Mathematics - 2015-04-24 (page 1 of 3)

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12:00 AM

@user perhaps noone knows the answer.

user143442

maybe :-(

@user have you posted on main?

There may be more topologists there.

user143442

I think it's just a detail, not so difficult to post it, and also they will ask what I tried and I don't have an answer to that because I don't even know how to start the proof

@user That is a problem with the current climate here. You can't have a question that you don't have any idea about.

user143442

Well I have an idea of what I have to proof, but I don't know how to prove it

12:13 AM

I hope that some day it will again be possible to ask a question about which one is totally baffled.

user143442

I don't know which method is convinient

@user Perhaps that could be made into something to add to the question.

We have $T=x^2e^y-xy^3$, where $x=\cos t$, $y=\sin t$ and we want to find $\frac{dT}{dt}$ using the chain rule.

Do we write at the beginning
$T(t)=x(t)^2e^{y(t)}-x(t)y^3(t)$ and then $\frac{dT}{dt}=\frac{\partial{T}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{T}}{\partial{y}}\frac{dy}{dt}$
or
$T(t)=f(x(t),y(t))$, where $f(x,y)=x^2e^y-xy^3$ and then $\frac{dT}{dt}=\frac{\partial{f}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{f}}{\partial{y}}\frac{dy}{dt}$ ?? @robjohn

Usually when I just sit and stare at something I get some sort of ideas (albeit totally incorrect approaches that don't help at all) but at least it's something

Disciple of Barney

@user Why don't you try (both methods even), I mean what is the point of doing a problem you know how to solve? Then you will better understand the problem and how to approach the problem or ask a better question.

12:17 AM

Both of those end in $\frac{dT}{dt}= \frac{\partial{T}}{\partial{x}}\frac{dx}{dt} +\frac{\partial{T}}{\partial{y}}\frac{dy}{dt}$. Is there a difference?

user143442

I tried but it's confusing, I don't get any conclusion, nothing I could post as a try.

@MaryStar $$\frac{\partial T}{\partial x}=2xe^y-y^3$$ and $$\frac{\partial T}{\partial y}=x^2e^y-3xy^2$$

So, both ways to formulate it are correct, right? @robjohn

Or is one of them better?

@MaryStar I don't see the difference.

@MaryStar They are both $f(x,y)$ with $x(t)$ and $y(t)$.

Sorry for all the trivial questions @Disc, but the elements of $S_3$ under multiplication are just all permutations using elements $\{1,2,3\}$, so we can write them all in cycle notation, for example: $(12)(3)$ or $(123)$ or $(1)(2)(3)=Id$

Disciple of Barney

12:31 AM

@Incurrence Yes. (although it is normally a convention to leave out the "1-cycles")

The theorem of the chain rule is the following:
We suppose that $c: \mathbb{R} \rightarrow \mathbb{R}^3$ and $f: \mathbb{R}^3 \rightarrow \mathbb{R}$. Let $h(t)=f(c(t))=f(x(t),y(t),z(t))$, where $c(t)=(x(t),y(t),z(t))$. Then $$\frac{dh}{dt}=\frac{\partial{f}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{f}}{\partial{y}}\frac{dy}{dt}+\frac{\partial{f}}{\partial{z}}\frac{dz}{dt}$$

So can we take at $\frac{dT}{dt}=\frac{\partial{T}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{T}}{\partial{y}}\frac{dy}{dt}$ "T" both at $\frac{\partial{T}}{\partial{x}}$, $\frac{\partial{T}}{\partial{y}}$ and at $\frac…

@DiscipleofBarney Yes that makes sense

@DiscipleofBarney Thanks

What is the generator of $S_3$?

Disciple of Barney

What do you mean? @Incurrence

There are lots of generating sets

And $S_3$ is not cyclic so there isn't a single generator that generates the group

Okay so let me just type out what I understand to make sure I am write so far

We have a group $G$ and then we can have a cyclic subgroup $H$ generated by some element $x\in G$ such that $H=\{\cdots, x^{-2}, x^{-1}, 1, x ,x^2, \cdots\}$ where $\langle x\rangle$ here is the generator

and since $S_3= \{(),(12),(13),(23),(123),(132)\}$

We need more than one generator

$\langle x,y,z | something\rangle$ probably

Disciple of Barney

"... and since $S_3= blah$ we need more than one generator" What do you mean by that?

12:41 AM

That all of that set $S_3$ can be made by the union of what is created by multiple generators

But I don't know what they are

Like the even $(\Bbb Z^+\cup \{0\},+)\subset \langle2 \rangle$

Under addition

Disciple of Barney

Does $\Bbb Z ^+$ mean integers under addition?

It means positive integers

Disciple of Barney

Well that isn't a group

(although you can extend the same idea to things other than groups)

Oh no inverse still

well

anyway I failed in my example

But my question was

Disciple of Barney

And it isn't a subset

12:45 AM

Can we have $G\not\subset \langle x_1\rangle$, $G\subset \langle x_1 \rangle \cup \langle x_2 \rangle$

Disciple of Barney

Yes

So what are the $\langle x_1 \rangle <x_2>$ etc needed to make $S_3$?

Disciple of Barney

Its known that all the transpostions generate the group. Or a transposition and a 3-cycle generate the group

Umm, so $(123)$ and $(12)$

Disciple of Barney

So $\langle (1,2), (1,2,3) \rangle=S_3$

12:49 AM

Now if we are to call these two elements(?) x,y respectively, we would have elements of order 2 and 3?

Disciple of Barney

Yes

and also, how is this written out as a set? Since normally we only have one thing $<x> = \{\cdots, x^{-2}, x^{-1}, 1, x^1, x^2,\cdots \}$

Disciple of Barney

So you know, using those generators you would have something like $\langle x, y \mid x^2,y^3,... \rangle$

Oh I see

One second I'll think about that

Disciple of Barney

For a lot of reasons, it isn't always easy to find a presentation or prove that the presentation you have actually presents your desired group, just so you know.

For example it is algorithmically not decidable to show that whether arbitrary presentations present the trivial group or do not present the trivial group.

12:55 AM

We compute permutations left to right, correct? $(123)(12): 123\mapsto 132$ right

Or do we do right to left

I always do right to left

Karl Kronenfeld

I flip a coin

$(123)(12): 123 \mapsto 321$

Disciple of Barney

You can do either, as long as you are consistent. It is not necessarily a universal convention

So there is no universal notation............

That is horrible in my opinion hahah

Disciple of Barney

12:56 AM

I think most people do right to left though

(but expected I suppose)

Right to left, okay

Sad thought

Left to right made my presentation almost on first try

But I guess I can just inverse the operations(?)

Yes indeed

So all of my elements so far $x^2=y^3=(),x=(12),(13),xy=(23),y=(123),yx=(132)$

Karl Kronenfeld

right to left should be the prevailing convention since permutations are, you know, functions.

Disciple of Barney

In group theory circles, I don't think it is uncommon (maybe even the majority, although I don't know) for left to right @KarlKronenfeld

Karl Kronenfeld

yeah, I recall martin Isaacs does it left to right (threw me way off at first) in the two books I have read some of

Disciple of Barney

Plus everyone knows the right to left convention for functions sucks (at least in written english)

Karl Kronenfeld

1:01 AM

:D

just draw arrow diagrams and there will be no problems :trollface:

Hi @PedroTamaroff

I wonder if functions would be less confusing to learn about if a left to right convention was in place @DiscipleofBarney. Indeed $(x)f$ looks a whole lot more like "take x and apply f to it" than does $f(x)$.

That is how cohn does it

He does $xfg=g(f(x))$

It is probably better, but not after learning solely using the right side of above for all of your prior education

Karl Kronenfeld

(left/right) Inverses actually actually would make sense; every once in a while I picture the actual operation taking place and get it wrong

@Incurrence That's the problem with conventions you disagree with. If you follow them then you are promoting them. If you don't then you confuse your readers.

@KarlKronenfeld Exactly

@KarlKronenfeld I can't exactly just use the (arguably better) left to right convention when talking to someone I have just met each time, since I would have to explain it for a few minutes at the start

So I can't spread it

Karl Kronenfeld

heh

Disciple of Barney

1:16 AM

Not sure, it might make it easier, although I think there are probably other more prevalent barriers for understanding functions, for example people making "the" jump from everything is a number and addition and multiplication to the more abstract ideas.

Karl Kronenfeld

@DiscipleofBarney Yeah, you're right. I sometimes forget about how ridiculous math must seem to be a beginner.

@KarlKronenfeld Hey Karl!

Kaaaaaaaarl.

YES got 95% for my functional assignment :))

::::::::::::::::)))))))))

@KarlKronenfeld I begun writing in a blog, see my profile. =)

Karl Kronenfeld

I'll check it out

Disciple of Barney

1:19 AM

Because $f(2)/f(5)$ we can cancel the $f$ and get $2/5$ :P @KarlKronenfeld

@PedroTamaroff It's very pretty xD

Disciple of Barney

I like that you commutative diagrams are hand written :D @PedroTamaroff (although not a big fan of the picture thingamajig homepage)

@DiscipleofBarney They are handwritten because I'm too lazy to learn tikZ, also because the latex supported by WP is poop.

@DiscipleofBarney I am thinking about changing to a setting that shows the post immediately.

I am not sure how to do it, though.

@PedroTamaroff Indeed. It is almost worth typing it all up elsewhere and posting a picture, but then the text isn't search available

So simple groups @Disc are just groups that only have themselves and the trivial subgroup as subgroups

E.g. any subgroup that is solely generated by a cycle is a simple group

Disciple of Barney

As normal subgroups

1:28 AM

Ahhhhh

Okay let me think about this

So if there is no quotient group for it, e.g. no cosets, then it is simple?

Disciple of Barney

Yes (other than the trivial quotients you already mentioned)

So $\Bbb Z / 3 \Bbb Z$ can't have a new quotient group taken from it, and hence it is simple

Disciple of Barney

Yes

infact any $\Bbb Z_p$ where $p$ is prime is possibly simple

Disciple of Barney

I think I might have to get on the math blog bandwagon...

1:31 AM

@DiscipleofBarney Hahaha

Disciple of Barney

Yes

@DiscipleofBarney I think it is nice to read other peoples progress for motivation, and other peoples expositions, and with wordpress, every time I login I can see all of my followed peoples posts

@Disc In fact we can prove my above statement by Lagranges theorem: Since $\Bbb Z_p$ has prime order, and a subgroup's order must divide the order of the main group, no subgroup has an order that divides the main group($\Bbb Z_p$)

(except the trivial subgroups)

Disciple of Barney

Yup @Incurrence

I am getting somewhere at last ahaha

Karl Kronenfeld

Some cute results @PedroTamaroff I didn't read your posts thoroughly since I want to look into them on my own. That said, my main criticism is you should decide on a good pace for your blog posts. Slow things down around your main results and throw in your opinions, etc., and keep a fast pace around your unimportant results.

1:42 AM

How do I pronounce the unfortunate name(when reading in an Australian accent atleast): Jacques Tits

Karl Kronenfeld

:)

Disciple of Barney

Not sure, I think I have heard Teets (so no better

Alright, I can just imagine if I presented on the simple groups and I mispronounced it like that and someone knew(how to pronounce it properly).......

Disciple of Barney

Sort of funny I had a wiki page opened on the Tits alternative

I was reading about the tits group en.wikipedia.org/wiki/Tits_group

I read her name like Jackie's tits as well, so (I probably have it even worse)

Disciple of Barney

1:45 AM

The pronouce guide does say teets like as in bee, Its a guy

Male

OH that's a relief then, not so awkward

I thought Jacques was a female name

Disciple of Barney

Well its french, so...

I don't know much French or any(?) French specific names

Karl Kronenfeld

it's their version of James

(if we can trust wikipedia on this one)

Disciple of Barney

Well I was wrong about that

1:48 AM

Jark

Jark Teets

@KarlKronenfeld Could you elaborate on what "a good pace" is?

Perhaps you mean avoid "spoiling things" too fast?

Karl Kronenfeld

Hm.

Disciple of Barney

Its sort of like "shock" but with a sort of J equiv to sh, like a jh I guess

Hi @Incurrence @PedroTamaroff

Karl Kronenfeld

It's more about writing style than the content. I'm still trying to think of a different way to say it.

1:52 AM

@KarlKronenfeld OK. I don't mind some tough love, though. =)

@Rememberme Who are you?

@PedroTamaroff Sayan Chattopadhyay

Karl Kronenfeld

@PedroTamaroff I don't mind dishing it out :P It'll just take some effort to explain I think.

Why did you give up on your monster avatars??

@KarlKronenfeld OK.

Karl Kronenfeld

Here are two ways of saying the same thing: "The glass is full." "I received a spectacular crystal glass from my aunt twenty years ago. Today I still use it. In fact, I just now filled it up with water."

1:57 AM

Hey. Does anyone know a good program for mathematical poster presentations

?

Disciple of Barney

@JulianRachman Do it in LaTeX with Beamer

Karl Kronenfeld

@PedroTamaroff The latter shows that I am taking interest in everything glass-is-full-related while the former shows that the fact that the glass is full is unimportant and merely contributes to a more important piece of the narrative.

2

@DiscipleofBarney What is Beamer?

@KarlKronenfeld Wait, aren't former and latter swapped there?

@JulianRachman what do you mean by poster representations

Karl Kronenfeld

1:59 AM

@PedroTamaroff I used it right. Former being the terse sentence, latter being the embellished description.

Disciple of Barney

@JulianRachman here, Beamer is sort of the "powerpoint/poster" package for latex

There are other ways, I am sure

@Rememberme I said presentations

Anyways what are they

@DiscipleofBarney Ok. Thanks! I want it to look like this: student.societyforscience.org/sites/…

Disciple of Barney

Definetly done in latex, probably with Beamer

2:01 AM

@KarlKronenfeld I don't see how the latter sentence shows you're interested in every glass-is-full-related.

In fact, it seems like you just said the glass is full, but you're also interested in the glass having a story.

And hence the fact the glass is full is not that relevant.

Disciple of Barney

@JulianRachman I think that is a template you could probably find floating around if you look enough

@DiscipleofBarney Ok. Thanks! I'll play around with it.

Karl Kronenfeld

@PedroTamaroff There are two questions I asked myself when I stated "the glass is full". What is in it? and Which glass is it?

@KarlKronenfeld OK.

I think I'd help me more if you gave me a concrete example of anything in the post.

Hey @Remem

2:06 AM

I am sometimes a bit obtuse when it comes to writing. I try not to be too poetic and iffy, which might make the post be slightly dull and to the point.

@Incurrence I want to know how much background do I require for incident combinatorics

@Rememberme I don't know what that is sorry

Karl Kronenfeld

I noticed that you are too willing to state related, interesting things that you don't need yet. I guess it is what you were saying about spilling details too early, but that has a different meaning to me.

@Incurrence no worries

Karl Kronenfeld

Something specific, do you reference the notion of polynomial identity rings after that first instance?

2:10 AM

@KarlKronenfeld Come again?

Karl Kronenfeld

Oh, right, I didn't say which post.

Yes, I know which post.

But I didn't understand your question.

Karl Kronenfeld

@PedroTamaroff After you point out that this means they are polynomial identity rings, do you bring up that concept again?

@KarlKronenfeld No. I mean, I define what it means for something to be a PI ring, but just for the record.

Karl Kronenfeld

It's a cool idea, but I'd leave it out since it is tangential to the blog post.

2:12 AM

@KarlKronenfeld OK.

Disciple of Barney

@KarlKronenfeld I haven't read what you are talking about but it sounds like your are saying "You are too willing to give your motivation and connections to other ideas"

@KarlKronenfeld It is mildly relevant in the sense it is a keyword related to the theorem.

What does the word invariant mean?

That the invariant thing doesn't change some conditions?

Karl Kronenfeld

@DiscipleofBarney ...around unimportant parts of the post. If it is the most important part (here: the changeover from ring theory to combinatorics) then do provide motivation/connections/etc.

Disciple of Barney

Not changed, although context is sort of important @Incurrence

2:14 AM

@DiscipleofBarney Karl is looking at this, I think.

What are some examples(if you can think of any)

@PedroTamaroff Indeed - I searched through all 4

@Incurrence Invariant subspaces of vector spaces, w.r.t. to some endomorphism, for example.

An endomorphism is a self homomorphism?

Disciple of Barney

@Incurrence Yes

@Incurrence An endomorphism is a morphism from a thing to itself.

So, yes.

2:15 AM

Why do you just say morphism in place of homomorphism?

Since an isomorphism to itself is an automorphism(I think)

Is that from category theory?

@Incurrence It's a wider reaching word.

Karl Kronenfeld

@DiscipleofBarney I am interested in whether you would keep that in though.

After that, I have to go. I am not sure I have helped @PedroTamaroff. :P But, I am glad you are writing a blog, and it already has some really cool stuff.

@KarlKronenfeld Well, I still appreciate the feedback.

It is boring to write and get no feedback.

Disciple of Barney

@KarlKronenfeld I would keep it

2:37 AM

So all the cosets are found by taking a subgroup of $G$, called $H$, and taking all elements $g\in G$ and operating(for a left coset) them on the left of the subgroup?

And those would be called left cosets of $H$ in $G$

Yes, the left cosets of $H$ in $G$ are the sets $gH = \{gh \mid h \in H\}$ for all $g \in G$.

And if these cosets are normal, then we can take a quotient group using them

Or what I mean is $G/ H$ iff $H$ is normal

Mmm, that doesn't quite make sense. It's not the cosets that are normal, but the subgroup that is normal. If the group $H$ is normal in $G$, then the cosets of $H$ in $G$ form a group.

Hmmmm okay

Ummmm

Okay so, $G/H$ is the set of all left cosets of $H$ in $G$ - so we are partitioning all the elements into these equivalence cases

Yes. But $G/H$ only makes sense as a group if $H$ is normal. Otherwise, you can't multiply equivalence class representatives in a well-defined way.

2:41 AM

Okay that helps

Thank you

Sure! Happy to help. Here for a bit if you have more questions.

They come often at the moment xD

Pretty much just reading up on all of the things I badly understand and whatever I can't work out from Dummit and Foote/Artin/Wiki I ask on here

Algebra can be really confusing during your first pass. :) Don't sweat it.

If $H$ isn't normal in $G$, then $G/H$ is not a quotient group, is it fair to refer to it as a quotient? Or is it meaningless

It's not meaningless, per se. The set of cosets of any old subgroup can be interesting in some ways in their own right. But you'll mainly be interested in the case when $H$ is normal.

2:48 AM

Fair enough

I must go to class now, I will talk later, thanks again for the help

For example, Lagrange's theorem (the order of a subgroup $H$ divides the order of the group $G$) is often proven using a coset argument, which works regardless of whether $H$ is normal.

Certainly! Talk to you later.

@Ted: Do exotic $\Bbb R^4$s carry almost complex structures?

Disciple of Barney

@Incurrence More generally cosets are important for studying groups acting on things

3:18 AM

@Ted: Nevermind, obviously yes.

Could you exaplin to me what it means that $y(x)$ is defined implicitly by $F(x,y(x))=0$ ? @robjohn

4:17 AM

@MaryStar, take for example the equation for the unit circle defined by $x^2 + y^2 = 1.$ here $y$ is defined implicitly by that equation.

4:56 AM

you guys gotta help

@DiscipleofBarney these are the best math jokes i have heard:-
You might be a mathematician if you think fog is a composition.
Old Macdonald had a form; ei /\ ei = 0

Disciple of Barney

@Rememberme Those are okay... short of...

My favorite one-liner:

Why did the mathematician name his dog "Cauchy"? Because he left a residue at every pole.
@DiscipleofBarney

5:15 AM

The best math jokes are the ones told awkwardly in lectures, although I guess then it's just technically math humor.

I still remember my number theory professor talking about primality testing. He was a bouncer at a nightclub, and only primes were allowed in.

But it took too long to check each number, so he wanted tests that were probably "good enough"

So to this day, I still imagine him bouncing composite numbers, and all the interesting primes he must have met...

Good one @pjs36

5:30 AM

According to algebraic topology can i make a circle into a square??

Does anyone know an answer to this math.stackexchange.com/questions/1249431/…

Disciple of Barney

Q: Why did the chicken cross the road?
A: The answer is trivial and is left as an exercise for the reader. @Rememberme

Really how do you think of them?

Disciple of Barney

5:45 AM

It was a math joke. Are you still not reading whole section? :P @Rememberme

@Rememberme You don't really know algebraic topology, so stop asking questions about it.

3

Wow! "For two centuries, until the recent discovery of the error in 2005, books, paintings and articles have incorrectly shown a side-view portrait of the obscure French politician Louis Legendre (1752–1797) as that of the mathematician Legendre." Source

Disciple of Barney

You can continuously deform a circle into a square, no algebraic topology needed @KarimMansour

Sorry @Balarka

Disciple of Barney

@pjs36 That is pretty crazy. Sort of strange that people didn't know it was a politician for such a long time.

@Rememberme You can continuously deform a circle into a square, no algebraic topology needed, sent it to the wrong person, lol

@KarimMansour ment to send that to "Rememberme"

5:50 AM

Well, it all started when I had to look up who the geometer Chasles, only to find his name was inscribed in the Eiffel tower, and browsed the inscribed names...

And couldn't help but notice that Lagrange looked like a banshee in his portrait...

Legendre**, whoops...

iwriteonbananas

@BalarkaSen lol

Interesting. I don't think existence of some Mayer-Vietoris long exact sequence is equivalent to excision, but I don't know of a homology theory in which excision doesn't hold but Mayer-Vietoris does, either.

Odd.

@BalarkaSen they are many questions in math overflow you can answer them about algebraic topology

hm?

Yes 41 questions in a week!!

5:57 AM

i don't know what you mean.

You were saying right you wanted to answer good questions about algebraic topology on the main?

yes.

MO's questions are overly good. I am not familiar with that much algebraic topology.

Oh.....

They are research-level, @remember, very unlike MSE.

@Balarka You know about incident Combinatorics

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