Is there a nice way to see that the center of $\langle x,y|x^2=y^3\rangle$ is $x^2$?

obviously $x^2$ is in the center

@user193319 let $f\in S_n$. If $f$ isn't the identity, it moves an element. Can you find a function $g\in S_n$ which doesn't commute with $f$?

NOT SO FAST, need $n>2$.

@Prototank Sorry, forgot to mention that $n \ge 3$; and to answer your question, I did that proof already. THe book I'm using wants me to prove it in the way i've attempted (in fact, this will be the second time this problem appeared in my book).

Hi. I have a question related to protocol. I posted a problem and my wrong answer at : math.stackexchange.com/questions/2652420/…

I know have what I believe is the right answer. Should I post the right answer there? or as a new question? or do nothing?

user193319 where is your proof?

There's really no positive to posting your own answer

You can either edit the question or else answer it and mark it community wiki.

Definitely NOT a new question.

unless it is a topic that other people are actually curious about

hi chat

@TedShifrin Hi Ted :(

I posted my correct answer.

So, my algebra pset has a problem involving a differential equation. It's an admittedly simple one, but I wonder if this is gonna be a problem for anyone since I'm not sure if everyone has seen this before

@Kasmir: Why do you put :( ... that is the opposite of a smile.

What kind of differential equation, Demonark? I mean, you guys should know some basic stuff! I don't care how fancy and highfaluting you are!

@TedShifrin Because I meant that ! -.- I have not been well for past week

fever and caught

:(

Oh dear ... Is the 'flu over there in Europe too?

But you directed it at me ... :(

its eveywhere :( mum got it too

Oh i only meant , Hi Ted, am sick :(

><

I'm sorry. I have excruciating pain in my neck, but luckily (so far) no 'flu.

I mean it's exceedingly basic, just solving $f'(x) = (\lambda - 1)f(x)$. It's just that you probably want to know that you have uniqueness in order to fully conclude what you want in the problem

Now this isn't really a question of fanciness

I hope you get better ! :D

You're supposed to know basics of separable differential equations from ordinary integral calculus, Demonark.

It's more, some people have only ever taken first year calculus and possibly never seen a differential equation before whatsoever

Divide by $f(x)$ and integrate.

You know $\int (1/u)du$ from calculus. Come on.

You do not need anything fancy or any theory of ODEs for this.

I mean I know how to do it of course, it's more if people are able to rigorously justify it (e.g. what if $f$ is 0? Can we divide then? etc)

Do you know the answer to that?

$f(x) = Ae^{(\lambda - 1)x}$

I meant the question you just asked me.

Well, what I did for that was just obtain $f$ by dividing naively, show that it works in hindsight, and then say well, we have uniqueness

So you should be able to argue rigorously that if $f(x_0)=0$ for some $x_0$, then $f$ is identically $0$.

Of course, if $\lambda = 1$, then $f$ must be constant.

Yeah, what I'm wondering about is how people are going to be able to approach this without knowing uniqueness. The context is that we have $V$ as the real functions of the form $f(x)e^x$ where $f$ is a polynomial of degree $\le n$. Differentiation is a linear map on this space, what are the eigenvalues?

@TedShifrin are you on any medications for that Professor?

The vibe I get is that in order to solve this you should know that a solution to the above ODE necessarily has to be of the form I mentioned above

No, @skull ... just lots of exercises ...

Demonark: Most students won't worry. You don't need anything fancy, though. Just a trick will prove it. Call that constant $\lambda-1 = \mu$ and consider $g(x)=e^{-\mu x}f(x)$.

exercise is good :)

\o @nitsua60

hiya

Ah that's slick

It's basically the integrating factor idea that shows up for more complicated first-order linear things.

hello @TedShifrin

What's the integrating factor idea? Doesn't ring a bell offhand

I was wondering if you have something to tell me about the following theorem.

hi Karim

Hey @Adeek!

A meromorphic function in the extended complex plane are rational functions.

@Daminark hey my man :)

Demonark: Suppose you want to integrate $y' + a(x)y = b(x)$. You say to yourself, "Self, what do I wish I had in order to get the left-hand side to look like product rule?"

Okay so we can prove it by subtracting poles both normal and infinity

is there a picture for the the theorem above ?

A picture? No ... but you're doing holomorphic maps $\Bbb P^1\to\Bbb P^1$.

I see

this seems to be an important theorem

in what context does it come up ?

So you want to say something like, $c(x)b(x) = c(x)y' + c(x)a(x)y$ where $c(x)a(x) = c'(x)$?

Right, Demonark. That $c(x)$ is the integrating factor.

Karim: Standard complex analysis qualifying exam questions are to classify various sorts of things. Bijective holomorphic maps. Holomorphic maps from the disk to the disk. Holomorphic maps from $\Bbb P^1$ to itself ...

I mean, classification is a global mathematical thing.

I see

Oh and $c(x)$ can be given by $c(x) = Ae^{\int a(x) dx}$

Yup, Demonark.

Anyone able to look at my group theory?

I'm trying to compute the center of $\langle x,y|x^2=y^3\rangle$

Which, I am pretty sure it is $\langle x^2\rangle$

You say group theory and Mathein descends immediately.

@TedShifrin haha

Wow, TIL "John Wallis can be considered as the inventor of the number line for negative quantities"

So we have $y = \frac{1}{c(x)}(\int c(x)b(x) dx + k)$ which can just be computed

And I guess you can just absorb the constant and say $c(x) = e^{\int a(x) dx}$

You didn't need the constant in the first place. We only needed a function.

Ah true

@TedShifrin I love complex analysis

It is awesome

I look back on my complex analysis fondly

Yeah, complex was fun

It is invaluable as an undergraduate course, too

yeah even grad complex analysis is so amazing

I am happy taking i

you get to talk about loops, analysis concepts obviously, manifold type ideas etc.

yeh

fractals/dynamics

special functions

geometry...

it is all there

number theory, too

oh yeah

@Prototank It's clear that $x^2$ is contained in the center. The quotient of that group by $\langle x^2 \rangle$ is $\langle x,y \mid x^2=y^3=1\rangle$ this quotient is the free product $C_2 * C_3$ of a cyclic group of order $2$ and order $3$ (random fact: that's isomorphic to the modular group $\operatorname{PSL}_2(\Bbb{Z})$ and it's easy to check that a free product of two nontrivial groups has trivial center

If there was any element in the center of $\langle x,y \mid x^2=y^3=1 \rangle$ in the center that is not contained in $\langle x^2 \rangle$, then this would be a nontrivial central element in the quotient

I like complex analysis, too. I was asked to TA it next semester, but I'll probably TA number theory instead

"As to a continuous body of water corresponds a continuous wetness, so to a continuous magnitude corresponds a continuous number."

...to find out who is the father of "number line" idea...

yes

you're right, that was typo

thanks for the help

I'm trying to understand why you can split a particular manifold along two different tori and get different decompositions

and it was related to this toy example that I cooked up

shouldn't the $S^1\times\lbrace 0\rbrace$ fiber have multiplicity 1, and all others have multiplicity $q$?

I can't help you with that. Unless you can somehow translate that question to group theory :P

I was being dumb

reading comprehension is important

^I want to roll that shape down a staircase

:O

Or up a staircase, for that matter.

On second thought, the wheelchair thing sounds like a bad idea

hi akiva

> there's no I in collaboraton - Ken M

Hi

collaboraton sounds like some friendly automaton

so friendly that he will only give you one definition of automaton

stay away from twitter :P

it'll destroy your spelling

@AkivaWeinberger why is that so scuffed

Scuffed?

@nitsua60 whatever happened to the "math.SE university" idea?

It had a promoter but not an audience I suspect

yeah, enthusiasm can only take you so far :P

your own enthusiasm won't, at any rate

I was not enthused

same

me three :-)

Same

a veritable crowd of disinterest, this

Probably didn't help the guy's case that he was established as a crank already and got really hostile really quickly when plans fell through

I'm curious what would happen if we defined a sequence of zeroes and ones with $a_0=0$, and with $a_n$ defined by: Train an LSTM neural network on $a_0,\dots,a_{n-1}$. Make it predict what $a_n$ is; define $a_n$ to be the opposite.

The most unpredictable sequence… to LSTMs, at least. It's always the exact opposite of what is predicted.

what trig identities are useful to integrate cosh(2x)*sinh^2(x) dx

@WDUK Think about how you would tackle normal trig functions

Maybe double angle or power reduction...

well i rewrote it to cosh^2(x)sinh^2(x) + sinh^4(x) dx but i would not say this made it easier

that was using cosh(2x) = cosh^2x + sinh^2x

sinh²(x) = (cosh(2x)-1)/2

so ((cosh(2x)-1)/2)*(cosh(2x)+1)/2) but then im still left with sinh^4(x)

oh i guess i could split that up

damn it i still don't see it

What's $\cosh(2x)$ when you expand it?

I mean honestly I'd just write it out in terms of $e^x$ and stuff

Or shove it into Wolfram Alpha

sec im writing it out ill show you what im trying

@WDUK Where are you getting sinh^4?

Replacing sinh²(x) with (cosh(2x)-1)/2 should remove all of the sinh's

cosh(2x) * sinh^2(x) = (sinh^2(x)+cosh^2(x))sinh^2(x) was what i first tried

does this look more correct

Well, your integrand is an even function and your result is odd. So that's as it should be

lol

for another check, at small $x$ your integrand is approximately $(1)(x)^2$.

what do you mean by small x?

He's doing asymptotic expansion at 0

not asymptotic

it's just the Taylor series of the integrand at $x=0$

Well, yeah, that

by contrast, to first order your result is (4x)/16+(x/4)-(2x/4)=0, which again is as expected.

or you can just differentiate the entire result to get $\frac14 \cosh(4x)+\frac14-\frac12\cosh 2x$ and check whether that's equivalent to what you started with

@0celo7 so, what sorts of cases were you wanting to visualize in the 2D case?

i should note that, if all you're wanting to see is $u(|x|)$ then the method pretty much works regardless of dimension

@Semiclassical Uhh I don't really have anything in mind. Ideally it would be something with absolutely no symmetry otherwise it's just the 1D thing again

right

@Semiclassical the best thing would be to define some $C^\infty_c$ function, like a shifted bump function, and then multiply it by some polynomial

That should make something with little to no symmetry

hmm

yeah, makes sense

I have trouble with

I don't understand Solution 1 (@ artofproblemsolving.com/wiki/…). There is no motivation given for why one has to square the ratio 3/5 and set it equal to 12/x.How would you even know that those two ratios are equal?

Very confused.

I personally did (3/5)=(12/x), and got x=20, which turned out to be wrong.

Hello?

@DarkRunner First, do you see why what really matters is the area of the triangle?

base, and height, and for an equilateral traingle, the area is always in the form of s^2sqrt(3)/4

What?

$\frac { { s }^{ 2 }\sqrt { 3 } }{ 4 } $

Okay. That's not what I asked.

ok

They asked for an answer that's a weight.

Why do I claim it's the area that really matters?

everything else is constant, I guess?

that's not very specific, but basically yes

@Semiclassical can you email me the new notebook please?

the weight is proportional to the volume, and since the thickness doesn't change that's proportional to the area

notebook(s)

You said you changed the method a bit

so we've got $w\propto A$

@0celo7 will do soon

@Semiclassical Oh, I see, so the weight is inversely proportional to the area? Shouldn't it be proportional? The greater the area, the greater the weight, no?

@DarkRunner $w\propto A$ is "w proportional to A"

oh ok, my fault

@Semiclassical no rush

but we're given the side lengths instead. so we should ask ourselves how $A$ and $s$ are related

Ah I got it!

yep

one thing I should note in passing is that all you really need to finish this is $A\propto s^2$. That's implied by what you had earlier, of course, but the constant in that relationship is really not relevant here

So we're asked for the weight of the triangle, and since both traignles are of same thickness, and the same triangle, they're areas will be proportional to their weights. Furthermore, a is related to the side length of a equilateral triangle by the mentioned formula!

Wow nice nice

Thanks @Semiclassical I get it

In particular, this argument would proceed exactly the same if you had squares instead of equilateral triangles

Got it, any regular polygon probably

really any plane figure that's described by a single length scale

Right thanks

@XanderHenderson just for you i.gyazo.com/7df2a370438a4f077d56ddee79d33f80.png

perfect brace size

I gave an honest effort for

I got 54 but it isn't even a choice; I did:

@Semiclassical crap, you know that feeling when you're writing up something you talked about with your advisor months ago

and were too stupid to take proper notes

ugggghhhh

yeah

there was a mistake in this proof that we fixed

I don't know where I wrote down the correct proof

if at all

The first thing I did was find the intersection of the three lines, which turned out to be:

Then, because it said the squares laid on lattice points, I found the GIF/LIF for each of the points, and got to:

The next thing I did was find the area of the right triangle ABC, which turned out to be 75/2 which I rounded down to 37. From there, I did the following:

Let GIF=greatest infeger function (I don't know how to write it in LATEX)

$GIF(\frac { 37 }{ 1 } )+GIF(\frac { 37 }{ 4 } )+GIF(\frac { 37 }{ 9 } )+GIF(\frac { 37 }{ 16 } )+GIF(\frac { 37 }{ 25 } )+GIF(\frac { 37 }{ 36 } )=54$

And that's how I got my answer

I honestly don't know where I went wrong with my reasoning

If anyone could help that'd be great

wow great

@TedShifrin i asked "what's the point of the pullback?" a few days back. The pullback is the point..

if anyone could take a look at my problem, I'd be really greateful

what is it

ok, do you know what the region looks like?

Yep

It looks like a right traingle

With vertex A at the origin

yeah.. so let's see

sure thank you for the help

no i dont think vertex A is at the origin

in fact no vertex is

Yeah, none of them are at the origin, but I found the nearest lattice point next to each vertex

Because the question asked for squares whose vertices lay on integer coordinates

i dont have a pen and paper, so let me try to guide you instead?

wolfram to the rescue -- wolframalpha.com/input/?i=y+%3D+pi*x;+y+%3D+-0.1;+x+%3D+5.1

here's what the region looks like.

going right to left -- whats the rightmost, bottommost point in the region? (i believe its (5, 0), but double check me)

how many 1x1 squares fit in between the slices x = 5; and x = 4 ?

how many between x = 4; and x = 3, and so on

then count how many 2x2 squares fit in

then 3x3, and you should be done? since 4x4, and 5x5 don't fit

Sorry I dozed off, OK, I see;

oh ok so you broke it into cases I guess

Thanks ansyway

@MatheinBoulomenos morning :D

@MatheinBoulomenos I got a question about modding out algebraic things

@Kasmir morning

Yeeey :D you are here =p

was about to leave

Ehm mathein, can you tell me how you understand "modding out process"

hmm

You throw away some information

modding groups, rings ,vector spaces etc

since the last two are speical cases of the first

can we start with a group case

and work on that

I think for the intution it doesn't really matter what we take

okay since am doing linear algebra now

say we have a vector space V

For example, consider $\Bbb{Z}/2\Bbb{Z}$. Here we take a whole number and we throw away all information about that number except the information whether it's odd or even. So in the quotient $\Bbb{Z}/2\Bbb{Z}$, we have two elements, one can denote them by $\overline{0}$ and $\overline{1}$, but one could also denote them by $\mathrm{even}$ and $\mathrm{odd}$

and W being a subspace

that particualar example I understand well

Z/nZ

So you know how the sum of two even numbers is even, an even number plus an odd number is odd, a product of even and odd is even etc.? These facts are all reflected in the ring structure in Z/2Z

I know that we divide elements of the big group Z , into classes

hmm am not sure what you mean about reflected

I mean we can see these facts in the operations of Z/2Z

oh yeah

[0]is addtive identity

and [1] multiplicative

the sum of two odd numbers is even, becaues $\overline{1}+\overline{1} = \overline{0}$ in $\Bbb{Z}/2\Bbb{Z}$

so it make sense if we add anything to 0 to get that hting back

right, exactly

Yes that particular thing is clear to me

but after that its not

if we take V vector space

and W being a subspace of it

V/W

what does this mean ?

@0celo7 GAH! Nononononononononononononono!

also, I'm going to bed; those braces are going to give me nightmares :(

@XanderHenderson hello to you to i guess

Okay, suppose V=R^2 and W is a one-dimensional subspace

okay so W is a line going thru the origin

right

Just like with Z/nZ we divide the elements of Z into classes, we divide the elements of R^2 into parallel lines to W

every element of V/W is a line that is parallel to W

so they are in some sense equal up to W

right

the elements in one coset v+W are equal up to a translation in the direction of W

it is very blurry to me how to make sense of it

let me reread what u said and see

so if we have say V= R^2

and W = <(1,0)^t >

ie the x-axis

the cosets are of the form v +W

grrrrrrr

yes, so these cosets are all the lines that are parallel to the x-axis

when we say v+W

do we mean to add the whole line?

ie c (1,0)^t

well, yes. You take the vector v and then you add the whole line W, so you get a line that is parallel to W, but which doesn't go through the origin, but it goes through v

okay i see that part now

so the elements of V/W are lines paralell to the x-axis

@MatheinBoulomenos hi

@Adeek hi

more abstractly how would you describe that process?

I did not find anything online that explains this well

@MatheinBoulomenos I am now committing 1 hour per morning on solving Allufi chapter 0 and reading it.

I mean I skimmed it before

but I think it would be good idea going through things carefully.

@Adeek really cool! that's one of my favorite books

Because I would like to start working on understand algebraic number theory by next year. So, I need to work on my rusty algebra skills.

yeah mine too @MatheinBoulomenos

@Kasmir So you know subspaces of a vectorspace, but there's also a more general notion, these are called affine subspaces of a vector spaces. They are like subspaces, but they can be translated by one vector. So if you look at 1-d affine subspace in a vector space, these are all lines, not just the lines that go through the origin. And if you look at 2-d affine subspaces, these are all planes, not just those through the origin etc.

@MatheinBoulomenos let us say I finish allufi + problems + MA + problems

then would I be prepared for that book Neurklish ?

the one that you gave me ?

MA?

waiting for algebra guys to get on soon

michael atiyah

hmm

yes, that would be sufficient

@KasmirKhaan if you take R^3/W where W is a 2-d subspace, so a plane trough the origin, then this will be all planes that are parallel to W