Mathematics - 2016-01-16 (page 1 of 3)

Superstar Monica

00:02

Despite the downvotes, I'll continute to post answers (just to be known - although they will be rare).

dzackgarza

00:26

@MaryStar Yep, and more generally, the eigenvectors of any diagonal matrix are just all of the standard basis vectors $e_i$

(By which I mean the vectors with zeroes everywhere and a 1 in the i'th coordinate)

Mary Star

I see... Thanks a lot!! :-) @dzackgarza

Kevin Driscoll

01:20

Nomenclature question for you guys, is there a name for taking function of derivatives, for example $$\cos{\frac{d}{dx}} \equiv \sum_{n=0}^{n=\infty} \frac{(-1)^n}{(2n)!} \frac{d^{2n}}{d x^{2n}}$$

I keep trying to search for "cosine of a derivative operator" and all I get are results for "derivative of a cosine"

WDUK

I am so confused by this. If i have to subtract two large numbers i do it like this:
(9*10^10) - (9*10^6) = 8.9991*10^10

But does it even sound right to say 9*10^6 is 8.9991*10^10 smaller than 9*10^10

or should i do 9*10^(10-6) = 9*10^4

Kevin Driscoll

01:44

@WDUK The first way is correct

$9*10^{10}$ is a LOT bigger than $9*10^6$. Think about it, the first has 10 zeros after the 9, but the second only has 6.

$9*10^{10}$ is 10,000 times bigger

WDUK

@KevinDriscoll but the answer 8.9991*10^10 means the difference is 8 with 10 zeros

its confusing to understand that =/

I was told to use significant figures that is given in the calculation so i would use 1 significant figure and get an answer of 9*10^10

which means i didn't subtract any thing

Kevin Driscoll

Yes, that's right

To 1 significant figure, the difference is 0

And the difference SHOULD have 10 zeros. Try writing it all out long form

$9*10^{10} - 9*10^6 = 90000000000 - 9000000$

when you see how much longer $9*10^{10}$ is, it makes more sense that the answer should be an 8 with 10 numbers after it

WDUK

So maybe i should not round the significant figures here

and keep to the precise number =/

Kevin Driscoll

02:16

If they say to use sig figs, then do

WDUK

they don't say to, was just told that the most accurate your answer can ever be is equal to the number with the lowest s.figures in the sum

Nathvi

02:32

Hello math people

Balarka Sen

08:08

@MikeMiller Hatcher clarifies in his exercise set why $S^1$ with two disks attached by degree 2 and 3 maps is homotopy eq to $S^2$. If I interpret his hint correctly, here's how it goes. $X$ be the CW complex obtained from attaching the two 2-disks by deg 2, 3 maps to $S^1$. This is the same as attaching disk wedge disk to $S^1$, one disk by deg 2 other by deg 3. I think this attaching map is homotopic to the one which attaches each disk in the edge by identity map to $S^1$.

Now prop 0.18 should give you a proof. I am not too confident about this though.

oh, actually, here's the correct way to do it. Attach one disk by degree 2 map. You get $\Bbb{RP}^2$. Then if you attach the next one by degree 3, as a degree 3 loop is homotopic to one with degree 1 ($\pi_1(\Bbb{RP}^2)\cong \Bbb Z/2$), you can homotope the attaching map to degree 1. Now apply prop 0.18.

Well, that wasn't so hard.

Mike Miller

@Balarka: I'm going to bed but don't forget I gave you a problem yesterday

Balarka Sen

Yep, haven't forgotten.

Will get to it after checking through some problems in Hatcher's additional exercises. The other half of (1) doesn't look as obvious as the first half.

Hi @iwriteonbananas.

iwriteonbananas

08:44

@BalarkaSen Good morning.

Balarka Sen

How's it going?

iwriteonbananas

Meh...doing exercises but I'm struggling

Balarka Sen

In diffgeo?

iwriteonbananas

yep

Balarka Sen

:(

iwriteonbananas

08:46

What're you up to?

Balarka Sen

Went through a few problems from Hatcher's additional exercises. Now doing exercises Mike gave me.

iwriteonbananas

Kewl, on what topic did Mike give you exercises?

Balarka Sen

Vector fields and stuff. See them yourself : chat.stackexchange.com/transcript/message/26870821#26870821.

iwriteonbananas

heh, nice

Balarka Sen

Stumbled upon the proof of (3) by accident. Apparently there's something topological going on. Want to think a bit more about it.

iwriteonbananas

08:52

Yes, there is

Balarka Sen

Don't reveal, then!

:)

iwriteonbananas

Oh, I didn't even think about revealing :P

Balarka Sen

09:04

@iwriteonbananas So, what have you learnt about smooth manifolds?

iwriteonbananas

That's a loaded question.

Balarka Sen

In what sense? :P

Too broad?

iwriteonbananas

I'm actually not so sure what I learned about smooth manifolds....I'm trying to process all the stuff we did.

We talked about Riemannian metrics, covariants derivates and curvature

Balarka Sen

Ah, ok, so Riemannian geometry. I don't know anything about that.

iwriteonbananas

But my lecture is void of examples or ideas, so I need to learn it on my own

Balarka Sen

09:12

Yikes.

iwriteonbananas

First world problems I guess :P

Balarka Sen

I wish you luck on self-studying Riemannian geometry.

(so that you understand it well enough to be able to answer silly question I'd ask when I learn it)

:)

Khallil

09:41

Hey, @BalarkaSen, @MikeMiller and @Huy :-)

Balarka Sen

Hi @Khallil.

Khallil

09:57

How's it going, @BalarkaSen?

Balarka Sen

10:09

@Khallil Alright. How about you?

Huy

morning

Khallil

Not too bad. Woke up fairly early today for a pleasant change ^_^

Morning, @Huy!

Huy

Hannah Murray?

Khallil

I think it is :-)

I just saw the image on Google images when I was searching for something else and it caught my eye

Alenanno

Hello all

Superstar Monica

10:25

I encourage you to post more answers here

1

A: Convergence of $\sin{\pi\sqrt{n}}$

To answer the point $iii)$, note that $$\lim_{n\to \infty} \sin ^2\left(\pi \sqrt{n^2-n}\right)=1,$$ since we have that $$\sqrt{1-\frac{1}{n}}\approx 1-\frac{1}{2n},$$ when $n$ is large. Q.E.D. (which you combine with $\sin{\pi\sqrt{n^2}}=0$ to show divergence)

100 points bounty!

(to be honest it will be hard to beat robjohn - but you never know until you try it)

Superstar Monica

10:39

BBL (kind of busy with serious research)

user276387

How do you use Abel's summation formula if you have something like $\sum_{n} (-1)^na_n\phi(n)$.

Abel's summation formula

In mathematics, Abel's summation formula, introduced by Niels Henrik Abel, is intensively used in number theory to compute series. == Identity == Let be a sequence of real or complex numbers and a function of class . Then where Indeed, this is integration by parts for a Riemann–Stieltjes integral. More generally, we have == Examples == === Euler–Mascheroni constant === If and then and which is a method to represent the Euler–Mascheroni constant. === Representation of Riemann's zeta function === If and then and The formula holds for It may be used to derive Dirichlet's t...

Superstar Monica

11:02

Lots of ways to do it.

user276387

Share one of them, please, @SuperstarMonica.

Superstar Monica

@user276387 Well, yesterday you were the user that talked about 'delusions of grandeur'?

I don't wanna help you, sorry (but I already did it saying there are lots of ways).

user276387

I'm baffled! What does that has to do with anything!?

Superstar Monica

For instance, does it help to know that $\cos(\pi n)=(-1)^n$?

BBL(back to my research)

Superstar Monica

11:20

(one more thing - that was a point besides redefining $a_n$ as being $(-1)^n a_n$)

Now I'm away for sure.

BBL

user276387

11:34

@SuperstarMonica $\cos(\pi n)$ did it for me. Many thanks.

Overflowh

Pardon, how is it called the double modulo operator? Like this: $\| y - y_0\| \leq b$

Khallil

I think we call it a norm, @Overflowh.

Overflowh

@Khallil Which is the perpendicular to $y-y_0$?

Khallil

I'm not sure what that means :(

I just meant that $\| \cdot \|$ is often called a norm.

Overflowh

@Khallil Actually I'm not use either because, in the theorem I'm studying, my book reports that $y-y_0$ thing surrounded simply by the modulo operator (single straight line), while wikipedia uses the double/norm version :/

Owatch

11:47

Morning all.

Huy

the absolute value function is a norm too

Owatch

Would you say: $y' + xy^{2} = \sqrt{x}$ is a non-linear differential?

I can't see how it fits the required: $y' + P(x)y = Q(x)$ formula

And then, what about $y' - \frac{1}{y} = \frac{1}{x}$. Can $P(x)$ be 1??

Overflowh

@Huy Which means the two notations are interchangeable?

Huy

no

just because every thumb is a finger doesn't mean the two notions are interchangeable

Overflowh

And this implies that either Wikipedia or my book is wrong

Huy

11:55

no

Overflowh

Why not?

Owatch

I have encountered the problem: $y' - 5y = x$. So I guess $P(x)$ may be $1$.

Owatch

12:23

Wolfram keeps giving me something weirdly different, but can anyone attempt to tell me what they get when solving: $x^{4} \frac{dy}{dx} + 4x^{3}y = sin^{3}x$ ?

Huy

$\frac{c_1}{x^4} - \frac{3 \cos x}{4x^4} + \frac{\cos 3x}{12x^4}$

why is this weird

Owatch

Nope... I think its my integral of $sinx^{3}$ that is the problem

Huy

what are you talking about

Owatch

I need to integrate sinx on the RHS

I think I did it wrong, which is why my answer is different.

Huy

ok

Owatch

12:27

I wrote it as: $\int{ (sinx)^{2} * sin(x)}$ and then as $\int{(1-cos^{2}x) * sin(x)}$.

And then as $- \int{(1-u)du}$

Huy

at least one factor of $\frac{1}{2}$ went missing

Owatch

I see a mistake already. I mis-wrote $\int{du}$ as $\int{u du}$

And I now get: $y = x^{-4}(-cosx + \frac{1}{3}cos^{3}x + C)$

Still wrong

Huy

are you just integrating both sides or what are you doing

Owatch

I've integrated both sides, I'm trying to solve for y..

Huy

that's not how solving differential equations works

Owatch

12:33

Is it not? Because that is what my book instructs.

Huy

how do you integrate $4x^3 y$

Owatch

What do you mean? The LHS is $\int{\frac{d}{dx}(x^{4}y)}$

Which doesn't integrate to anything.

Just $x^{4}y$

Huy

what

Owatch

Yeah.

Huy

ah I see what you're trying to do

still don't know where the problem is

what is your result for $\int \sin^3 x \, dx$

Owatch

12:40

It's $-cos(x) + \frac{1}{3}cos^{3}(x) - C$

Huy

wait a second

your sub before was wrong

how did you proceed after $\int (1-\cos^2 x) \sin x \, dx$

Owatch

u = cosx

du = -sinx dx

Huy

yes

so you get $-\int (1-u^2) \, du$

Owatch

Yes.

And then I made it: $-1 (\int{du} - \int{u^{2}du})$

So: $-1(u - \frac{1}{3}u^{3} + C)$

Huy

hm

seems alright after all

sorry my mistake

Owatch

12:45

Or: $-u + \frac{1}{3}u^{3} - C$. Then subbed back in cos.

Huy

yeah I think all is alright

you probably just need to apply some weird trig identity to get the same result as on wolfram

Owatch

I hope so.

With my luck, I'd be asked exactly this, and the prof would check with wolfram or something, IDK.

Huy

cool

Owatch

Thanks.

Huy

np

weird profs you have over there

I doubt we'd have a prof check this kind of stuff

Owatch

12:47

Well, I was speculating as to what sort of bad luck I would probably get. I don't know how the profs grade.

Huy

ok

Owatch

I'd be skeptical too, the prof we have makes lots of mistakes on the board.

And I know he posts answers in the form of hand-written scans.

Huy

cool

better than not at all

Owatch

Yes. I have another professor who refuses to give out anything.

Answers to homework? Denied. Can we get help before-hand from the TA's? Denied. Can we see the answers to the old-exam paper? Denied (He had to give us one as the school requires they make an example available, but he decided the answers aren't included)*

Can we see our exams after the mid-terms? Denied!

Huy

very good

Owatch

12:52

I remember the mid-exam started with a multiple-choice problem section before going on to the main part (Problems). And I was writing down my answers on scrap paper so I didn't have to go back to the cover page and fill them in over and over. The supervisors came and took it away. :( No writing answers like that down I guess.

Huy

interesting school

Owatch

No other exams did that. It's not good.

It's not the school, it's the professor in question who makes things hard.

He tells us the reason he doesn't provide answers for homework and whatnot (After it's been done for weeks even) is somehow because we need to be motivated to study or something. I don't know. It's hard to imagine how to study well if you never know what you're doing is correct.

He seems a bit paranoid to me.

He asked in the lecture once if we had covered boolean algebra or something. And someone stupid said we had done a bit of it in the last block. So he said okay, and skipped to the end of the whole section and decided since we apparently knew it all (Obviously not) none of it would be covered at all.

Huy

your fault for not screaming out "NO WE DIDN'T"

Owatch

He also decided to make the homework harder, since he felt people were getting higher grades than they ought to. (He specifically said he felt people must be cooperating even thought he could not prove it, since they are pretty un-varied answers), and so the difficulty would be increased to a level where you would need multiple people to solve it.

I probably ought to have said we didn't. But I still feel he is being a bit aggressive about it.

At your last comment.

Owatch

13:20

I've ran into $\int{sin(e^{x})e^{x}}dx$.

I can't think of a good way to approach this. I cannot substitute, by parts will only make it a bigger problem, what do I do?

I can't think of any identities either.

Huy

what's the derivative of $\cos(e^x)$

Owatch

$-e^{x}sin(e^{x})$

Huy

almost

Owatch

Plus see

Huy

no, you don't add a constant if you differentiate

Owatch

13:27

So you

Oh oops

Huy

$(\cos(x))' = -\sin(x)$

but usually I fix sign problems after trying out differentiation functions

Owatch

Oh, right.

Huy

so there's your integral

Owatch

Thanks.

But what way other than intuitively can this be solved? Was it possible to do it with substitution or IBP?

Huy

substitution is chain rule reversed

this was chain rule reversed

Owatch

13:30

I do see how it works this way of course.

oh..

Huy

substitute $t = \cos(e^x)$, so $dt = - \sin(e^x) e^x \, dx$, so the integral becomes $-\int dt = -t + C = -\cos(e^x) + C$

imo it's easier to "just spot the chain rule" instead of trying to apply substitution here though, even though it's the same

User1

hey

anyone have a min to help me work out somthing rather easy?

its regarding this proof ; proofwiki.org/wiki/Banach-Alaoglu_Theorem

user276387

13:53

How do you apply Weierstrass substitution to something like $\displaystyle \int \frac{\sin{nx}}{2\sin(3x)+4}\,{dx}$

That's just a random example, btw. But I'm wondering how to use that substitution when your numberator contains $\sin{nx}$.

Huy

15:19

@DanielFischer: a very simple question: when showing that $\mathfrak{sl}_2(\mathbb{R})$ is simple, we considered the usual sl2-triple $e,f,h$. assume $V$ is an ideal of $\mathfrak{sl}_2(\mathbb{R})$. note that we can restrict $\operatorname{ad}_h: V \to V$. now the argument is that assuming $V \neq \{0\}$, one of the three eigenvectors of $\operatorname{ad}_h$ must be contained in $V$, i.e. $h,e$ or $f$. why?

I understand well that those are eigenvectors to $-2, 0, 2$ on $\mathfrak{sl}_2(\mathbb{R})$ but why does one of them have to be in $V$ too then?

I mean on $R^3$ I could have some eigenvectors $e_1, e_2, e_3$ and a subspace spanned by $e_1 + e_2$, which contains none of those eigenvectors. I assume it has something to do with the ideal property, @DanielFischer ?

Khallil

@Huy, do you know if this is the place to be discussing measure theory?

Huy

@Khallil: last time I checked measure theory indeed was a part of mathematics

Khallil

I was thinking about posting in the set theory room but I wasn't too sure ^_^

Huy

no

Khallil

Do you know why it's important that we use the Borel probability measure for integrals over the surfaces balls?

Huy

15:26

probably because it works out fine, but no, I have forgotten almost everything I ever knew about probability

Alenanno

@AsafKaragila hey, did you receive my ping? Just making sure.

Khallil

When you say it works out fine, is that in the sense that's it gives a normalised result for a unit sphere?

Huy

can you elaborate where you exactly encountered that, @Khallil?

Khallil

Yep, I was looking at the mean value property of harmonic functions - If a function's harmonic over a ball, then the value at the centre is the same as the average over the surface of the ball.

Huy

mostly, the Borel measure implies it's unique

Khallil

15:27

Uniqueness was mentioned.

Huy

honestly I wouldn't care too much about it if you haven't even seen Borel measures yet

Khallil

Ah, ok!

Huy

maybe it's important at some point in the proof of the property but frankly I don't remember

can't be very important if I don't remember, right? ducks

Khallil

Haha! I don't recall seeing it in the proof explicitly. It may be hidden

It probably has more impact in a volume sense.

Huy

harmonic on B means with respect to the usual Laplacian in $R^n$?

Khallil

15:37

I believe so yep! The usual sum of single second partials

iwriteonbananas

@MikeMiller Hey, are you here?

Huy

hm

I feel like you should have to divide by something

like the volume of the ball or something like that

Khallil

Some normalisation factor maybe?

Huy

is that included in your $\mathrm d\sigma(\xi)$?

Daniel Fischer

@Huy Something with that indeed. $V$ is an $\operatorname{ad}_h$-invariant subspace.

Huy

15:39

@DanielFischer: yes, because $V$ is an ideal

I still don't quite see the connection

Khallil

It was only mentioned in passing that $\sigma$ would be the unique probabilistic Borel measure :-/

Clarinetist

I can't find anywhere where it would be appropriate to ask this, but I might as well try here...

Does anyone have recommendations for abstract algebra and topology texts ? Assume I have a good grasp of introductory real analysis (without topology), i.e., Bartle & Sherbert. I don't like Munkres or Dummit & Foote.

Here's what I look for in a textbook:

- I don't like extra fluff. A great example of this is teaching combinatorics in probability. As much as I love probability (heck, I'd maybe like to do a Ph.D. studying probability some day), combinatorics is not at all essential to understanding undergraduate probability.

Daniel Fischer

@Huy You have a linear operator $T \colon E \to E$ with $\dim E$ distinct eigenvalues. You have a $T$-invariant subspace $F\subset E$. If $F\neq \{0\}$, does it have to contain some eigenspace of $T$?

Huy

yes, otherwise it wouldn't be invariant -_-

@Clarinetist: Munkres. :P

why do you not like it?

Munkres is all about topology

Daniel Fischer

@Huy (and @Khallil) That's what is done. You divide the surface measure by the area of the sphere and you get a positive measure such that the sphere has a total measure of $1$, i.e. a probability measure.

Huy

15:48

@Clarinetist: about the surjective iff right-inverse thing: that's something I have seen in freshmen analysis, for example. most people seriously attempting to learn topology would very likely have studied analysis before, so already know those kind of results

Clarinetist

@Huy First chapter is extremely dense (axiom of choice, orderings, etc.). I don't have the expertise to know what to skip on a first read.

Sigh, I didn't even know the axiom of choice existed until the senior year of my undergrad

Khallil

I still fail to see why a probabilistic measure would be important in defining such a property as the mean value one, @DanielFischer. Could we not have chosen another measure instead?

Clarinetist

and we used it once

Huy

@Clarinetist: that's not too uncommon if you haven't had any topology or algebra

I think outside of those courses and functional analysis, we've only used it to prove existence of basis

Clarinetist

@Huy I've considered just relearning calculus entirely

But there's no way I would relearn calculus without at least adding some analysis to it

I.e., I am not going to learn entirely from Stewart

I've considered doing a mix of Spivak, Rudin, and Stewart simultaneously

Huy

15:51

@Clarinetist: can you send me the table of contents of munkres?

I think you need AoC only for Tychonoff, but I could be wrong

Daniel Fischer

@Khallil You take an average, that means you integrate the function and divide by the total measure of the space. If you take a (positive) measure $\mu$ on a space $X$ and then consider the measure $\sigma = \frac{1}{\mu(X)}\cdot \mu$ to avoid writing $\frac{1}{\mu(X)}$ all the time, you get a positive measure with $\sigma(X) = 1$. Such measures are called probability measures. Simple as that. It hasn't really anything to do with probability.

Clarinetist

@Huy Yeah, I'll send you an e-mail

Paulo Cereda

@Huy Where, where?! Oh, you meant the verb. :)

Alenanno

@PauloCereda :D

Paulo Cereda

We ducks are watching the network. :)

Huy

15:53

-_-

Clarinetist

@Huy Sorry, didn't catch that fast enough. Would your gmail be fine?

Huy

sure

Clarinetist

Thx

Huy

but then google knows that you're studying Munkres

Paulo Cereda

By the way, hi friends at Math.SX!

Huy

15:55

hello duck

Clarinetist

Dang Machine Learning.

Hmm

Huy

@Clarinetist: just out of curiousity, why do you want to study topology?

is that necessary for statistics?

Clarinetist

@Huy If I ever want to do a Ph.D. statistics, I think doing well on the Math GRE Subject Test would benefit me

Huy

ok

no idea about the American system

@Clarinetist: but don't you wanna do one after the other? I mean you also have to catch up on some serious linear algebra, right?

Clarinetist

@Huy Calculus has a 50% weight on the exam. I'd like to do analysis + calculus + topology simultaneously to knock that out, and then proceed to linear algebra, and then abstract algebra

Huy

15:59

but linear algebra is super important :P

and you even use it in your fancy stats proofs

Clarinetist

@Huy Trust me on this one, I have it taken care of. I bugged the professor and he's given me the Ph.D. course notes for the class I'm taking

Huy

why split up calc and analysis btw?

do you actually have to do a lot of calculus at that GRE test?

Clarinetist

@Huy It's been a while, but I remember a lot of calculus. The questions that really got me, though, were some of the analysis + topology questions. I still don't know what a compact or connected set is, for example

Alenanno

@Clarinetist via email? Because if so, I can already guess how long his email was. :D

Clarinetist

@Huy For the TOC to Munkres, click here, click the picture of the book, and scroll down

Huy

16:02

@Clarinetist: compact = every open cover admits a finite subcover. connected = the only clopen subsets are the whole set and the empty set

now you know :P

Khallil

Oh, I see. That makes a lot more sense. Thanks, @DanielFischer!

(Also, sorry for the late reply.)

Clarinetist

Now I need to learn what an open cover and finite subcover are @Huy :P

Huy

@Clarinetist instructions unclear, my head is stuck in the microwave

Clarinetist

@Huy

Huy

yeah, just do chapters 2 and 3 to start with @Clarinetist

look up what you need in chapter 1 when necessary

that would be the very basics of topology

Mary Star

16:05

A plane through the origin and perpendicular to the unit vector (a,b) is $(x,y)\cdot (a,b)=0$, right?
This stand when $a,b\in \mathbb{R}$, right?
What if $a,b\in \mathbb{C}$ ?

Clarinetist

K thanks @Huy!

Huy

and don't skip the quotient topology part

Clarinetist

TIL what a topology is -_-

Huy

(Y)

iwriteonbananas

@MikeMiller What you describe here is essentially the following, isn't it? Given $X$ of $TM$, extend it to a section of $T \Bbb R^n$ and via the usual linear connection on $\Bbb R^n$ (directional derivatives), we get a section of $\Gamma (T\Bbb R^n\otimes T^*\Bbb R^n)$. Orthogonally project it onto the subbundle $TM\otimes T^*M$ to get a section of $\Gamma(TM\otimes T^*M)$

Clarinetist

16:07

The definition reminds me of things I've seen in measure theory

iwriteonbananas

Then we want to show that the so defined map $\Gamma (TM)\to \Gamma (TM\otimes T^*M)$ coincides with the LC connection

Huy

@Clarinetist: it should remind you of something else :P

Clarinetist

@Huy It's almost like a $\sigma$-algebra

Balarka Sen

More useful than learning what a topology is learning what a metric is if you're beginning to learn.

Clarinetist

almost

@BalarkaSen Oh, I know what a metric is

iwriteonbananas

16:09

By the uniqueness statement in the fundamental lemma of Riemannian geometry, it suffices to show that the so-defined connection is compatible with the induced metric $g$, and that it is symmetric.

Huy

@BalarkaSen: he insisted on learning topology and analysis simultaneously rather than first analysis then topology :P

Balarka Sen

@Clarinetist Then try proving why a metric space is a topological space.

@Huy: I'd better not make any comments on that because I learnt what analysis is after I learnt what a topology is...

Oops, I just did.

Mary Star

@Huy do you maybe have an idea about my question above? ( chat.stackexchange.com/transcript/message/26882410#26882410 )

Huy

@BalarkaSen: I can tell, you can't even solve SuperstarMonica's integrals

Clarinetist

golfclap

Balarka Sen

16:11

@Huy -_-

Huy

^_^

Balarka Sen

I like this smiley.

Huy

of course, it's like my face and everybody likes my face

@MaryStar depends on the definition of perpendicular, but probably yes. note that you're then in $\mathbb{C}^2 \cong \mathbb{R}^4$

Clarinetist

@BalarkaSen Hmm, I know what a metric space is but I'm not seeing the connection here to a topological space. What "collection $\mathcal{T}$ of subsets" would I be working with?

Huy

@BalarkaSen: don't answer it

Balarka Sen

16:13

That's for you to figure out :)

Clarinetist

K

Huy

:D

Clarinetist

Lol

Balarka Sen

@Huy I didn't plan to :)

Huy

@Clarinetist: that's also why I said "it should remind you of something else" just before

Clarinetist

16:14

Well, this is going to bug me all day. :P

I'll have to think about it

Balarka Sen

bugged by interesting problems is good

Mary Star

So, in the complex case it is the same as in the real case?
We have that $a=a_1i+a_2$ and $b=b_1i+b_2$, then
$(x,y)\cdot (a,b)=0 \Rightarrow (x,y)\cdot (a_1i+a_2,b_1i+b_2)=0 \Rightarrow a_1xi+a_2x+b_1yi+b_2 =0$ right? @Huy

Balarka Sen

I wonder if I normalize $\vec{X}(x, y) = (-y, x)$ it'd remain irrotational on $\Bbb R^2-\{0\}$.

Huy

@BalarkaSen: I know that $\pi_1( X \times Y) \cong \pi_1(X) \times \pi_1(Y)$. can I somehow decompose $\pi_1(G/H)$ too? (I feel like I've asked this question before)

iwriteonbananas

What're $G$ and $H$?

Balarka Sen

16:19

@Huy Well, if $X$ is a topological space, $G$ acts on $X$ properly discontinuously and freely, then there is a SES $1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 1$.

So depending on $G, H$, yes.

Huy

Lie groups

Balarka Sen

Ok, does H act properly discontinuously freely on G?

Huy

urm

I always need to look up this action terminology

Balarka Sen

actually, if not, then I think you get a fibration $H \to G \to G/H$. I don't know much about Lie groups to confirm. If there is a fibration, then you can still do it.

iwriteonbananas

@BalarkaSen Is that true?

Huy

16:21

it's transitive for sure

it's not free

Balarka Sen

@iwriteonbananas what makes you think it isn't? we have talked about this before.

it's the Galois short exact sequence.

iwriteonbananas

Right, nevermind me

Alenanno

@iwriteonbananas Does it work? I mean, writing on bananas.

Balarka Sen

Well, ok, $X$ needs to be locally path connected and semilocally simply connected.

iwriteonbananas

@Alenanno Try for yourself. You'll be amazed.

Alenanno

16:22

@iwriteonbananas adds to to-do list

iwriteonbananas

Write the to-do list on a banana

Alenanno

@iwriteonbananas That sounds too recursive.

Balarka Sen

@Huy If my guess that $H \to G \to G/H$ is a fibration is true, then there is still a SES $1 \to \pi_1(H) \to \pi_1(G) \to \pi_1(G/H) \to 1$, I think. Quotient of Lie groups is Lie, isn't it? And second homotopy groups of Lie groups are 0, iirc.

But ask Mike. He'd know all this.

Clarinetist

It looks like I'm going to have some good reading this weekend

Balarka Sen

Ok, I gotta go.

Huy

16:25

@BalarkaSen: I think if you quotient by an open group, you can end up losing Hausdorffness

Mary Star

In the complex case is $(x,y)\in \mathbb{C}$ ? @Huy

Huy

@MaryStar: I don't know what is meant by that exercise

Mary Star

@Huy

Huy

@MaryStar: it says $a \in \mathbb{C}, b \in \mathbb{R}$. you wrote $a, b \in \mathbb{C}$

Krijn

@BalarkaSen Hey, I had some questions on (co-)homology, are you here?

Clarinetist

16:32

I'm pretty sure $(a, b) = a + bi$

@Krijn I think you just missed him

Krijn

Ah well, such is life.

Huy

@Clarinetist don't skip the exercises

but I'm sure you know that already

Clarinetist

Yep, definitely not going to

iwriteonbananas

@Krijn Out of curiosity, what're you currently learning about?

Mary Star

@Huy So, is it $(a_1+a_2i)x+by=0 $ ?

Or is $x$ or/and $y$ also complex? @Huy

Mike Miller

16:50

@iwriteonbananas: Yes, the thing you said was correct.

Superstar Monica

@user276387 welcome

Let me see what news we have for my bounty

iwriteonbananas

@MikeMiller Ok, compatibility with $g$ is clear enough then, let me see if I can prove symmetry.

Superstar Monica

No news.

1

A: Convergence of $\sin{\pi\sqrt{n}}$

To answer the point $iii)$, note that $$\lim_{n\to \infty} \sin ^2\left(\pi \sqrt{n^2-n}\right)=1,$$ since we have that $$\sqrt{1-\frac{1}{n}}\approx 1-\frac{1}{2n},$$ when $n$ is large. Q.E.D. (which you combine with $\sin{\pi\sqrt{n^2}}=0$ to show divergence)

It's a mystery though that people stay away from taking a bounty. I mean at least a marvellous solution can be written there.

Maybe 100 points is not that attractive? Maybe. Maybe I should have granted 500 points.

Mike Miller

@iwriteonbananas: This fact is not particularly useful (at least in Riemannian geometry) but I do think it's charming.

iwriteonbananas

@MikeMiller It's useful to me in that it helps me understand the LC connection better :P

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$Mathematics$ Mathematics