Mathematics - 2018-01-12 (page 3 of 4)

Balarka Sen

14:05

boi Tame Impala is so bad and corny it makes me puke

this is like pop rock on LSD

King Gizzard & The Lizard Wizard - Rattlesnake (Official Video)

►

this is what I call psychedelic rock

rattle schnek boiz

Eric Silva

@Semiclassical I'm doing my e&m pset rn and still very not used to physics yet

Semiclassical

Lol

Eric Silva

It's like almost like doing math but also it's kind of not like doing math

So weird

@BalarkaSen ahh

Semiclassical

Sounds right

Michael.P

14:39

Hello everyone

I have a soft question, regarding studying process, can anyone help?

Faust

14:52

im fairly sur ei have this order right PID's ED's UFD's

all PID are ED's but not all ED's are PID's

so no

@MatheinBoulomenos i probably wouldn't understand any current research in the area but im fine with like the latest and greatest from 50 years ago kind of thing,

Michael.P

Guys, what do you do when you stumble upon a problem you cannot solve? Especially when it's from a handbook, and you cannot proceed without

Faust

15:18

i stare at it for awhile

if its for hw thats due in a finite amount of time i will eventually ask for help

if its just an exercise i may leave it until i get it on my own

start by reviewing the definitions in the related section thats always a good place.

Alessandro Codenotti

16:03

@Faust the other way around

Michael.P

16:21

@faust, what do you do meanwhile

John11

16:39

Hey guys! Can anyone confirm if my following thinking is correct? : Start with the most simple normal subgroup, the trivial subgroup, then the second you add a non-trivial element you have to include all of its conjugacy class for the subgroup to remain normal.

Alessandro Codenotti

Yes

John11

@AlessandroCodenotti thank you! :)

Leaky Nun

Define $d : \bigwedge^* (\Bbb R^3) \to \bigwedge^* (\Bbb R^3)$ a linear map that sends $\bf v \mapsto \bf v \wedge (e_1 + e_2 + e_3)$. This gives the following exact sequence: $$0 \longrightarrow \Lambda^0 (\Bbb R^3) \overset d \longrightarrow \Lambda^1 (\Bbb R^3) \overset d \longrightarrow \Lambda^2 (\Bbb R^3) \overset d \longrightarrow \Lambda^3 (\Bbb R^3) \longrightarrow 0$$

It resembles the de Rham complex of $\Bbb R^3$

@BalarkaSen @MatheinBoulomenos

also, the three maps in the middle resemble grad curl div

Semiclassical

Seems as if you’d have the same thing if you replaced e1+e2+e3 with just e1

Balarka Sen

You can define a chain complex on forms by defining your $d$ operator as $d\omega = \omega \wedge \eta$ for some chosen $1$-form $\eta$. It wouldn't be a very interesting chain complex; as you said, it's an exact sequence, so has homology 0. Boo.

Semiclassical

16:49

I feel like I brought this up a looong time ago

Balarka Sen

You did

:)

Semiclassical

Lol

Akiva Weinberger

De Rham is my school's mascot

Wait, no, it's just "Ram"

Leaky Nun

1 min ago, by Semiclassical

Seems as if you’d have the same thing if you replaced e1+e2+e3 with just e1

but they won't resemble grad curl div

Balarka Sen

I wonder what the sequences look like if $\eta$ is not a 1-form. You would break exactness then

In any case this doesn't seem very interesting or natural

DISAPPOINTED

Leaky Nun

16:53

@BalarkaSen you would break the codomain

Balarka Sen

Just shift degree up

sent Omega^i to Omega^(i+k)

Leaky Nun

then where is the sequence

it's still a complex though, since x wedge v wedge v = x wedge 0 = 0

Balarka Sen

Omega^1 --> Omega^k --> Omega^2k --> ... you dumbo

yes it is a chain complex that's why i suggested it

exact chain complexes are boring as fuck boi

Leaky Nun

wait

v wedge v doesn't need to be 0, if v isn't a 1-form

so, omae wa mou shindeiru

Balarka Sen

take it to be odd graded

then it holds

Leaky Nun

16:57

mmhmm

but the main point is that it looks like grad curl div

Balarka Sen

That does not seem like a very interesting description to me

If you have something resembling the de Rham chain complex you should have nonzero homology.

In this case you don't so it's all bull

Semiclassical

the discussion I'm thinking back to starts around here:

Jun 23 '17 at 4:24, by Semiclassical

Had a diff. forms question which I wondered if you'd know the answer to. (It's a terminology question more than anything)

skullpatrol

how's the wind chill factor?

Semiclassical

Pretty brutal

skullpatrol

breaking any records?

Eric Silva

17:02

it went from 60 to 11 in a day here what is this

Leaky Nun

@Semiclassical nice

@MatheinBoulomenos (or anyone) so Hom(-,B) is contravariant in the category of modules. Hom preserves limits, so does that means it preserves finite products?

Mr. Xcoder

Hello @LeakyNun!

Leaky Nun

hi

Mr. Xcoder

;-; Nobody knows better ways than by-hand search of natural numbers ;-;

2

philmcole

Hello

Can somebody give me a quick hint? I want to show, that a complex number has exactly $n$ disctinct roots. I've shown that $\sqrt[n]{w} = \sqrt[n]{r}e^{i\frac{\theta + 2\pi k}{n}}$ with $0 \le k \le n-1$ are roots. But how can I show that they are distinct?

skullpatrol

17:27

@Mr.Xcoder grade 7?

12 year olds?

Mr. Xcoder

@skullpatrol Yes.

@skullpatrol 13

Rick

0

Q: Implicit solution to first degree ODE

Im a bit confused. Lets take $$\begin{cases} y\, y'=-x\\ y(0)=1 \end{cases}$$ Does this equation have as solution: the whole circle $x^2+y^2=1$ or just the positive part $y(x)=\sqrt{1-x^2}$?

differential-equations

Is the whole circle the solution or just the sqrt?

Leaky Nun

@philmcole their Args are different

philmcole

@LeakyNun Ok, it's not possible, that two of them have the same arg for different $k$?

Leaky Nun

ja genau

Semiclassical

17:42

one way to see what's happening is to assume that two such roots are actually the same, i.e. $\sqrt[n]{r}e^{i(\theta+2\pi k_1)/n}=\sqrt[n]{r}e^{i(\theta+2\pi k_2)/n}$

if you divide both sides by the right-hand side, you get $e^{i 2\pi (k_1-k_2)/n}=1$

But when does a complex exponential equal 1?

(If you're not sure, think in terms of Euler's formula.)

Lozansky

@Rick Won't the solution be unique on the open interval $(-1,1)$?

philmcole

@Semiclassical if the arg is zero and therefore $k_1=k_2$. Thanks!

Lozansky

Write $f(x,y) = -x/y$. Clearly $f$ and $\partial f /\partial y$ are continuous on the rectangle $(-1,1) \times (0,1+\delta)$ which contains $(0,1)$. Then your solution is unique on an interval $(-\epsilon, \epsilon) \subset (-1,1)$

Semiclassical

@philmcole not quite. You could also have $e^{2\pi i}=1$

Lozansky

@philmcole Think of the unit circle :)

philmcole

17:53

@Semiclassical then if it's an integer multiple of $2 \pi$?

Semiclassical

Right.

Lozansky

$2 \pi i$

Semiclassical

But that implies that $k_1-k_2$ is an integer multiple of $n$

and what did you assume about k1, k2?

philmcole

That they are smaller or equal to $n-1$.

Lozansky

@Rick My point is, you can't allow $-\sqrt{1-x^2}$ since your solution to the IVP is unique

philmcole

17:55

@Semiclassical and positive

Lozansky

@philmcole Also $k_1 \neq k_2$ I would imagine

philmcole

Then together $k_1 = k_2 l n$ for some positive integer $l$, but $k_1 \le n-1$, so a contradiction.

Vrouvrou

hello, what is the method to find the adherent value of $w_n=\sin(n\pi/2)\sin(n\pi/2)$ on (R,|.|) ? please

Semiclassical

another point is that, if both of them are drawn from 0 through n-1, then the farthest they could possibly be apart is n-1. so the only way they can differ by a multiple of n is if they're actually equal, and therefore not distinct roots

philmcole

Yeah, true! Thank you. I see it now more clearly :)

Vrouvrou

17:58

@AkivaWeinberger hello

Akiva Weinberger

Hi

I have no idea what an "adherent value" is

MatheinBoulomenos

@LeakyNun preserving limits means a different thing for contravariant functors. $\operatorname{Hom}(-,X)$ turns colimits into limits

Vrouvrou

the limit of the sequence or the limit of subsequences @AkivaWeinberger

Leaky Nun

@MatheinBoulomenos yes, that's what I mean

MatheinBoulomenos

yes, that's true and it follows from the covariant case. $\operatorname{Hom}_C(-,X) = \operatorname{Hom}_{C^{op}}(X,-)$

Leaky Nun

18:05

@MatheinBoulomenos how do you prove that it preserves limits?

wait, it's a functor to itself o.O

MatheinBoulomenos

well, $\operatorname{Hom}(X,-)$ preserves limits basically by definition of a limit

not sure what you mean by "a functor to itself"

Rick

4

Q: Solve $(y+ x^3y + 2x^2)dx + (x + 4xy^4+ 8y^3)dy = 0$

Basically I tried to group the terms ydx + xdy is a perfect differential but I couldn't think a method for ($x^3$y)dx + (4x$y^4$)dy . Also dividing by xy didn't help me.

differential-equations

What about cases like this ? @Lozansky

Leaky Nun

@MatheinBoulomenos hmm...

MatheinBoulomenos

The isomorphism $\varprojlim \operatorname{Hom}(X, X_i) \cong\operatorname{Hom}(X, \varprojlim X_i)$ works as follows: if you're given a set of maps that define a cone over $\varprojlim \operatorname{Hom}(X,X_i)$, the universal property of the limit (i.e. the definition) gives you a map $\operatorname{Hom}(X, \varprojlim X_i)$

Hi @Ted

Ted Shifrin

Hi @Mathein, Leaky, DogAteMy

Balarka Sen

18:11

Hi @Ted

Leaky Nun

hi

Ted Shifrin

oh, and hi @Balarka

MatheinBoulomenos

For the other direction you just compose with your morphism $\varprojlim X_i \to X_i$ for every $i$

Lozansky

@Rick Eh not too sure about uniqueness/existence on nonlinear DEs

Vrouvrou

are you there @AkivaWeinberger?

MatheinBoulomenos

18:12

the uniqueness part of the universal property gives you that these are inverse bijections

Akiva Weinberger

I need to go now, sorry @Vrouvrou

Vrouvrou

ok

Leaky Nun

@MatheinBoulomenos so it works in every category?

MatheinBoulomenos

yes

Leaky Nun

:o

even if we define Hom differently?

MatheinBoulomenos

18:14

well, I'm assuming your Hom is always a set

Leaky Nun

like for modules, Hom is equipped with the structure of a module

Lozansky

@Rick Nevermind, if you write it like $dy/dx = f(x,y)$ you can just apply the existence/uniqueness theorem

Vrouvrou

@TedShifrin bon soir, savez vous comment on fait pour obtenir les valeurs d'adherences de $w_n=\sin(n\pi/2)\sin(n\pi/2)$ sur (R,|.|) ?

Ted Shifrin

@Vrouvrou: Alors, quel est le valeur de $w_n$, exactement?

Lozansky

So if you have an initial value $y(x_0) = y_0$ and can find a rectangle $(a,b) \times (c,d)$ which contains $x_0, y_0$ and where $\partial f /\partial y$ and $f$ are continuous, then you for sure have a unique solution on some subset $(x_0-\epsilon, x_0 +\epsilon) \subset (a,b)$, @Rick

Vrouvrou

18:18

ca depend de n, si n=0, pi, pi/2,...

@TedShifrin

Ted Shifrin

Oui, je sais. Mais dites-moi les valeurs ...

Leaky Nun

@Vrouvrou n n'est pas entier?

Vrouvrou

je n'ai pas calculer

Alessandro Codenotti

Hi @Ted

user193319

I know that the product of two normal topological spaces is generally not normal, but are there any conditions on the two spaces that will guarantee that the product is normal?

Ted Shifrin

18:19

@Lozansky @Rick: It seems to me that in the implicit case we're talking about flows of vector fields, not solutions for ODEs of functions.

Vrouvrou

@LeakyNun ah oui c vrai n est dans N

Ted Shifrin

Hi @AlUssandro

Leaky Nun

@Vrouvrou alors quels sont les valeurs possibles?

MatheinBoulomenos

@LeakyNun no wait, I've been talking nonsense. What I said holds for limits in the category of sets. As the forgetful functor $R-\mathbf{Mod} \to \mathbf{Set}$ is conservative (=reflects isomorphisms) and preserves limits (but not colimits), you can conclude that it holds in $R-\mathbf{Mod}$ as well (and likewise for rings, groups, etc. (not fields though))

Alessandro Codenotti

I blame that on my phone's keyboard and my big hands!

Leaky Nun

18:20

not fields?

isn't a field just another ring?

MatheinBoulomenos

no, the category of fields is really strange

Ted Shifrin

@user193319: I don't know a necessary condition, but certainly it works if they're metrizable, for example.

MatheinBoulomenos

I was talking about the respective categories

Alessandro Codenotti

In that case the product is even metrizable. Works with countably many spaces too

user193319

@TedShifrin Ah! Yes! That's all I need. Thanks!

Alessandro Codenotti

18:22

I just found out today that there those things called ultrametric spaces

Leaky Nun

@MatheinBoulomenos oh, lol

Lozansky

@TedShifrin In that case, I'm way out of my depth

Leaky Nun

can you give me an example how it doesn't preserve limits in fields? @MatheinBoulomenos

Ted Shifrin

But see this, for example. Product of metric and normal needn't be normal.

@Lozansky @Rick: I think the point in your question is to tell when such a 1st order ODE $M\,dx + N\,dy=0$ is exact or has an integrating factor. That's a different flavor question.

@Alessandro: When do you retake your test? I need to know when to go back to hiding.

MatheinBoulomenos

a lot of limits in the category of fields don't even exist

Alessandro Codenotti

18:24

I don't know yet, haven't been driving in a while

I'm dealing with math tests in uni for the time being

Julius

I'm dealing with $\sum_{k=0}^K a_{k,K}$ where $a_{k,K}$ are nonnegative. Suppose I know that all $a_{k,K}\propto K^{1/2+\delta}$, $\delta\geq 0$, terms are dominated by something summable and that those $a_{k,K}\to 0$ as $K\to\infty$. Hence, I'd like to show that $\sum_{k=0}^{?} a_{k,K} \sim \sum_{k=0}^{\infty} a_{k,K}$, where in the first sum the upper limit is "up to order $K^{1/2}$".

How to do that formally with the dominated convergence theorem (to show that the rest of the series is zero) when there is no first term of $K^{1/2}$ order? A verbal argument perhaps is enough?

Semiclassical

So field theory is weird in both math and physics

Balarka Sen

@Ted I tried to write an h-principles answer to the dz = xdy question you answered earlier

Semiclassical

can't say I'm surprised

Balarka Sen

I flailed

Ted Shifrin

18:25

OK, whew, @Alessandro. I'm safe for a while, then ;)

Alessandro Codenotti

Yup, got to think about functional analysis and commutative algebra now

Ted Shifrin

Interesting @Balarka. But it's exactly this sort of non-integrable thing you were convincing me you could do?

Lozansky

@TedShifrin Originally, the question was if the solution to $yy' = -x, y(0)=1$ is valid for the whole unit circle or just the positive part $y=\sqrt{1-x^2}$

MatheinBoulomenos

@Semiclassical not really. It's just that if you apply category theory naively to fields, then you get strange phenomena. I'd say field theory on its own is not that weird (but I'm an algebraist, so maybe I'm just weird myself)

@Alessandro what are you doing in commutative algebra right now?

Ted Shifrin

@Lozansky: Right. Well, that's phrase specifically in terms of differentiable function $y=y(x)$ ... It's not written as an implicit ODE $x\,dx + y\,dy=0$. For the latter, I would say that circles centered at the origin are the solution curves.

Balarka Sen

18:27

@TedShifrin Not really. What I can give you is a curve joining $(0, 0, 0)$ and $(x, y, z)$ which is $\epsilon$-close to being tangent to the distribution $D$, for arbitrarily small choice of $\epsilon$.

But that works for any non-vertical distribution $D$ whatsoever

Ted Shifrin

Oh, right, @Balarka. Can you prove that the achievable endpoints form an open/closed set or not?

Alessandro Codenotti

@MatheinBoulomenos nothing new, the course has ended, I need to study for the exam

MatheinBoulomenos

ah I see

Balarka Sen

There is something special about this specific distribution that upgrades $\epsilon$-tangency to actual tangency

Ted Shifrin

Right.

Balarka Sen

18:28

Eric and I were having conversations about it.

I think it relates to the fact that there is no h-principle for Lagrangian submanifolds, even though there is an h-principle for epsilon-Lagrangian submanifolds

I'll need to learn more symplectic/contact geometry to be able to see this

Ted Shifrin

In general, it probably depends on how many iterated brackets it takes to generate everything.

This is sub-Riemannian geometry ... (I got the name screwed up yesterday.)

Balarka Sen

mm true

Ted Shifrin

@Balarka: So I was going to write something on this question. But what's the least fancy way to argue that a line bundle on a manifold is trivial iff its restriction to the $1$-skeleton is trivial? No characteristic classes ...

Balarka Sen

Give a cell decomposition, and extend the trivial line bundle on the 1-skeleton over the 2-cells, and so on, I think

The obstruction theory argument, but parsed elementarily

Ted Shifrin

There should be something easier.

Like the orientation-reversing loop argument ...

Warlock is comfortable with smooth manifolds, not with topology.

Balarka Sen

18:34

Hm.

Ted Shifrin

He wants to go directly to $n$-forms, which isn't going to work. But just working with the line bundle $\Lambda^n T^*M$ should suffice.

So the usual monodromy argument should prove that if $\mathscr L|_Y$ trivial for every embedded circle $Y$, then $\mathscr L$ is trivial on $M$.

Balarka Sen

Right. I guess what you should be able to do is to say $Y$ is homotopic to some loop on the 1-skeleton

Over which $\mathscr{L}$ trivializes...

Vrouvrou

@LeakyNun $0, 0,70..,0,-0,70,...$

Ted Shifrin

@Vrouvrou? Comment?

Leaky Nun

@Vrouvrou quoi?

Daminark

18:38

Hello nerds

MatheinBoulomenos

Hi @Daminark

Abcd

Hi.

Balarka Sen

@TedShifrin Give the line bundle a Riemannian metric; asking it's triviality is the same as asking whether the unit $S^0$-bundle (the double cover) is disconnected or not.

Leaky Nun

our algebra topology lecturer is a category theorist in disguise lol

2

Eric Silva

sup chat

Vrouvrou

18:39

pour n paire w_n=0, pour n impaire w_n= (-1)^n (0,70)

@TedShifrin

Abcd

Just need verification of my answer.
Give examples of two functions $f: N \to Z$ and $g: Z \to Z $ such that $g \circ f$ is injective but $g$ is not injective.

Balarka Sen

But since the cover is trivial over the 1-skeleton, any loop lifts trivially in the cover

Abcd

My answer: $f(x)= \dfrac{1}{x}$ and $g(x)= x^2 $. I feel its correct.

Ted Shifrin

Je ne comprends pas le $(0,70)$?

MatheinBoulomenos

sup @Eric

Vrouvrou

18:40

racine(2)/2

Ted Shifrin

@Abcd: Domains and ranges don't match.

Vrouvrou

@TedShifrin

Balarka Sen

(Take a loop $\gamma$ with basepoint $x_0 \in M^{(1)}$ on the 1-skeleton; $\gamma$ is homotopic to a loop $\gamma'$ in $M^{(1)}$, which lifts trivially in $\widetilde{M}$, hence so does $\gamma$)

That guarantees that the unit $S^0$-bundle is disconnected

Therefore the line bundle is trivial

Abcd

@TedShifrin made a typing error. Edited.

Daminark

They have just emailed me being like "Oh yeah please do these tasks on workday so we can pay you for grading discrete"

Ted Shifrin

18:42

Your $f(x)$ doesn't take values in $\Bbb Z$, does it? @Abcd

Abcd

@TedShifrin No

Ted Shifrin

@Vrouvrou: La fonction est $(\sin(\pi n/2))^2$???

So make it way easier, @Abcd.

Eric Silva

@Daminark i hated filling out workday stuff

Ted Shifrin

heya @Antonios

Semiclassical

If your function doesn't map to integers, then it's not an example of what you want.

Eric Silva

18:43

they were way late on paying me for TAing cause the system is kinda dumb

Abcd

@TedShifrin is my answer incorrect?'

oh yes.

Ted Shifrin

Of course it's incorrect. Your function $f$ doesn't have values in $\Bbb Z$.

Semiclassical

The codomain isn't Z, so nope

Abcd

Lets say $f(x)= \mathbb{floor({x})}$

Ted Shifrin

You could change the codomain (range) or you could change the function. As I said, you made it too hard.

Semiclassical

18:44

^

Ted Shifrin

So your $f$ is just the $0$ function?

I doubt $g\circ f$ will be injective, then.

MatheinBoulomenos

@Abcd the idea is that you make the points where the injectivity of $g$ fails not be contained in the image of $f$

philmcole

@Semiclassical Hey Semi, can you give me a hint how to show that $\sum_{k=0}^{n-1} e^{i\frac{2 \pi k}{n}} = 0$

Ted Shifrin

Hush, @Mathein. Damn.

Abcd

@TedShifrin now?

MatheinBoulomenos

18:44

I was only giving a hint

Ted Shifrin

@Abcd: So that's just $f(x)=x$?

Abcd

yes, identity function.

Ted Shifrin

Inclusion function.

Semiclassical

@philmcole Let $\omega=e^{i 2\pi/n}$, so that your sum is $S=1+\omega+\omega^2+\cdots+\omega^{n-1}$ and $\omega^n=1$

Ted Shifrin

Not quite identity because domain and codomain are different.

Eric Silva

18:45

@philmcole just multiply your sum by something

Ted Shifrin

Yes, then it works fine. That was what I had in mind for your easier example.

Eric Silva

see what happens

Semiclassical

shhhhh

Ted Shifrin

We have people giving away too many hints here!

Semiclassical

quite so

Balarka Sen

18:46

I can give you some audio-visual hints

philmcole

Thanks guys I'll see where it gets me :D

Semiclassical

pretty sure all of these count as 'visual' :P

Eric Silva

@BalarkaSen zip zip ... zip back where u started

Daminark

@Eric luckily I did do this before when I graded the REU first year, so the process took a couple minutes

Eric Silva

that's good

Balarka Sen

18:47

Hint: pckotmp :flailing hands: flabbergastlybloof :sharp hand movement:

that's my hint to you

Eric Silva

i did it over summer for bootcamp

MatheinBoulomenos

@philmcole compute the field trace of $\omega$ wrt to the extension $\Bbb Q(\omega)/\Bbb Q$ (you can read that of the minimal polynomial)

Semiclassical

my other hint would be: How does the center of mass of a wheel change as you turn it?

Eric Silva

wow such symmetric

Balarka Sen

what if the wheel is really fucked

a dumpling shaped wheel

:3

Daminark

18:50

@Balarka so it turns out my prof had in mind the norm based solution to the $\mathbb{Z}[\omega]$ business

Balarka Sen

@Daminark aha

Ted Shifrin

What about the square wheel that rolls along a piecewise catenary path?

Balarka Sen

@TedShifrin Oh I love that

Semiclassical

hate you alllll

3

skullpatrol

:(

Ted Shifrin

18:51

@Semiclassic: I missed the question, but if we're asking about the sum of the vectors to the vertices of a regular $n$-gon ...

Semiclassical

yeah. i wasn't being precise in my 'hint'

Balarka Sen

the computation is pchank pchankg pchang

floop and done

Daminark

@Balarka we're not all fluent in topology

Tone it down to English for a moment

skullpatrol

Prof @TedShifrin did you see Francis Su's joke on fb: 6,000 mathematicians at #JMM2018 walk into A-bar. The set A-bar says “Sorry, I’m closed.”

2

Ted Shifrin

Nope.

Balarka Sen

19:02

JMM must be fun

Ted Shifrin

I am having dinners with people who are at it, but I'm not attending. Did see a bunch of people I knew downtown last night, though.

Vrouvrou

@TedShifrin $w_n=\sin(n\pi/4) \sin(n\pi/2)$

Balarka Sen

Ah

Ted Shifrin

@Vrouvrou: Ça, ce n'est pas ce que tu avais écrit!!

skullpatrol

@TedShifrin His "math flourishing" speech is going to be a book next year

Ted Shifrin

19:04

Mais, tout de même, il faut explicitement trouver les valeurs et résoudre la question.

Vrouvrou

Je trouve 3 valeurs si n=2k, alors w_n=0, si n=2k+1, w_n=(-1)^k(racine(2)/2)

Danu

Hey @Ted!

It's been ages!

skullpatrol

> Excited to announce that 'Mathematics for Human Flourishing', my speech at the Joint Mathematics Meetings last year, will be extended to a full-length book to be published by Yale University Press in 2019.

Danu

@skullpatrol Are you Francis Su?

skullpatrol

nope

philmcole

19:11

@Semiclassical So I've used your hint and applied the geometric sum. This got me
$$\sum_{k=0}^{n-1} e^{i \frac{2 \pi k}{n}} = \sum_{k=0}^{n-1} \omega^k = \frac{1-w^n}{1-w} = \frac{1-e^{i 2 \pi}}{1-e^{i\frac{2\pi}{n}}} = \frac{1-1}{1-e^{i\frac{2\pi}{n}}} = 0$$

We can do this, because $\omega \neq 1$ because $n \ge 2 $ is given. Is this what you had in mind?

skullpatrol

@Danu how goes the transition to pure math pal?

Danu

Is that speech not by Francis Su? (cf. this blog mathyawp.wordpress.com/2017/01/08/…)

skullpatrol

9 mins ago, by skullpatrol

@TedShifrin His "math flourishing" speech is going to be a book next year

I was giving evidence :P

Danu

Ah, you forgot to put the other post in a quotation. Do you know how to do that?

(I can do it for you if you like---it might not be possible to edit anymore for you)

skullpatrol

please do

Danu

19:14

Done.

skullpatrol

thanks pal

his prison inmate story is awsome

Chris White volunteered some time to teach inmates Physics and Math.

Semiclassical

19:28

@philmcole yeah

a very similar idea is to note that $\omega^n=1$ implies $$1+\omega+\cdots+\omega^{n-1}=\omega^n+\omega+\cdots+\omega^{n-1}=\omega(1+ \omega+\cdots +\omega^{n-1})$$

So the sum is unchanged upon multiplying by $\omega$. Only way this can happen is if $\omega=1$ or the sum is zero, and we know it's not the first.

(This really isn't different from your argument: It amounts to the fact that $\omega^n-1=(\omega-1)(\omega^{n-1}+\cdots+\omega+1)=0$)

Akiva Weinberger

19:44

$n\ne1$

philmcole

@Semiclassical I like that line of thought, it's so simple. Thanks!

Semiclassical

to relate this to the wheel observation, consider the set of points (1,omega,omega^2,...,omega^n)

that's like a wheel with n equally spaced points on it. let me put masses there

if I rotate that wheel through 2pi/n, then the shape returns to itself

philmcole

What do you mean by the shape returns to itself?

Semiclassical

Take a wheel with spokes.

If I turn it by the right amount, then each spoke moves to the same place as another was

so I can't tell the rotated wheel apart from the original one

philmcole

19:59

Ah because the $n-1$th becomes the first one again, meaning $e^{i\frac{2\pi (n-1)}{n}}$ becomes $e^{i\frac{2\pi (n-1)+2\pi}{n}} = e^{i\frac{2\pi n}{n}} = e^{i 2 \pi} = 1$

Beautiful geometrical thinking!

Semiclassical

Right

philmcole

I would have never thought of connecting this with the above sum.

But it's beautiful indeed.

Semiclassical

in math language there's a Z/n action on the nth roots of unity

anyways. symmetry is fun

philmcole

What's a Z/n action?

Semiclassical

"In mathematics, an action of a group is a formal way of interpreting the manner in which the elements of the group correspond to transformations of some space in a way that preserves the structure of that space."

so all I really mean is that multiplying by $\omega$ shifts all the exponents up by one taken mod n

philmcole

20:04

Is that group theory?

Semiclassical

ya

philmcole

Cool

we didn't have this yet

Semiclassical

sure

philmcole

I'll keep this in mind though

Thanks again for the help!

Abcd

Why is number of binary operations given by $n^{n^2}$?

on a set with $n$ elements...

MatheinBoulomenos

20:18

how many possibilities are there for each pair?

Abcd

possibilities of?

MatheinBoulomenos

If you want to define a binary operation $\star$, how many possibilities do you have to define $x \star y$ for each pair of elements $(x,y)$ from your set?

Are you sure?

Abcd

so every pair has two possibilities..

the binary operation may be defined on it or not...

MatheinBoulomenos

A binary operation is always defined for every element

I meant how many possible values can you give to $x \star y$

Abcd

You mean the range of the function?

it's $1 $ to $n$.

MatheinBoulomenos

20:24

right, so $n$ different values are possible for each pair

and there are $n^2$ pairs

Abcd

is $(x,x)$ also a cartesian product pair?

i mean something like $(1,1)$. Is that included too?

MatheinBoulomenos

sure

Abcd

but what does "number of binary operations" even mean?

I don't get it.

MatheinBoulomenos

If $X$ is your set, then a binary operation is just a function $X \times X \to X$

For example, addition and multiplication are binary operations on $\Bbb N$

Abcd

but we can define infinite functions...

MatheinBoulomenos

20:31

not on a finite set

Abcd

How?

MatheinBoulomenos

If $X$ is finite, then so is $X \times X$

Abcd

but we are talking about functions.

not sets.

MatheinBoulomenos

If I have two finite sets $X$ and $Y$, then for every $y \in Y$, there are exactly $|X|$ possibilities for $f(y)$ and there are $|Y|$ elements in total

Abcd

There can be infinite functions. for example I may define a new function $++$ which gives $x+1$

MatheinBoulomenos

20:34

Okay, let's try a concrete example. If I have $X=Y=\{0,1\}$

Abcd

yes..

MatheinBoulomenos

then the only possible functions are $0 \mapsto 0,1 \mapsto 0$ and $0 \mapsto 0,1 \mapsto 1$ and $0 \mapsto 1, 1 \mapsto 0$ and $0 \mapsto 1, 1 \mapsto 1$

Abcd

yes

MatheinBoulomenos

so there are only 4 functions

Abcd

right.

MatheinBoulomenos

20:38

You can think as a binary operation on $\{1, \dots ,n\}$ as an $n\times n$ grid of numbers where each number is in $\{1, \dots ,n\}$

the entry in the $i$-th row and the $j$-th column tells you the value of the binary operation on the pair $(i,j)$

Daminark

20:52

It is but a matrix :P

Leaky Nun

1

Q: The fundamental group of $S^3-S^1$

How can I calculate the fundamental group of $S^3-S^1$? I believe it might be similar to the fundamental group of $S^2-S^0$ which is $ \Bbb Z$ but I'm having trouble showing it. Is this the right direction or am I way off?

algebraic-topology

@Daminark @MatheinBoulomenos

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$Mathematics$ Mathematics