Mathematics - 2015-04-10 (page 1 of 5)

Ted Shifrin

00:00

I guess you need to prove this recursively. Right? If you take $Y=P(E)$, then you have the tautological line bundle. I haven't thought about this in almost 40 years.

Incurrence

Is the commutator always defined as $[a,b]=ab-ba$? and what does it mean to speak of a 'commutator subgroup'?

Ted Shifrin

You need to actually know the algebra structure of $H^*(Y)$ as an $H^*(X)$-module, most likely.

No, @Incurrence. For groups, that makes no sense. That's for lie algebras.

For groups, it's the subgroup generated by $xyx^{-1}y^{-1}$ for all $x,y$.

Incurrence

Oh wait they have a wiki page, I was on the commutator page before

@TedShifrin Ahhh okay, that looks familar

Michael Albanese

Yeah. I think this is a non-trivial part of the argument. I will have to look into this. Thanks.

JC574

@Incurrence it's also the smallest normal subgroup such that when you quotient by it you get an abelian group

Ted Shifrin

00:03

@MichaelA: It looks like you'll keep me busy thinking once I've retired. That's either good or bad :P

Incurrence

@TedShifrin Good haha

@TedShifrin You will stay here forever, and enjoy it

Ted Shifrin

not necessarily ...

JC574

@Incurrence so $G/N$ is abelian if and only if $N$ contains the commutator

Incurrence

@JC574 That's interesting thanks

Ted Shifrin

@MichaelA: Did you do well on your orals?

Incurrence

00:05

I'll have to explore that since I am not overly comfortable with quotient groups

Michael Albanese

I haven't had them yet. I will do them in August/September.

Ted Shifrin

yikes, @Incurrence ... how much algebra have you had?

ah, ok, @MichaelA

Incurrence

@TedShifrin Linear A I&II, Abstract algebra I

@TedShifrin 50 pages of axler, 60 pages of cohn

Ted Shifrin

axler isn't helpful for this ... I don't know Cohn ... Did you do group theory for Algebra I? Our first course is rings.

Michael Albanese

By the way, when I ask you to look at a question, you should feel free to say no.

Ted Shifrin

00:07

@MichaelA: Our interests overlap quite a bit. I'm just removed from thinking about a lot of the stuff, sadly.

But I hope all my grad course notes I sent you might be of some help somewhere :P

Incurrence

@TedShifrin I was still in the engineering mindset so I did badly in Abstract algebra I

Ted Shifrin

ohhh, @Incurrence

Incurrence

@TedShifrin I am still trying to recover :\

Michael Albanese

They have been so far. I'll take a look.

Ted Shifrin

Quotient constructions are highly important in math ...

Incurrence

00:08

@TedShifrin Which sucks since Algebra is my favourite area right now

Ted Shifrin

That won't help for your question here, @MichaelA.

ᴇʏᴇs

@Incurrence But if you don't know algebra how do you know it's your favorite

Incurrence

I am fine with modulo arithmetic and isomorphisms of $\Bbb F_p$

Ted Shifrin

OK, I'm gone for a while. Have fun.

Michael Albanese

The quotient constructions or your notes?

Ted Shifrin

00:08

LOL, good point, mr eyeglasses

my notes, @MichaelA :P

I would suggest looking at Hirzebruch, though.

Incurrence

@ᴇʏᴇs I am fine with linear operators and jordan form stuff

Ted Shifrin

That's where I first learned the splitting principle.

That's not algebra, @Incurrence. Linear algebra is quite distinct.

JC574

If $D$ is a degree zero divisor on an elliptic curve, and $b$ is a point of our curve, then by Riemann-Roch we see $l(D+b) = 1$ and so $L(D+b) $ is just the constant functions

Incurrence

Fine, I am good at linear algebra at least :)

Ted Shifrin

we've been here before, @JC574

JC574

00:10

;)

If $D = \sum n_i p_i $

Michael Albanese

What exactly is the Hirzebruch reference? I've been struggling with the splitting principle as I've only seen it done for the tautological line bundle on an infinite grassmannian. I imagine the general splitting follows from this, but I haven't seen why yet.

ᴇʏᴇs

@Incurrence Don't engineers like linear algebra

JC574

suppose we have at least two $n_i>0$

Incurrence

@ᴇʏᴇs I would make a good engineer :)

Ted Shifrin

Topological Methods in Algebraic Geometry is the title, I believe, @MichaelA.

JC574

00:11

then let $f$ be meromorphic with its only poles being simple at two of these $p_i$

Michael Albanese

Thanks.

JC574

Now $(f)+D+b \ge 0 $

this is a contradiction, so we must have $D = p_1 -p_2$

Ted Shifrin

Some sign is wrong.

Wait, I'm confused.

Where is the contradiction?

JC574

maybe i've made a mistake, um $f$ isn't constant

but it's in $L(D+b)$

Incurrence

@ᴇʏᴇs @Ted You guys are demotivational lol

Ted Shifrin

00:14

I am a rat, @Incurrence.

a mean rat.

Jasper Loy

Hi @ᴇʏᴇs @Incurrence @TedShifrin.

ᴇʏᴇs

Hi @JasperLoy

Ted Shifrin

I see, @JC574. So, if $D=p_1-p_2$, then we must have $b=p_2$.

Yes, you're right.

Incurrence

Ted Shifrin

rehi @Jasper

Incurrence

00:15

@JasperLoy Hey Jaspen

ᴇʏᴇs

@Incurrence What classes are you gonna take next semester

JC574

I'm not sure we have $b = p_2$

Incurrence

@ᴇʏᴇs Algebraic methods of physics, honours advanced algebra, graph&design theory, PDE's

JC574

if $p_1 = b$ would something go wrong?

Incurrence

I found out yesterday because I started university in the second semester, even though I will have done all of the required math courses for either of two majors, I can't graduate until halfway through next year :\

So I will be taking research in Nov and in Feb

Ted Shifrin

00:17

$(f)=0$, if $f$ is constant, @JC574, so you need $D+b\ge 0$.

If you really want to learn some serious math, @Incurrence, being in a hurry to graduate isn't good. Take more courses.

JC574

yes!

Incurrence

@TedShifrin Wasn't in a hurry, I want to do honours afterwards

@TedShifrin and drag it out :)

Ted Shifrin

So $b$ has to cancel out $-p_2$, @JC574 ...

ᴇʏᴇs

Yea @Incurrence I'm trying to draw out my graduation as much as possible that my financial aid would cover

Incurrence

Lecturers give way more attention to honours students

Ted Shifrin

00:19

oh, just saying courses instead/plus research would be advisable, @Incurrence.

JC574

hmm i'm not sure about my earlier argument. The reason I didn't see that was because i forgot the definition of $\ge$ here

Incurrence

@TedShifrin Oh yes of course, I will be doing half research half courses I think

Jasper Loy

@ᴇʏᴇs @Incurrence I would need to do some very embarrassing things to resolve some of my mental problems, and I pray for the courage to do them when the time comes.

Incurrence

@JasperLoy Embarrasing?

Ted Shifrin

Night for now, y'all ...

Incurrence

00:20

@TedShifrin Good night

Jasper Loy

@Incurrence Yes.

Incurrence

@TedShifrin Bonne nuit

JC574

night @TedShifrin! I've fixed my argument, thanks for all the help

Jasper Loy

@Incurrence Most people in the world will not be able to do them.

Incurrence

@JasperLoy Any examples?

Jasper Loy

00:23

@Incurrence I cannot describe them to you, it's too complicated.

Incurrence

@JasperLoy I hope you may succeed!

Jasper Loy

@Incurrence I will do all those things in the month of May. For now, I need to prepare myself mentally.

Hi @Owatch is that you in the picture?

Hi @robjohn you look handsome today.

r9m

00:41

@JasperLoy (-_-)

Incurrence

@r9m Why are you making faces at JL?

Jasper Loy

@r9m I see. Too bad, I was planning to date her.

r9m

@JasperLoy ?? :P You can't date sarah greene ?!

Incurrence

@r9m Why is your blog empty?

Jasper Loy

@Incurrence I am really scared I won't be able to do them. I guess only I can help myself now.

Incurrence

00:44

How do you define embarrasing in this case?

Jasper Loy

@Incurrence People will know that I am crazy.

r9m

@Incurrence sorry (I changed my domain name last night .. I was fedup with an Aztec God :P) try again ..

Incurrence

Random people?

Jasper Loy

@Incurrence Strangers I need to speak to.

Incurrence

@r9m Yes that works now

@r9m Robjohns thing doesn't make your math render

r9m

00:49

@Incurrence I am sorry once again .. if you are using firefox .. the latest updated version has been troublesome when it comes to disabling security .. if you are using google chrome .. just disable the security option :)

Incurrence

@r9m Oh yes of course, I had encountered this long ago on my blog

@r9m How many views do you average a day would you say?(non-unique)

r9m

@Incurrence lemme check .. :)

Jasper Loy

@Incurrence Anything else you would like to say to me about this matter?

Incurrence

@JasperLoy I don't have enough information to say much about it

Jasper Loy

@Incurrence I guess I just have to muster the courage somehow.

Incurrence

00:56

@JasperLoy What strangers will you be talking to? I usually don't get embarrased with strangers, since I don't care what they think of me

Owatch

@JasperLoy chat.stackoverflow.com/transcript/message/22596008#22596008

Incurrence

As opposed to people I know well of which I consider important

Jasper Loy

@Owatch Too bad.

Owatch

It's also in mockery of another user.

Incurrence

How can I show $G$ is a subgroup without being told the operation...?

Jasper Loy

00:58

@Owatch Are you the user called Oracle?

Owatch

@JasperLoy If you're interested, search for Sarah Greene.

No, I'm not affiliated with Oracle.

Jasper Loy

@Owatch Sarah Greene is not going to come to this chat to talk to me.

Incurrence

@JasperLoy That is the user Fermé somme now(once Oracle)

Owatch

Afraid not.

Jasper Loy

@Incurrence If I don't succeed in May, I don't know what I would do.

Incurrence

01:00

@JasperLoy Start working on mathematics with me :)

Jasper Loy

@Incurrence I must succeed in May, so I must prepare myself mentally to do the difficult things the next few days...

@Incurrence It is very hard for me to imagine a future. Because I think I only have a 1 per cent chance of getting well and a 1 per cent chance of getting into grad school...

r9m

@Incurrence 88 views in total this month (uptil now) .. before that 46, 57, 105, 168, 134, 22, 104, 67. :P

(total number of monthly views since August 2014)

Incurrence

You have gotta get those blog role buddies I think haha @r9m, get those view counts up

Jasper Loy

@r9m Blog views mean nothing.

Incurrence

The user Eric stucky has a really interesting blog, and he and I link to eachother to share interested traffic

(although he is much more advanced in math than me, and is on the end of a thesis now)

Jasper Loy

01:08

@Incurrence I have finally heard The Three Tenors 1990, 1994, 1998 and 2002.

Incurrence

@JasperLoy I haven't

Jasper Loy

@Incurrence I guess there is nothing else for me to say about next month. I will need to spend time in solitude to muster the necessary strength and courage to do what is necessary.

Incurrence

Indeed

Or you can share more information about it in email if you wish

Disciple of Barney

I am pretty sure spending time in solitude will just make it more difficult to talk to people...

Jasper Loy

@DiscipleofBarney That applies to normal conversations, not abnormal ones.

Incurrence

01:13

Usually if I don't talk to people for awhile(5 hours), I become highly antisocial until people talk to me for 20ish min, so I don't know what would happen if the isolation dragged on for days.

Fortunately I have a girlfriend who breaks me out of my study isolation, otherwise people would think I am weird

Disciple of Barney

I think it applies to abnormal ones too

Jasper Loy

Well, I am going to trust myself on this one.

Karim Mansour

@DiscipleofBarney I think spending time in solitude however makes you better mathematician

Disciple of Barney

@KarimMansour Maybe, probably depends on the person and how they are spending time with people. Collaborating can be very stimulating...

Incurrence

@KarimMansour I have read studies that show strong evidence to support the theory that intellegience is heightened in an environment of continuous social interaction.

Jasper Loy

01:19

I do not think much of these studies, which have many underlying assumptions that may not apply. Social science is different from hard science.

One day, one study says this and the next, another says the opposite.

Many are just trying to publish for the sake of publishing.

You look at the numbers of studies published and you wonder how many of them are rubbish.

Incurrence

Sure, but mostly in non-reputable journals

Jasper Loy

I know I sound like a crank now, but it is OK what you think of me.

Maybe 90 per cent of research is rubbish.

Incurrence

There is a study about this :P

Karim Mansour

working with people is alright, but studying with people I don't think that works really well.

lolz

@Incurrence

Jasper Loy

@Incurrence Maybe I will end up as a beggar on the streets in a few years, still mentally ill.

Incurrence

01:26

@JasperLoy Maybe I will be wealthy enough to get you over here in a few years

Disciple of Barney

@Incurrence Lol, Your planning on going to grad school in math, right?

Jasper Loy

@Incurrence If I had a gf, maybe she could be a great motivation for me to get well, but I don't have one.

Incurrence

@DiscipleofBarney Yes, but my girlfriend is very kind and will be starting on 110k next year

@DiscipleofBarney And grad students in australia get 60k as tutors

Karim Mansour

lol @Incurrence solving h.w for money

haha

Incurrence

She has a mining job with the largest mining company in the world already lined up

Disciple of Barney

01:29

@Incurrence You should rename your account "Gold DIgger" ;P Plus having Jasper over will break your relationship

Incurrence

And tutors here get $60 an hour

Karim Mansour

hahahahaahahaha

Disciple of Barney

Or "Gold Miner"

Incurrence

Hahaha, she said she will go back to uni and do something better after I start earning money

Jasper Loy

I think only @ᴇʏᴇs and @Incurrence understand me well enough in this chat.

Incurrence

01:30

And I might sell out, you never know

Karim Mansour

what about me @JasperLoy ?

I understand you !

Incurrence

@JasperLoy What about ol' skullpatrol?

Jasper Loy

So there isn't really any point in sharing my problems with anyone else.

@Incurrence No, he doesn't understand me. He just says random deep sounding stuff.

Incurrence

@JasperLoy Yes indeed haha. This is called pretending to be wise and there is a really funny chapter in a book I like about htis

Jasper Loy

@Incurrence Many people in my country pretend to be wise but they are actually extremely stupid.

Disciple of Barney

01:33

@JasperLoy I don't think that is restricted to your country

Incurrence

"Well, sounding wise wasn't difficult. It was a lot easier than being intelligent, actually, since you didn't have to say anything surprising or come up with any new insights. You just let your brain's pattern-matching software complete the cliche, using whatever Deep Wisdom you'd stored previously."

Sayan Chattopadhyay

Hi superman@JasperLoy

Incurrence

'"Headmaster," Harry said solemnly, "I would rather not define myself by my enemies."

Somehow, even in the midst of all the whirring and ticking, there was a kind of silence.

That had come out a bit more Deeply Wise than Harry had intended.

"You may be very wise, Harry..." the Headmaster said slowly. "I do wish... that I could have been defined by my friends." The pain in his voice had grown deeper.'

Jasper Loy

@SayanChattopadhyay Hi, calculus is hard. Take your time to master it.

Incurrence

'Harry's mind searched hastily for something else Deeply Wise to say that would soften the unintended force of the blow -

"Or perhaps," Harry said more softly, "it is the foe that makes the Gryffindor, as it is the friend that makes the Hufflepuff, and the ambition that makes the Slytherin. I do know that it is always, in every generation, the puzzle that makes the scientist."'

Sayan Chattopadhyay

01:35

I am taking my time........@JasperLoy

robjohn

@JasperLoy what, my orange is more orangey?

Incurrence

@robjohn I should become the nice square :)

Jasper Loy

I am waiting to install Debian 8 on Apr 25.

Incurrence

@JasperLoy Debian, I have heard of this from my crpytography class

robjohn

@JasperLoy Is that when it is released?

Jasper Loy

01:37

@robjohn Yes, probably. With Debian, you can never be sure, but that's the current date set.

Incurrence

Debian had DES

Jasper Loy

@Incurrence The best linux distro in my opinion.

@Incurrence What is DES?

Incurrence

An old symmetric key algorithm for the encryption of data

robjohn

@JasperLoy nvm en.wikipedia.org/wiki/Debian#Release_timeline

Jasper Loy

@Incurrence Maybe if I were well, I would have won the Fields medal by now, or maybe I would have gotten into an accident and died. Life is unpredictable.

@robjohn Yes, like I said.

Incurrence

01:39

@JasperLoy If I were pretending to be wise, I would say "Yes indeed, life is a mystery, the only way to know what will happen is to let it happen"

Jasper Loy

@Incurrence LOL.

Incurrence

"Let yourself be free Jasper, the only way to free the mind, is to free the heart"

Jasper Loy

@Incurrence LOL

Yes, many "wise" people say that.

Incurrence

I made that up and when you said that, I thought it was an actual thing already so I googled it, but yep, I am original :-)

"When you know yourself truly, then you will know the world"

Jasper Loy

Sometimes, context matters too.

Incurrence

01:42

I am a generator for this garbage :)

Sayan Chattopadhyay

@JasperLoy is there any kind of a problem if I do linear algebra

Incurrence

@SayanChattopadhyay Linear algebra can be done with only set theory as a prerequisite

Jasper Loy

@Incurrence youtube.com/watch?v=bWtUM1J9eSc always inspires me.

@SayanChattopadhyay Finish calculus first.

Sayan Chattopadhyay

I cannot do it that means?@Incurrence

Incurrence

@SayanChattopadhyay You have not done set theory?

Sayan Chattopadhyay

01:44

Done and dusted

Disciple of Barney

@SayanChattopadhyay What @Incurrence means is: do you know what union, intersection, elements, and what it means to be a subset or equal set?

Jasper Loy

I tell him to finish calculus first because he already has enough books.

Sayan Chattopadhyay

No I just have two in which I have finished one @JasperLoy

Incurrence

@SayanChattopadhyay Linear algebra is very pleasant, probably one of the most pleasant subjects

Sayan Chattopadhyay

The rest are soft copies

Jasper Loy

01:46

@SayanChattopadhyay Then finish the other one first, full stop.

Sayan Chattopadhyay

Ok....that means I have to revise continuity and do its exercises again......

Jasper Loy

If you jump here and there, you will never finish anything, and you would have learnt nothing.

Sayan Chattopadhyay

Ok.....

Jasper Loy

Kids these days like to study algebraic geometry without knowing how to solve quadratic equations.

Incurrence

@disc What is the commutator subgroup of $GL_n(\Bbb R)$ I don't understand this concept yet, is it all elements that satisfy: $xyx^{-1}y^{-1}=$ what?

Disciple of Barney

01:48

@Incurrence it is the group generated by those

Incurrence

@DiscipleofBarney Generatred by $xyx^{-1}y^{-1}$?

With what $x,y$?

Disciple of Barney

Every $x,y \in GL_n(\mathbb{R})$

Sayan Chattopadhyay

i taught myself Quadratics when i was in 5 grade @JasperLoy

Incurrence

Can you walk me through some trivial example @Disc

Disciple of Barney

An example of a commutator subgroup? For $GL_n$ or just any group

?

Incurrence

01:51

For any group

I just want to understand what you mean

Jasper Loy

I am going to bed @Incurrence, I will see you in my dreams.

Disciple of Barney

First if we have a group $G$, then the commuator subgroup $[G,G]= \langle xyx^{-1}y^{-1} \mid x,y \in G \rangle$

Semiclassical

@JasperLoy one sort of funny thing: you'd never find a physicist with a knowledge of field theory under their belt not knowing to solve quadratics, if only b/c gaussian integrals are such a huge part of that (and that relies on computing the square)

Sayan Chattopadhyay

Great night

Jasper Loy

It's 10 am here. I will probably go out for a walk tonight.

Sayan Chattopadhyay

01:54

Oh its 5 am here

Jasper Loy

And come back in the morning.

Incurrence

@JasperLoy 10pm you mean?

Oh wait nvm

It is 10am right

Jasper Loy

@Incurrence It is 10 am here now.

Incurrence

You mean a nap now? Or you never slept?

@SayanChattopadhyay Great night is very weird to say lol

Jasper Loy

Well, I don't differentiate between sleep and nap. My sleep is intermittent anyway.

I sleep a while, think a while, and so forth.

Incurrence

01:55

@JasperLoy Alright Good night Jaspy

@DiscipleofBarney I don't get it, I actually have to try every element??

Disciple of Barney

That is what the group equals, I am not sure what you mean by try every element.

I will see if I can find a suitable example

Incurrence

Well if I want to actually build the group I mean

Like enumerate the elements

Disciple of Barney

What do you mean build?

Incurrence

Find all the elements of

Disciple of Barney

basically, yes. Depending on the group though there might be more structure to take advantage of to make it easier

Incurrence

01:59

Or atleast find the classes as arbitrary matrices for example

Disciple of Barney

or to describe the elements

I am fairly sure it is also unsolvable in general, by algorithm, to say whether or not some element is in the commutator subgroup.

Incurrence

Hmmm that is unfortunate

Disciple of Barney

For matrices it probably is not difficult, not 100% sure. For example all such matrices will have det=1

zeror9m

@robjohn does usernames of an account ever change on their own ? :o

Incurrence

@zeror9m Indeed, I changed it my bad

zeror9m

02:03

@Incurrence oo !! how did you do that ? :D

@Incurrence now it won't let me change the username for another 30 days ! :O

Incurrence

@zeror9m Sorry :\

zeror9m

:| .. I'm glad you kept that r9m part atleast :P

Incurrence

@zeror9m Yeah I didn't actually mean to change it for the 30 days, but I didn't change it back fast enough

My method gives me a minute I am fairly sure(but haven't verified)

zeror9m

sigh :|

Incurrence

@zeror9m The names not that bad anyway

zeror9m

02:15

@Incurrence :) that was a relief :P

Incurrence

@zeror9m just making you feel better :)

zeror9m

lol

Sayan Chattopadhyay

@Incurrence you know lie groups.......sorry this is the question out of my reach because the name always makes me feel really weird about it......what are they.....

Incurrence

@SayanChattopadhyay I don't know to be honest. I have worked with them, but I have never worked with manifolds

Sayan Chattopadhyay

They are groups on manifolds?

Incurrence

02:25

@SayanChattopadhyay I don't even know what a manifold is sorry

@SayanChattopadhyay They are topological spaces of some kind, but I have never worked with them

Sayan Chattopadhyay

Oh......

Incurrence

Yeah, all I know is, Lie algebras have something to do with a differentiable topological space

And they make for interesting exercises in algebra :P

Sayan Chattopadhyay

Lol....

Incurrence

Oh, don't mistake me for an expert if that is what you have done

I am a third year student lol

Sayan Chattopadhyay

For me everyone who learns math with passion is an expert

You can change names.....how?

Disciple of Barney

02:39

@Incurrence mathb.in/31641

Karim Mansour

@DiscipleofBarney given a group G what is there a way to figure out which topology one can assign to it to make it a topological group ?

Disciple of Barney

@KarimMansour The discrete topology always works, but I suspect it is difficult (maybe impossible), to classify all the topologies one can give.

Karim Mansour

I see that would be an interesting question

Disciple of Barney

Did you take a look at that paper I posted a while ago (we were talking about these sort of thing, I believe with David)

@KarimMansour

Karim Mansour

no @DiscipleofBarney

@DiscipleofBarney do you still have the source maybe I want to have a look on it

Disciple of Barney

02:50

You should, the first couple of pages describe what they are answering and some things that are known about these sort of problems (in a reasonably non technical way), plus it could give you some ideas of what to study further.

Karim Mansour

yeah

Disciple of Barney

@KarimMansour arxiv.org/abs/1210.7895

Karim Mansour

Thank you @DiscipleofBarney

user196651

Yo I have a math question.

So the question is to utilize the uniqueness of prime factorization in $\mathbb{Z}[i]$, show that if $\tan z_1 = 1$ and $\tan z_2 = 2$, then $z_1$ and $z_2$ are linearly independent.

My work so far: observe that $z_1 = \text{arg}(i+1)$, $z_2 = \text{arg}(1+2i)$. Suppose that there exists $c_1, c_2 \in \mathbb{Z}$ such that $c_1z_1 + c_2z_2 = 0$ which is true iff $(2+i)^{c_i}(1+i)^{c_2} = k$ for some $k \in \mathbb{Z}$, which clearly isn't possible.

I am confused what to do next??

Incurrence

@DiscipleofBarney Thanks, I'll read over it properly after lunch

Disciple of Barney

02:56

alright it is just a simple example of commutator subgroups.

user196651

Can anyone answer my question???

robjohn

03:55

@zeror9m they don't, but if someone has an objectionable username, the moderators may change it.

zeror9m

04:14

@robjohn Incurrence seems to have changed my username somehow .. :| 'zeror9m' I don't quite like the ring to it :P is there a way I can revert it back to r9m before 30 days ?

Disciple of Barney

@zeror9m Well zeror9m is an extremely derogatory word, so they will probably have to change it... ;)

zeror9m

:P lol ... okay

cxseven

there's a nice formula for the minimum value of a quadratic polynomial with positive leading coefficient.. is the same true for critical values of quintics?

if not the minimum, I wonder if substituting the roots of the derivative, for which there is an explicit formula, into the original polynomial will simplify

zeror9m

04:31

@DiscipleofBarney is this considered derogatory ? :P

Disciple of Barney

@zeror9m Haha, I did find a post sort of about that though, meta.math.stackexchange.com/a/8889/29123

ADG

04:53

wassup?

@robjohn you remember the a,a^2,b,b^2 question?

Disciple of Barney

05:13

If I turned super meat boy into a drinking game, and had a shot every time I died, I would die.

cheeseinthetrap

Hello, I'm new to this community and I've posted a question on Math.Stackexchange. If anyone is interested, can anyone give me advice regarding my question? math.stackexchange.com/questions/1227983/…

It's about the importance of exercises for self-studying

Disciple of Barney

@cheeseinthetrap This question seeems relevent (I didn't read your whole question) and there are a lot of other similar questions on the site you might want to browse.

Many of those questions can be seen on the "side bar" related on the far right of the screen, for your question and the one I linked @cheeseinthetrap

cheeseinthetrap

Thank you very much @DiscipleofBarney

Disciple of Barney

05:30

@cheeseinthetrap It sort of seems like the guy was maybe having the opposite problem you are having, wondering if they should lay off the exercises a bit.

cheeseinthetrap

05:40

@DiscipleofBarney Yes, it is interesting. I guess both extremas are not optimal. I just hope I didn't join the party too late :)

Balarka Sen

05:53

@KarimMansour you can endow stuff with profinite topology.

Karim Mansour

can you explain I am not familiar with the terms @BalarkaSen

Balarka Sen

it's not very interesting as a topology, so i dunno if this is what you want but let me try :

Singh

0

A: Eight points are in/on the circle of radius 1cm. Show that distance between some two points is less than 1cm.

I agree with your brother. You have shown that your particular placement works. You have not shown that any placement works. This might work: In your heptagon, since you are putting 8 points into 7 segments, there must be a segment with two points. If you can show that the diameter of each segme...

Kindly see this.

Balarka Sen

$G$ be a group. the topology obtained by considering the basis of left cosets of subgroups of finite index is called the profinite topology on $G$.

so, for finite subgroups, the discrete topology coincides with profinite topology.

Karim Mansour

oh I see

Balarka Sen

06:00

a nontrivial example of a profinite group is $\mathbf{Z}_p$

are you familiar with inverse limits?

Karim Mansour

yeah

Balarka Sen

$\mathbf{Z}_p$ is defined to be inverse limit of $\mathbb Z/p^n$s for all $n$, with obvious bonding maps.

It's called the group of p-adic integers.

What's more, the profinite topology of $\mathbf{Z}_p$ is actually ultrametrizable. Google "p-adic norm"

Karim Mansour

that is awesome

Balarka Sen

my favorite example of a profinite group is $Gal(\overline{\Bbb Q}/\Bbb Q)$, the group of automorphisms of the algebraic numbers fixing rational numbers pointwise.

you'll get to know about them when you study galois theory.

Karim Mansour

that is awesome I am doing that class next year

next fall I should say

but I will study them independently over the winter.

Balarka Sen

06:06

nice.

@DiscipleofBarney I have kind of wondered if there is a Uryshon-like theorem for ultrametrizability of profinite spaces. You know anything about them?

I just remembered that I wanted an answer for it a long time ago.

Disciple of Barney

@BalarkaSen Maybe I am misunderstanding you, but all ultrametric spaces are metric spaces which are all normal

Maybe I am thinking of a different theorem

Balarka Sen

I am aware of that, but it doesn't say anything about my question.

Uryshon says if a space is Hausdorff and second countable, then it's metrizable. A profinite space is both of them.

So every profinite space is metrizable -- doesn't necessarily mean it's ultrametrizable.

Disciple of Barney

Okay, I see what you are asking now.

robjohn

07:45

@DiscipleofBarney why is your avatar not purple?

Oh, it is... I hadn't looked close enough

@zeror9m no sweat. Chat is all one thing; a mod on any site is a mod on all chat sites.

zeror9m

@robjohn ? but is Incurrence is not a mod is he ? :o he seems to know a way of changing my username without letting me know :|

Jasper Loy

@robjohn Hi, it is Friday today.

robjohn

@zeror9m The record looks to say that you changed your name.

zeror9m

@robjohn :O I didn't .. not to my knowledge :( chat.stackexchange.com/transcript/message/21016454#21016454 did you see the deleted message after the linked one ? :o

robjohn

@zeror9m there is no way they could have changed it.

Jasper Loy

08:00

@zeror9m Nobody can change unless they have access to your account.

zeror9m

I was surprised ..

Jasper Loy

However, I can change my username before 30 days is over.

robjohn

@zeror9m That doesn't say anything

Jasper Loy

@zeror9m If that really happened, he has your password.

zeror9m

@robjohn oo .. holy cow !

Jasper Loy

08:05

@robjohn You look as delicious as an orange.

robjohn

@zeror9m he names a script. I have no idea what that script does, but I doubt it changed your username.

zeror9m

@robjohn could you tell me what it was ? :o

Jasper Loy

@zeror9m If he really did that, he is very smart.

zeror9m

mmm .. okay !

Anthony

08:22

@robjohn You there?

robjohn

@Anthony what's up?

Anthony

Do you know very much functional analysis?

robjohn

@Anthony not a lot.

Anthony

Ah, nevermind then.

Here's a different question.

How do I minimize the maximum of $|\sqrt(x)-ax-b|$?

Where $x$ is in the unit interval, and a and b are free to vary?

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