Mathematics - 2015-01-28 (page 2 of 4)

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12:09 PM

@BalarkaSen I see you haven't been following politics closely.

I reckon I missed something (interesting).

@BalarkaSen Just saying that in politics, you can find things more idioter than Numberphile.

Oh.

Hehe.

Offtopic whine : I've been smacked in my algebraic topology classes today. There was an exercise in Hatcher asking to find simplicial homology of a 2-simplex with three vertices identified. I did it by showing that it is homeomorphic to the real plane minus two points the fundamental groups of which is Z * Z, so abelianization gives H^1(X) \cong Z + Z. :P

Such unfair.

@BalarkaSen What was unfair? The idea is to get experience with determining the homology from triangulations. Besides, I expect that the theorem saying that the first homology is the abelianisation of the fundamental group wasn't yet proved.

Yeah well simplicial homology is boring. Count dis, count dat, dat is da rank, plug up, etc.

that I have been smacked might have been a little overstatement.

12:21 PM

@BalarkaSen It's not idiotic. You may not like it, and there surely are poor videos, but many are enjoyable.

Never said it's idiotic.

Idiotest.

There are a lot of poor videos. Most of what I saw were poor. The zeta(-1) thing is more than enough reason to say that it's stupid.

Oh, LOL, sorry, I misread what you wrote, I'm actually in a boring (because of my professor) Latin class. I see what you mean, but Frenkel's video on thay compensates for the other who, which were poor. At least, to me. Dunno if it's generally gotten worse recently.

*on that

**other two

Hi everyone

hi @JohnJack

How goes it? @huy

A Beautiful Mind

12:34 PM

@robjohn You seem very quiet these days. You must be busy with work.

@JohnJack: Trying to study a bit of diffgeo. Not doing a very good job.

you?

You studying for a course or interdependently? Busy looking for some papers about variational inequalities, wish to be done with this course @huy.

@JohnJack: for an exam, but I don't know how on earth I'll pass it. I didn't really invest enough time.

Good luck with that, when is the exam? @huy

@JohnJack: Tomorrow morning. :P

12:42 PM

Hello =) At http://math.stackexchange.com/questions/1122481/the-diophantine-equation-y2-x37-has-no-solutions/1122506#1122506 in the first step, to show that $x$ cannot be even, can we also do it using $\pmod 4$ ???

$$x=2k \Rightarrow y^2=8k^3+7 \equiv 3 \pmod4$$

Since
$$y^2 \equiv 0,1 \pmod4 \equiv 3 \pmod4$$

$x$ cannot be even. Is this also correct???

Tough times :) @huy

@JohnJack: I'll survive it. I can still repeat it and do much better next time if all fails.

Would like to study some diffgeom some time also, quite interesting. @huy

Hi.

@JohnJack: I'm not going to lie: I expected it to be much more interesting.

12:49 PM

@huy Yeah but you are writing tom, you are bias to hating it at the moment.

@huy I can't really say though, I don't know much about it.

@huy have just read through some intros.

@JohnJack: Idk, I found QM and GR really interesting while I was studying it. Functional analysis too.

(studying for an exam)

What is «Course grade breakdown», what I am expected to write here in a course description form? Can anybody clear this up?

Yeah I would like to do more functional analysis and maybe applications of functional analysis in QM. That would be interesting. Do you have interest in topology at all? @huy

@JohnJack: Not really, I had an introductory class on topology and didn't really like it nor do very well on the exam. I took DG mostly for its usage in GR.

1:16 PM

Anybody?

@DanielFischer: I want to show that $O(n)$ is a Lie group by showing $\mathbb{1}_n$ is a regular value of $$f: M^{n \times n} \to M^{n \times n}, \, A \mapsto A^T A.$$ Can you help me understand what $df_A$ looks like?

@Huy Look at $f(A+H)$, and take the term linear in $H$. What do you get?

@DanielFischer: $A^T H + H^T A$?

@Huy Yes, so $df_A \colon H \mapsto \;?$

@DanielFischer: I still haven't really figured out how to actually compute $df_A$ in general, using our definition with the directional derivative.

@DanielFischer: Actually, do you maybe know some notes which have some insightful examples of finding $df_p$? I think that's the main problem, for me, because I am stuck at how to applying the definition.

1:36 PM

@Huy With the directional derivatives, look at $\gamma_{ij}(t) = f(A + t\cdot E_{ij})$, where $E_{ij}$ is the matrix with a $1$ in the $j$-th column of the $i$-th row, and zeros everywhere else. These form a pretty natural basis of the tangent space. Take $\gamma_{ij}'(0)$ to get $\frac{\partial f}{\partial x_{ij}}(A)$.

@Huy Ask Ted, my experience with differential geometry consists mostly of looking in the other direction.

@DanielFischer: So $\gamma'_{ij}(0) = A^T + A$? Or do I have to derive $A^T A$ as well (I took it to be constant)?

@Huy $\gamma_{ij}'(0) = A^T E_{ij} + E_{ji}A$

Yes, I thought something was missing.

But the coordinate-free version is more enlightening.

@DanielFischer: How do I get that one?

1:46 PM

@Huy $df_A \colon H \mapsto A^T H + H^T A$.

I have a question about an exercise of a diophantine equation. Can you help me?

@DanielFischer: Why is this surjective? Clearly the identity matrix is in there, but what else can we say about the map?

@Huy Since $A$ is invertible for $A \in O(n)$, we can write $H = A\cdot K$. If you look at the map in terms of $K$, $K \mapsto A^T(AK) + (AK)^TA$, it is pretty easy to see the kernel.

Morning.

Morning, @MikeMiller: More trivial problems have come up.

1:54 PM

Good night @Mike ... Yikes, it is night there.

hi @DanielF, @Huy

Hi, @TedShifrin.

Actually, I'm going back to bed. I just wanted to beat Huy today.

@MikeMiller: Beat me where?

Andrew Thompson

@TedShifrin, if I recall correctly, you are a lecturer/professor, right? Mind me asking some quick "career advice"-questions?

To saying good morning, @Huy.

1:55 PM

@TedShifrin: Do you maybe know some notes which have some insightful examples of finding $df_p$ for some $f: M \to N$?

Now... good night.

@Andrew: I am not around for long now, but sure.

@MikeMiller: I've been up since 9 hours...

Not offhand, @Huy. This is in every multivariable analysis type book, if you mean on $\Bbb R^n$. On manifolds, you work in local coordinates, or you note that $f$ is the restriction of a smooth map on the ambient $\Bbb R^n$ and compute there.

Hi @Ted. I'll hand the differential geometry questions over to you now.

1:56 PM

@DanielF: This isn't differential geometry. It's calculus!

I am leaving shortly, however.

Andrew Thompson

Alright, so last semester I took a normal Norwegian courseload of three subjects; Real Analysis, Topology and Calc3. Seeing as I got very limited time to read Calc3 I ended up with a C (Norwegian grades: pass from A-E, fail F). I am now taking quite a few subjects, including one graduate course. Should I spend time out of an already tight schedule to fix the Calc3-grade or will a seemingly impressive courseload make up for that?

@Ted He doesn't want to calculate Jacobians. He's worried about doing it in the abstract setting.

@TedShifrin Well, but for the DG course.

Andrew Thompson

I'm kind of uncomfortable with a C on my record.

@DanielFischer: I'm a bit confused. Am I supposed to simplify your map to $K \mapsto K+K^T$ which has skew-symmetric matrices as its kernel?

1:58 PM

And he seems to have some objection to coordinates.

Oh, I missed part of your message. I really will go back to bed.

@Huy Yes. Then you know the dimension of the kernel, hence the dimension of the image, and you also know the dimension of the codomain.

Well, @Huy, you need to show that $df_A$ is surjective for any $A$ with $f(A)=I$.

@Tedshifrin: Yes, that part I completely understand.

@DanielFischer: IIRC, the kernel's dimension is $n(n-1)/2$ and thus the one of the image $n(n+1)/2$.

@Andrew: I can't speak to the graduate committees in your part of the world. Here, C's aren't wonderful, but if you have strong letters of recommendation and strong grades in more advanced courses, it would not be a disaster.

Yup, @Huy.

2:00 PM

OK, I'm heading off to the office. See y'all later.

@TedShifrin: Laters.

@robjohn how are you doing?

@DanielFischer: But the codomain's dimension is $n^2$, so why is the map surjective? Am I missing something?

A Beautiful Mind

Hello @TedShifrin.

Hello @user130018

@Huy The codomain's dimension is not $n^2$. $f(M)$ is a symmetric matrix.

2:03 PM

@DanielFischer: Oh. I missed that.

Let me quickly look up the definition of a codomain.

Ok.

@Huy Well, you could take $\mathbb{R}^{n\times n}$ as the codomain. But then the derivative isn't surjective.

I'm trying to work out the fourier transform of $f$, where $f = 1-|x|$ for $|x| \leq 1$ and $0$ otherwise. I'm getting the same answer every time, and that is $\frac{2}{k^2}(1-\cos k)$, but this answer isn't correct. I don't suppose anyone would be able confirm that my answer is wrong, and then if I post what I've done help me find where I've gone wrong.

@DanielFischer: Exactly, that's what I thought.

@DanielFischer: So how do we know what the codomain is? It should be the tangent space at $f(A)$. Why is that the symmetric matrices?

@Huy If you view $f$ as a map $M_n(\mathbb{R}) \to \operatorname{Sym}_n(\mathbb{R})$, it's clear that the tangent space (at every point of the codomain of $f$) is $\operatorname{Sym}_n(\mathbb{R}$. If you view it as a map $M_n(\mathbb{R}) \to M_n(\mathbb{R})$, the tangent space at each point of the codomain of $f$ is $M_n(\mathbb{R})$, and the derivative of $f$ isn't surjective. As a rule of thumb, you take the codomain to be the smallest (nice) submanifold containing the range of $f$.

Here, $f(M)$ is always symmetric, and the space of symmetric matrices is a linear subspace, so a very nice submanifold. Worth taking a look at.

@DanielFischer: Ok. So it is definitely wrong writing $f: M_n \to M_n$, as was written in the lecture notes.

I think I'll head back to the differential again and see if I can work out some examples. I just don't have any little intuition yet and struggle finding the differential.

2:14 PM

@Huy If you want a surjective derivative, it is wrong.

But you don't necessarily need a surjective derivative to show something is a submanifold.

@DanielFischer: No, I don't necessarily need a surjective differential, but the exercise was supposed to be an application of the proposition that if $f:M\to N$ is smooth and $q\in N$ a regular value, then $f^{-1}(q)$ is a smooth submanifold of $M$.

@Huy And "regular value" was defined as "the derivative is surjective at every point of the preimage"? Then oops, mistake.

So I have $$\int_{-1}^1 e^{-ikx}(1-|x|)dx = \int_0^1 (1-x)e^{-ikx} dx + \int_{-1}^0 (1+x)e^{-ikx} dx.$$ Are these integrals correct?

@DanielFischer: Yes, exactly.

For the function I posted above

2:24 PM

I guess it was just a typo in the notes.

@user112495 Yes, that's correct so far.

@Chris'ssis Fine. Just got up. I have to proctor a midterm today.

Off to walk the dog now

@Huy Slip of attention rather than typo, but it belongs to the class of "minor mistake".

Yeah, I guess.

Rather confusing for someone not at all familiar with notation etc. yet though.

@robjohn btw, sorry for my questions, they become harder and harder. OK.

2:27 PM

@DanielFischer Using IBP, I then get $$\frac{1}{ik}-\left[\frac{1}{k^2}e^{-ikx}\right]_0^1 + \left( \frac{-1}{ik} + \left[\frac{1}{k^2}e^{-ikx}\right]_{-1}^0\right) = \frac{2}{k^2}(1-\cos k)$$

@user112495 There should be a $-\frac{1}{ik}e^{-ik}$ and similarly for the other integral.

$$\left[ -\frac{e^{-ikx}}{ik}\right]_0^1$$

@DanieFischer Just a yes or no : Is $\Bbb RP^n/\Bbb RP^{n-1}$ homeomorphic to $S^n$?

2:43 PM

That reminds me, I've seen Whitney's theorem stating that every secound countable smooth $n$-manifold can be embedded in $\mathbb{R}^{2n}$. As example, it is given $\mathbb{R}P^2 \to \mathbb{R}^4$. I thought $\mathbb{R}P^n$ are the lines through the origin in $\mathbb{R}^n$. Why is $\mathbb{R}P^2$ then a $2$-manifold and not just of dimension $1$ (lines)? @DanielFischer

RP^n is not lines through origin of R^n.

@BalarkaSen: Hm, I must be mistaking this notion then.

It's the space of all equivalence classes of lines through the origin of R^n.

i.e., R^n with identification (x, y) \sim (ax, ay)

@BalarkaSen: Which relation?

Ah.

You can easily see that RP^2 is not a 1-manifold then.

2:46 PM

@BalarkaSen Here $X/Y$ means to identify the subspace $Y$ to a single point?

Yes, @DanielFischer

@BalarkaSen: Do you mean 1 as opposed to 2?

@Huy $KP^n$ is the space of lines through the origin in $K^{n+1}$, so that you get something $n$-dimensional.

@BalarkaSen Look at a closed hemisphere and its equator.

Yes or no, @DanielFischer.

If you'll give me a no, I'll try to figure out why not.

@BalarkaSen You answer that, after looking at what I told you to look at.

2:49 PM

sigh

doing the Mike again

hehe

Well, I don't get your hint.

Closed hemisphere of what.

hi

I got a question..

@DanielFischer Are you saying that it's a "yes" and asking me to prove it?

If so, I can.

A complex vector is a pair of complex numbers or just 1 complex number?

2:52 PM

RP^n is a disk D^n with identifications x \sim -x at the boundary.

And the boundary is precisely S^{n-1}

@Mircea: Completely out of context, I'd assume the former, but not restricted to it being a pair.

@BalarkaSen Good start, continue.

So the antipodal identification at the boundary gives you a copy of RP^{n-1} embedded in RP^n

@Huy alright thanks!

If you punch the whole RP^{n-1} to a point, you're really punching all of the boundary of D^n to a point

Which gives you S^n

I was simply asking you to make sure I'm getting at a sensible result.

2:54 PM

@BalarkaSen Right. At some point, you need to actually verify that the bijection you get is indeed a homeomorphism, but that's pretty easy here, since the spaces are compact.

OK, phew. I can now compute homology of RP^n. Hurrah.

I will log out and be projective.

My analysis professor just got back from a stay at the Institute for Advanced Study :O

@DanielFischer: Is it true that for $n \leq 3$ for any topological manifold we can find a smooth structure to make it a smooth manifold?

Yes, @Huy, and uniquely so.

hi

3:09 PM

@MikeMiller Uniquely up to diffeomorphism. </annoying pedant>

You've gotten good at doing the Mike, @DanielF.

A Beautiful Mind

What's the Mike?

8-D

A Beautiful Mind

What's 8-D?

@ABeautifulMind Helping by "being completely unhelpful".

A Beautiful Mind

3:11 PM

You guys are speaking in mysterious tongues.

@ABeautifulMind A laughing guy with sunglasses (for coolness).

A Beautiful Mind

@DanielFischer Ah, OK. Also, D less than the number 8.

@MikeMiller: The amount of high schoolers in my maths class will reduce by three soon. That makes me sad.

@Huy Why, are you putting down the students you don't like?

@MikeMiller: No, but their physics teacher is being rather mean to them, imo.

(I'm not their physics teacher)

3:19 PM

So their physics teacher is rubbing them out.

Pretty much, yeah. Two of them were pretty lazy anyways, so they should have avoided the situation by working more, but one of them was a really hard worker and improved significantly in literally every subject at school but the physics teacher still did his best to make her fail, successfully. :(

The imaginary part of a+bi is b or bi?

b

alright, thanks again

How would you convert 11010.11 from base 2 to base 8?

3:28 PM

@DemCodeLines Collect groups of three bits, and transform each of these groups into base 8: $(011)(010).(110) \leadsto 32.6$

Good less night, @Mike

morning, @Ted

3:46 PM

People saw that we were here and immediately stopped chatting, @Ted. We must be scary.

Well, then it's a good thing I'm growing scarcer and scarcer.

iwriteonbananas

good morning ted

A Beautiful Mind

@mike @ted I am here, start talking.

I'm about to go teach class :P

good morning, bananas :)

@DanielFischer Sorry, I had skipped a couple of steps in my working. My full working is as follows:

$$\int_0^1 (1-x)e^{-ikx} dx + \int_{-1}^0 (1+x)e^{-ikx} dx$$

$$ = \left[ \frac{x-1}{ik}e^{-ikx}\right]_0^1 - \int_0^1 \frac{1}{ik}e^{-ikx} dx - \left[\frac{x+1}{ik}e^{-ikx}\right]_{-1}^0 + \int_{-1}^0 \frac{1}{ik}e^{-ikx} dx$$

$$ = \frac{1}{ik} - \left[\frac{1}{k^2}e^{-ikx}\right]_0^1 - \frac{1}{ik} + \left[\frac{1}{k^2}e^{-ikx}\right]_{-1}^0$$

$$ = -\left(\frac{1}{k^2}e^{-ik} - \frac{1}{k^2} \right)+ \left(\frac{1}{k^2} - \frac{1}{k^2}e^ik\right) = \frac{1}{k^2}\left(1 - e^{-ik} + 1 - e^…

3:58 PM

@MikeMiller: I still don't quite understand how the Möbius strip is supposed to be a subset of $\mathbb{R}P^2$. Can you help?

iwriteonbananas

i can help

Sure.

go bananas! :)

iwriteonbananas

write down the polygonal representation of $\mathbb{R}P^2$

I have never heard of a polygonal representation of $\mathbb{R}P^2$, unfortunately.

(or of anything)

iwriteonbananas

3:59 PM

now draw two vertical lines in the square

boom mobius strip in there

:D

You have to do a bit more teaching than that, bananas.

iwriteonbananas

what else do u want me to say ted?

Well, @Huy only knows the projective plane as a disk with its boundary antipodally identified. You have to show him in that picture or you have to teach him to think of it the way you think of it. But I'm off to teach.

iwriteonbananas

ok shouldnt bee to hard to show that those are the same

have fun

@user112495 Okay, that looks correct. What makes you think it isn't?

4:02 PM

@Huy: Do you know how $\Bbb{RP}^2$ is a quotient of $S^2$?

@MikeMiller: Yes.

@DanielFischer It was a question in an online quiz we were set, and it's not accepting the answer as correct.

Look at what happens to a cylinder around the equator under the quotient map.

@DanielFischer I'll probbaly email the lecturer to make sure then.

@user112495: Online quizzes are rubbish.

@MikeMiller: I'm trying to.

4:04 PM

I've made a mistake here in math.stackexchange.com/review/suggested-edits/344055 but it looks like there's nothing to help me retract/fix/comment on the edit... I was thinking of making a post on meta.SE?

Also if someone would be kind enough to decline it... :)

@user112495 It maybe uses a different convention for the Fourier transform. Is it mathematics or engineering? If mathematics, you should probably multiply with $\frac{1}{\sqrt{2\pi}}$. Or maybe use the $\int f(x) e^{-2\pi i kx}\,dx$ convention.

@Huy Thank you...

@DanielFischer It is technically a physics module I guess. But the questions were all set by the lecturer, and so I used the convention he used in the lectures, and that worked for the other questions.

although so now it needs a minimum of 6 characters changed for my edit to go through... it was supposed to be Z and Q and I accidentally put R and Q

@user112495 Yes, you should use the convention used by the lecturer, whichever that one was.

4:09 PM

@Huy thank you for the fix

150 users with 200+ rep (currently 194 users with 200+ rep)
• 10 users with 2,000+ rep (currently 16 users with 2,000+ rep)
• 5 users with 3,000+ rep (currently 12 users with 3,000+ rep)

wtf so I am only one of 16 users with over 2k rep?

Or 1 in 12!

that's really low

Where are you getting that stat?

iwriteonbananas

@Huy i.imgur.com/Um5Zjgn.jpg that's the polygonal representation of $\mathbb{R}P^2$

if u draw two vertical (or horizontal, doesnt matter) lines in that square, u can recognize a mobius strip

4:25 PM

@Nahiyan data.stackexchange.com/math/queries?order_by=featured

Haider

@Nahiyan Hello angry Pikatchu

@DonLarynx any idea ?

0

Q: Uniform convergence to 0

Let $(f_n)_\mathbb{N}$ be a sequence of continuous functions converging to $0$. The functions are such that for all $x$, $(f_n(x))_\mathbb{N}$ is decreasing. How can one show that the $f_n$ converge uniformly to $0$ ? What I got so far : Let $x_n$ such that $f_n(x_n)=\sup\{f_n(x)\}$, we sup...

functions convergence

What do you get when you multiply a matrix and a vector?

A Beautiful Mind

A vector is a matrix.

And how do you do it? Do you just turn the vector into a column matrix?

A Beautiful Mind

When you multiply two matrices you get a matrix.

Usually a vector is written as a column matrix.

4:36 PM

"written as" is not the same as "is"

A Beautiful Mind

Well, your question lacks context. Where do you want to multiply a vector and a matrix?

first property

In mathematics, a complex square matrix U is unitary if where I is the identity matrix and U* is the conjugate transpose of U. In physics, especially in quantum mechanics, the Hermitian conjugate of a matrix is denoted by a dagger (†) and the equation above becomes The real analogue of a unitary matrix is an orthogonal matrix. Unitary matrices have significant importance in quantum mechanics because they preserve norms, and thus, probability amplitudes. == Properties == For any unitary matrix U, the following hold: Given two complex vectors x and y, multiplication by U preserves their i...

@Hippalectryon Any progress on my question? :-)

@Chris'ssis Not yet :c

It's just applying the matrix to the vector. Matrices are linear transformations.

A Beautiful Mind

4:39 PM

It's also just matrix multiplication, ain't it?

I didn't work with bost vectors and matrices at the same time so far

both*

I didn't work with bost vectors and matrices at the same time so far

A Beautiful Mind

Well, in this case, the vector is a matrix.

@Mircea What's the point of a matrix if you can't apply it to a vector?

so you just write it as a column and do the multiplication?

A Beautiful Mind

Yes, you just multiply matrices.

4:40 PM

@MikeMiller So far I've only multiplies matrices with other matrices or scalars and used them for solving linear equations

@ABeautifulMind alright, thanks

A Beautiful Mind

@Mircea Actually, I should not be answering you, I might be wrong. I have forgotten 99 per cent of math.

@Chris'ssis Is there any progress on your love life? =)

@MikeMiller It's good for MM to answer M.

@Mircea One can define linear transformations and vectors without matrices, of course. But both can be represented by matrices.

@ABeautifulMind I am too confused about someone teaching matrices without teaching how they represent linear maps, which is the entire point of matrices. :/ I guess it doesn't matter if one learns it eventually, e.g. now...

A Beautiful Mind

@MikeMiller Well, I think poor quality teaching and curriculum is everywhere in the world.

(I am not criticising you, @Mircea, in case it sounds like it.)

@Hippalectryon my knowledge in real analysis is slowly but surely decaying. I'd be glad to answer any questions related to discrete maths/number theory

4:48 PM

@MikeMiller i know

@MikeMiller but matrices are also very useful when calculating systems of linear equations

A Beautiful Mind

@Mircea Yes, that is true.

When you're finding solutions to those systems you're really considering them as linear maps. They just haven't told you that.

@MikeMiller also, the highschool curriculum doesn't even cover vector spaces so it would be silly for it to cover linear maps... i think

A Beautiful Mind

@MikeMiller Well, she is only in HS.

@ABeautifulMind he*

4:50 PM

Not at all, they're just linear maps between $\Bbb R^n$ and $\Bbb R^m$, instead of between abstract vector spaces.

A Beautiful Mind

@Mircea Oh? I thought you were a girl, LOL.

@ABeautifulMind I don't see how that's relevant, though.

@ABeautifulMind it's ok, 90% of foreigners do

Mircea is usually a male name, no?

A Beautiful Mind

I don't know where I got the wrong impression.

4:52 PM

@MikeMiller yes

A Beautiful Mind

Maybe I have associated you with Chris's Sis.

@ABeautifulMind well it ends with ea

A Beautiful Mind

@mike My high school crush married a guy called Mike. When I was in high school, I prayed to God that he would take away all my happiness and give it to her, it seems my prayer came true. I want my happiness back.

lol

@JMoravitz: thanks :)

5:13 PM

Hello!! What does the symbol $\mathbb{P}^1(\mathbb{Q})$ mean??

Sayan Chattopadhyay

Hi guys

My linear Diophantine solver is complete!!! Give me any three integers and I will find you a solution.

@ABeautifulMind Why would you want your happiness taken away?

A Beautiful Mind

@DonLarynx To give it to someone I loved.

@ABeautifulMind You should only love those who love you back.

i.e. Don't be a doormat

A Beautiful Mind

@DonLarynx Well, I know better now.

iwriteonbananas

5:23 PM

@MikeMiller the group $SL_2(\mathbb{R})$ operates isometrically on $(\mathbb{H}, d_{ \mathbb{H} })$ by $\sigma(z) = \begin{bmatrix} a & b \\ c & d\end{bmatrix} z = \frac{az+b}{cz+d}$

what does this mean?

does it mean $d_{\mathbb{H}}(z,r) = d_{\mathbb{H}}(\sigma(z), \sigma(r))$ ?

@DanielFischer @TedShifrin Hello!!! :-)
How could we show that the diophantine equation $x^p+y^p=z^p, p \in \mathbb{P} \setminus{2}$ has no solution?

A Beautiful Mind

@DonLarynx Are you going to reconcile with your ex?

@ABeautifulMind no

I've lost all my trust and respect for her, so no.

A Beautiful Mind

@DonLarynx Does she still like you?

Yes.

@ABeautifulMind

A Beautiful Mind

5:37 PM

Aww.

@Huy Give me any three integers.

Sayan Chattopadhyay

Hi @DonLarynx

@DonLarynx: 10000!, 1000000000!, 100000000000000000000000000000000!.

Sayan Chattopadhyay

My lord such huge numbers

@Huy: The numbers cannot exceed 2.147 * 10^9

5:40 PM

@DonLarynx: Sorry, I can't help you then.

Sayan Chattopadhyay

@DonLarynx are cubes fine

one hundred nonillion

Sayan Chattopadhyay

Don can u find huge cubes

I am out. My task is complete.

Sayan Chattopadhyay

Don can u find huge cubes for me once

5:42 PM

@iwriteonbananas It means either that it preserves the Riemannian metric or that it preserves the hyperbolic metric on the upper half plane.

Not the Euclidean metric.

We have that $z^2$ is odd. Does this mean that $z$ is also odd??

Sayan Chattopadhyay

Hi @MikeMiller

@DonLarynx did u proceed anywhere on my question

iwriteonbananas

@MikeMiller ok, so in other words for any matrix in $SL_2(\mathbb{R})$ the mapping $\sigma$ from above is an isometry with respect to the hyperobloc metric?

aye

Sayan Chattopadhyay

Why aren't people talking to me on this room

iwriteonbananas

5:48 PM

aye aye captain

@MaryStar You are asking whether $z = 2k+1$ for some $k \in \Bbb{N} \cup \{0\}$.

Sayan Chattopadhyay

Hi @DonLarynx

@MaryStar Suppose $z$ wasn't odd. Then $z = 2k, k \in \Bbb{N}$. But then $z^2 = 4k^2$ which is clearly even. hence $z$ is odd

Does this answer your question?

Sayan Chattopadhyay

@DonLarynx is there any form for cousin primes

I mean general form

@MikeMiller: I watched The Imitation Game today. Was rather good, as a movie.

5:55 PM

Yeah. Not particularly historically accurate, but a good movie.

@MikeMiller: I wouldn't know about the accuracy of the actions themselves, since I didn't know much about them before watching the movie. Anything not character-related that you know was shown in a false way?

They certainly knew to check for phrases they knew would be in the messages from the start - that's one of the most obvious ideas there is. Turing's (and the other mathematicians') contributions were mainly in optimizing.

the thing about diagonal bands or whatever is probably more in line with history than realizing messages end with heil hitler

@MikeMiller: That part I thought too, it's a really obvious idea.

Turing wasn't, to my knowledge, the socially awkward (semi-autistic?) fellow he was portrayed as.

@MikeMiller: I guess they switched the roles of those two ideas for better mainstream-understanding?

5:58 PM

Probably. A movie where the key revelation is that their algorithm could be made $O(n\log n)$ instead of $O(n^2)$ probably isn't very exciting.

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