much the proof

user43758

23:19

Do we say n-uple ?

If there is anybody you speaks french here, how do we say in english n-uplet ?

I found it we say: n-tuple

Arkamis

From the department of things that make no fucking sense: You are doing a least squares regression to compute a power-law fit of multiple samples of data for multiple tests taken over a long period of time, and for each sample you fit an exponent of exactly 1/2

This is random data. This does not make sense.

Charlie

@skull you can call me Chucky

skullpatrol

@Charlie Great! thanks :-D

Arkamis

Wait, ok, no I'm computing based on pre-conditioned data, that makes more sense.

Charlie

@skullpatrol no problem

skullpatrol

23:32

@Charlie :D

Charlie

;)

skullpatrol

;)

Peter Tamaroff

Yello.

user43758

How do you express the solutions of a non-homogenous system of the form AX=B in terms of the solutions of the homogenous system ?

@PeterTamaroff Hello

Peter Tamaroff

@user43758 The solution space is $A+s$

Where $s$ is a particular solution of the homogeneous system

And $A$ is the space of hom. sol.

user43758

23:44

@PeterTamaroff Merci

Peter Tamaroff

@user43758 That is most certainly in your book, now isn't it?

user43758

Well, my problem is actually a step by step proof of the theorem

Peter Tamaroff

You want a proof?

user43758

if you mean proof of the expression then yes

but in the question they just tell me "express all solutions of the non-homogenous system AX=B in therms of the set of solutions of the homogenous system"

Peter Tamaroff

@user43758 Oh, I see.

I take it $A$ is $n\times m$, $X$ is $m\times 1$ and $B$ is $n\times 1$?

user43758

23:49

yes

Peter Tamaroff

OK. Let $p$ be a solution of $AX=B$, and let $S$ be the subspace of solutions of $AX=0$. You want to show that $S+p=\{s+p:s\in S\}$ covers all solutions of $AX=B$.

user43758

I have to show this ?

Peter Tamaroff

@user43758 That is what you asked, to determine all solutions of $AX=B$ in terms of the solutions of $AX=0$.

user43758

Yes

@PeterTamaroff

Peter Tamaroff

I changed the notation a little.

user43758

23:53

what is S ?

Peter Tamaroff

"...let $S$ be the subspace of solutions of $AX=0$..."

user43758

Oh sorry

Peter Tamaroff

=)

user43758

tired

and p is ..

Peter Tamaroff

4 mins ago, by Peter Tamaroff

OK. Let $p$ be a solution of $AX=B$, and let $S$ be the subspace of solutions of $AX=0$. You want to show that $S+p=\{s+p:s\in S\}$ covers all solutions of $AX=B$.

user43758

23:55

Oh God...

I am so sorry

Peter Tamaroff

So what you want to show is the following: If $q$ is any solution of the system, then $q\in S+p$, while if $m\in S+p$ then $m$ is a solution.

The proof goes two ways, since we want to show a set equals another.

OK, can you prove either way? @user43758

Transcript for

$Mathematics$ Mathematics