Mathematics - 2016-11-30 (page 3 of 5)

Null

13:02

Assumtion: $\mathbb{F}_n^n$ has a number of elements, lets call it $x$. How can i prove this?

(x is dependent of n obv)

Kaj Hansen

$\mathbb{F}_n^n = \mathbb{F}_n \times \mathbb{F}_n \times \cdots \times \mathbb{F}_n$ (n times)?

1

A: Order of a direct product of finitely many groups equal to the product of the orders of the groups

Your proof is fine, but you can make the induction step a lot shorter. To wit, we can write $G_1 \times G_2 \times \cdots \times G_n \times G_{n+1} = H \times G_{n+1}$, where $H = G_1 \times G_2 \times \cdots \times G_n$. You can now immediately arrive at the conclusion you want since $\display...

abenthy

@KajHansen do you happen to know if local homeomorphisms of manifolds with boundary preserve the interior points / boundary points?

Kaj Hansen

define local homeomorphism @abenthy

abenthy

$f \colon X \to Y$ is a local homeomorphism if each $x \in X$ has an open neighborhood $U$ in $X$ such that $f(U)$ is open and $f : U \to f(U)$ a homeomorphism.

Null

@KajHansen sorry, with $F_n$ we notate $Z_2$. But yeah that's it basicly i think

Kaj Hansen

13:08

That's a good question

I'm not sure

And don't immediately see a proof either way

Null

induction maybe? @KajHansen

Kaj Hansen

lol

abenthy

Let $f \colon M \to N$ be a local homeomorphism of manifolds with boundary. Take $x \in M^\circ$. We can find a neighborhood $U_1$ such that $U_1 \cong \mathbb{R}^n$ and $U_2$ such that $f$ is a homeomorphisms restricted to $U_2$.

Set $U = U_1 \cap U_2$, then ...

Null

@KajHansen actually i think it's $n^n$ elements

Kaj Hansen

Right

Null

13:11

but induction is maybe not the right choice i agree

since $(n+1)^{n+1}$ is not friendly

Kaj Hansen

Induction is fine

$F^{n+1} = (F^n) \times F$

Use induction hypothesis on F^n and then base case on $F^n \times F$

All of those F's above are supposed to be $F_n$

Null

ah

but technicly $F_n^n$->$F_{n+1}^{n+1}$ has to be shown or not?

Kaj Hansen

Yeah it does

F^n can be treated as one single vector space that has $n^n$ elements by the induction hypothesis

Null

i think a grid argumentation works better, but maybe im naive hehe

Kaj Hansen

It's a simple baby combinatorics problem without induction; doesn't really matter what path you choose.

Null

13:17

alright got to go, we'll talk us later @KajHansen =)

Kaj Hansen

I'm going to sleep soon as well

abenthy

Let $f \colon X \to Y$ be a local homeomorphism. If $x \in X$ and $U$ is an open neighborhood of $x$ s.t. $f(U)$ is open in $X$ and $f : U \to f(U)$ is a homeomorphism, is the same true for any smaller open neighborhood of $x$ contained in $U$?

Hiro

13:38

Any idea how to solve this: math.stackexchange.com/questions/2037297/…

s.harp

14:00

When talking about "non-commutative geometry" a name like "Alan Cong" or something was mentioned to me with connection to $C^*$ algebras

does anybody know what this name could be?

arctic tern

alain connes

s.harp

the sentence was something like "you should probably understand the term non-commutative geometry in the way that Alan so and so meant it"

thanks!

s.harp

14:12

> Given a countably generated measure space X, the linear space of square-integrable (classes of) measurable functions on X forms a Hilbert space. It is one of the great virtues of the Lebesgue theory that every element of the latter Hilbert space is represented by a measurable function,

I don't understand why this is a virtue and not a tautology of the definition

Are those sentences erroneous or is there a way of understanding the thing I cannot see?

Semiclassical

15:02

hi chat

meow-mix

15:15

good morning, in computers class atm

Semiclassical

hi @AntonioVargas

Antonio Vargas

@Semiclassical yo, how's it going?

Semiclassical

eh, alright. need to work on grading today and not get distracted

(and yes, I did just type that on this chat. so obviously i'm doing a great job at that already)

Antonio Vargas

perfect time to jump in the MSE chatroom :)

heh

Semiclassical

quite

Did that Hermite polynomials calculation work out for you?

Antonio Vargas

15:19

yeah actually

Semiclassical

Nice

Antonio Vargas

I kind of justified it to myself

I hadn't really read about limit relations between OPS before, it lead me down a rabbit hole about the Askey table, pretty interesting

but today I am a random matrix

Semiclassical

Neat.

Did the shifting trick prove helpful?

Antonio Vargas

yeah definitely, helped me recognize that the limits were shifted Hermites: $H_n(w+t)$ or somesuch

Semiclassical

Ah

Antonio Vargas

15:27

my cup underfloweth with coffee

2

Semiclassical

Java underflow error = "I need coffee now"

2

DHMO

finally someone who really knows how to use -eth and -est

Semiclassical

store.dieselsweeties.com/products/fucking-coffee-mug

I kinda want to get that cup eventually

but I probably don't want a mug with profanity on it.

Antonio Vargas

the drawbacks of seeming professional

Semiclassical

ya

Jessy Cat

15:43

So, this is the problem that caused me to have another meltdown last night.

0

Q: Polar Laplace equation on partially bounded domain - what am I doing wrong?

I am trying to solve the following problem: Let $\Omega$ denote the region $\{ (r, \theta) : r>1, 0 < \theta < \pi \}$. Find the bounded solution to the problem: $\begin{align}\nabla^{2} u(r, \theta) = 0 && \text{in}\,\Omega \\ u(r, \pi) = u(r, 0) = 0, && r>1 \\ u(1,\theta) = 1 && \text{...

I can't get the same solution that's in the back of the book, and I have no idea what's wrong with it.

Semiclassical

Eh, you're not going to like this, but you made an error in your separation of variables

Hey-men-whatsup

0/

hi

Semiclassical

you got $\displaystyle \frac{r^{2}R^{\prime\prime}(r)}{R(r)} + \frac{rR^{\prime}(r)}{R(r)} = -\frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)}$ @JessyCat

Jessy Cat

@Semiclassical yes, that's in there.

isn't it?

@Semiclassical why?

Semiclassical

...because I'm stoopid. Ignore me.

Hey-men-whatsup

15:50

no, he's not syoobid

i'm syoobid

Jessy Cat

Syoobid?

Antonio Vargas

@JessyCat, sanity check: your solution should be even in $\theta$ about the point $\theta = \pi/2$, so you should have $u(r,\theta) = u(r,\pi-\theta)$, and this isn't true if you include the terms $\sin(n\theta)$ in your sum with $n$ even.

Jessy Cat

@Antonio, how do you know it should be even in $\theta$ about the point $\theta = \pi/2$?

Antonio Vargas

Because of the symmetry of the initial conditions

Jessy Cat

@Antonio weird I've never heard of that.

Do you know of any place online I could read about this?

Ooh wait a minute

That's just a basic property of symmetric intervals.

meh

Semiclassical

15:53

That doesn't sound quite right. The ODE $y''(x)=0$ with boundary conditions $y(-1)=y(1)=0$ has both even and odd solutions.

Jessy Cat

Okay, so @AntonioVargas if I go back and fix that, it should work?

Antonio Vargas

@Semiclassical, but we're talking about the laplace equation here, it loves symmetry

oh that is the laplace equation

Semiclassical

ya

Jessy Cat

But it doesn't love me.

Antonio Vargas

well in 2d

I'm just visualizing what the solution should look like given the boundary conditions

Semiclassical

15:54

If you've got zero boundary conditions on a hypercube $[-1,1]^n$, then Laplace's equation still allows even/odd stuff.

So I'm skeptical.

Jessy Cat

So @AntonioVargas will changing $u(r,\theta)$ to $u(r, \pi - \theta)$ fix the problem?

Antonio?

Semiclassical

I see two errors near the end, @Jessy

Antonio Vargas

@Semiclassical, if you instead have $y(-1) = y(1) = 0$ plus $y(0) = 1$ do you get the even solution only?

Semiclassical

One, $\cos(n\pi)=(-1)^n$

That's what leads to the even solutions dying off.

Antonio Vargas

that's the closest thing I can imagine to a 1-D analogue of the problem in the link

Semiclassical

15:57

@AntonioVargas Yeah, since an odd solution would have $y(0)=-y(-0)=0$.

Jessy Cat

Thanks @Semiclassical

Semiclassical

Two, I don't think you should have $r^n$ in your integral for $A_n$.

Jessy Cat

@Antonio, is that all that needs to be done?

Semiclassical

You're evaluating something on the boundary, so it should have $r=1$ there.

Antonio Vargas

@Semiclassical, so the $y(0) = 1$ condition is the analogue of the condition $u(1,\theta) = 1$ in this problem.

Semiclassical

15:58

@AntonioVargas Hmm, fair.

Jessy Cat

@Semiclassical no I should. That's the way you find the Fourier coefficients, isn't it? Although it doesn't really make a diffefrence. The $r$ has no impact on the outcome ofthe integral.

Semiclassical

I think those together fix the problem.

Uh

Yes, it does. It leads to you having a factor of $r^{-2n}$ in your final result rather than $r^{-n}$.

(ignoring the whole even/odd thing for the moment)

Jessy Cat

I'm supposed to have $r^{-2n-1}$

Actually, it is $1$ in the integral.

Semiclassical

?

Jessy Cat

I put the $r^{-n}$ back once I'm writing up the solution.

Maybe I made a typo, hold on. Because in my handwritten version, the $1$ is in there.

Okay, nevermind. I see what you're saying. YOu mean in the denominator.

Semiclassical

16:01

You've got $$\displaystyle u(r,\theta) = \sum_{n=1}^{\infty} A_{n}r^{-n}\sin (n\theta )\implies A_{n} = \frac{2}{r^{n}\pi}\int_{0}^{\pi}\sin(n \theta)d\theta$$ in your posted work

Jessy Cat

@Semiclassical yes, I see what you're saying.

Semiclassical

Okay.

That means that your final result would have $r^{-n}A_n$. If you restrict that to even values of $n$, that's $r^{-2n-1}A_{2n+1}$.

And I think that fixes everything.

And, hah, Mattos just added a comment to the effect of this to your question

Jessy Cat

@AntonioVargas at what point should I add in the $u(r, \theta) = u(r, \pi - \theta)$ part and start applying it?

Semiclassical

I don't think you -need- to put it in: It appears by itself once you see that the $A_{2n}$ terms vanish

Jessy Cat

(Antonio never wants to answer me directly)

@Semiclassical okay, that's good.

Antonio Vargas

16:04

@JessyCat, my comment regarding that was only to indicate how you can realize that the solution you obtained couldn't be right. It's not a method of fixing it.

Jessy Cat

He speaks! ;)

Antonio Vargas

Fortunately I am not chained to my keyboard :)

Jessy Cat

Well, what @Semiclassical added to the mix kind of helped in that regard, too. So, I give you both credit.

@AntonioVargas it's too bad. That's no way to live.

Semiclassical

It might help to note how I spotted the errors. First, I noticed that your solution included the even terms and the book's answer didn't; that suggested looking back at your integral, and that revealed the $\cos n\pi$ error.

Jessy Cat

@Semiclassical yes! I'm learning so much by asking you guys questions.

Semiclassical

16:07

I then realized that that wouldn't -quite- fix everything, because the $r$-dependence wouldn't be right. The extra $r^{-n}$ dependence showed up from $A_n$, so that had to be the issue.

Fairly simple reasoning, but effective.

FYI, an approach that might help in the future: Post the question by itself, and at the same time provide your partial work as an answer.

It's partly a matter of taste, but it means that if you figure out the answer yourself---or get pushed in the right direction for it---you can then edit the answer and accept it yourself.

Jessy Cat

@Semiclassical good idea. It's just that in the meantime, before I get to post the answer, some idiot might downvote because he thinks I'm trying to get people to do my work for me.

Semiclassical

Eh, that's not really true: You can submit both a question and your own answer to it at the same time.

Jessy Cat

@Semiclassical that's pretty wicked. I had no idea.

All right then, Imma go and fix this, and then start on the next question, then eat some lunch, then get ready to go to class.

Semiclassical

Yeah. If you look at the submission page for questions, there's the checkbox below it for answering it as well. That'll give a second area in which to write that.

Jessy Cat

Thanks to you, and to @AntonioVargas

Mahmoud

16:14

Hi. $:)$

Semiclassical

One last cute thing: If you interpret each $\sin n\theta$ as the imaginary part of $e^{i n\theta}$, then your final answer is the imaginary part of a geometric series in $e^{i \theta}/r$

Which means it can be resummed, and a closed-form solution can be written. @JessyCat

If you do a bit in your course on conformal mapping, you'll see that concept provides a direct method to obtaining the solution in that form.

Antonio Vargas

@JessyCat Sure thing.

Semiclassical

okay, back later

NaCl

16:28

Let $a,b\in(\mathbb{Z}\setminus\{0\})$ and $d\in\mathbb{N}$ be the smallest number for which exist $l,k\in\mathbb{Z}$ such that $d=l\cdot a+k\cdot b$. How can you show using $\text{mod}$ that $d\vert a$ and $d\vert b$?

Jessy Cat

@NaCl mod means the remainder that's left over after division, right? So, $d \vert a$ means when you divide $a$ by $d$, the remainder is $0$. So, $a = 0 \mod d$. Same idea with $d\vert b$, Now, that equation you wrote $d = l\cdot a + k \cdot b$ looks an awful lot like what happens when you do the division algorithm...

NaCl

Yep, it does

Jessy Cat

So, if $a = 0 \mod d$, that also means $\exists m \in \mathbb{N}$ such that $a = d\cdot m$

And if $b = 0 \mod d$, that means $\exists n \in N$ such that $b = d \cdot n$

@NaCl Now, try to manipulate those arouond somehow so that you try to extract something out of $d = l \cdot a + k \cdot b$ that looks like them :)

that should get the job done

Anyway, I'm out.

Astyx

17:38

Hi everyone

NaCl

17:53

@Astyx hi

So I get $a=nla+nkb$ and $b=mla+mkb$

Astyx

How are you ? :)

NaCl

I'm good, really. I feel very good, switched math courses

How are you?

Astyx

I'm fine thanks

What are you studying now then ?

NaCl

I'm studying computer science mainly, but we have to choose a secondary field of study, which was math before

Which didn't work well in my opinion :D

Astyx

Is it applied comp sci or theorical ?

NaCl

17:56

Kinda both

But more on the theoretical side

Astyx

What kind of theory ?

NaCl

Nothing specific yet

Astyx

And do you not have a planning of how the year will go ?

NaCl

That is basically the first year :)

Danu

Stupid combinatorics question: If I have two sets of, say, $N$ and $M$ elements, and I want to count all permutations of the total set of $N+M$ elements that involve switching elements from both sets. So if I exchange two elements of the first or second subset, that doesn't count. But if I interchange an element of the first witht an element of the second, then it counts (I'm ordering powers of operators). How do I do this?

NaCl

18:00

After that I'm more or less free to specify for things (e.g. cybersecurity)

Astyx

@Danu I don't get it

NaCl

What is $N+M$? Is it $N\cup M$?

Astyx

@NaCl ah right

Danu

How do I enumerate all inequivalent orderings of $A^p B^q$ where $A$ and $B$ don't commute?

Astyx

@Danu So you want permutations of both your sets where at least one element form the first is mapped on an element of the second ?

Krijn

18:03

@Danu Why not count the full set of permutations and take out the set you dont count

I.e. $S(N)\times S(M)$

Danu

Right. Sure.

I'll do that

Thanks Krijn

I knew the question had to be trivial :P

Writing out an arbitrary analytic function of two matrices sucks :P

Semiclassical

18:17

okay, I'm both proud and amused at John Hughes' comment here: math.stackexchange.com/questions/2037675/…

Hiro

18:31

Any idea how to approach this: math.stackexchange.com/questions/2037297/…

Antonio Vargas

18:54

@Hiro, could you just check if the 4th power of the adjacency matrix has any nonzero elements on the diagonal?

Maybe not, been a while since I thought about graph theory, that might count walks with repeated edges. Dunno.

user193319

Would someone mind checking the correctness of my proof given here: math.stackexchange.com/questions/2035711/…

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