Mathematics - 2016-10-22 (page 2 of 4)

Simple Art

15:00

Is there a simple explanation for why?

DHMO

@SimpleArt you can use Green's Theorem to reduce the integral to around the poles

Simple Art

I ought to learn multi-vari calculus

DHMO

@SimpleArt you might like to start at $\displaystyle \int_{-\infty}^\infty e^{-t^2} \mathrm dt = \sqrt{\pi}$

Simple Art

with a half circle with radius $R\to\infty$?

DHMO

@SimpleArt no, not that approach. I haven't tried that, so tell me if it works

Simple Art

15:05

Oh wait

:/

Yeah, residue theorem doesn't work there, right?

Or wait

DHMO

I thought you are talking about multi-variable calculus

instead of residue theorem

Simple Art

The Laurent series?

Oh, idk tbh

DHMO

yesterday, by DHMO

Let $I = \int_\Bbb R e^{-x^2}\ \mathrm dx$.
Then $I^2 = \int_\Bbb R \int_\Bbb R e^{-x^2-y^2}\ \mathrm dx\ \mathrm dy$.
$I^2 = \int_\Bbb {R^+} \int_0^{2\pi} r e^{-r^2}\ \mathrm d\theta\ \mathrm dr$.
$I^2 = \int_\Bbb {R^+} 2\pi r e^{-r^2}\ \mathrm dr$.
$I^2 = \int_\Bbb {R^+} 2\pi e^{-r^2}\ \mathrm dr^2$.
$I^2 = 2\pi\left( e^{-\infty^2} - e^0 \right)$.
$I^2 = 2\pi$.
$I = \sqrt{2\pi}$.

(contains a typo)

the end result should be $I=\sqrt{\pi}$

Simple Art

I haven't learned real analysis yet

DHMO

$\int_\Bbb R$ just means $\int_{-\infty}^{\infty}$

Simple Art

15:07

wow, that was beautiful

DHMO

I just found the proof online

Simple Art

still

Oh, so if I tried the half circle method, the integral along the entire thing is $0$.

Well, not doing that

DHMO

Let $\displaystyle I = \int_\Bbb R e^{-x^2}\ \mathrm dx$.
Then $\displaystyle I^2 = \int_\Bbb R \int_\Bbb R e^{-x^2-y^2}\ \mathrm dx\ \mathrm dy$.
$\displaystyle I^2 = \int_\Bbb {R^+} \int_0^{2\pi} r e^{-r^2}\ \mathrm d\theta\ \mathrm dr$
$\displaystyle I^2 = \int_\Bbb {R^+} 2\pi r e^{-r^2}\ \mathrm dr$
$\displaystyle I^2 = \int_\Bbb {R^+} -\pi e^{-r^2}\ \mathrm d(-r^2)$
$\displaystyle I^2 = -\pi\left( e^{-\infty^2} - e^0 \right)$
$\displaystyle I^2 = \pi$
$\displaystyle I = \sqrt{\pi}$

corrected version

Simple Art

lol

Say

Could you explain why the following:

$$(1-e^{2\pi iz})\int_0^\infty\frac{v^z}{(1+v)^2}dv=\oint_C\frac{v^z}{(1+v)^2}dv$$

DHMO

Did you type that out by yourself :o

Simple Art

15:13

Obviously, because I missed the second parenthesis...

DHMO

@SimpleArt what is $C$?

Simple Art

There we go

DHMO

@SimpleArt what is $z$?

Simple Art

An independent variable?

DHMO

alright

Simple Art

15:15

I believe we are to take the limit as the radius of $C_2\to\infty$ and the radius of $C_4\to0$

And those two come out to be $0$

DHMO

So the only pole is $v=-1$?

Simple Art

Yeah

The contour integral is easy, but I can't grasp why the $1-e^{2\pi iz}$ is there

DHMO

@SimpleArt so you evaluated the right hand side? what is it?

Simple Art

I believe its because of where $C_1$ and $C_3$ are located, and that they are going in opposite directions

The RHS is...

Well, if you divide the $1-e^{2\pi iz}$ over, the RHS becomes $\frac{\pi z}{\sin(\pi z)}$

...Or maybe not with the $z$ in the numerator

I'm going off memory

And if you are wondering, this is the Gamma reflection formula

DHMO

Isn't the contour integral just the integral around the pole?

Simple Art

15:19

I guess yeah. I can easily take it with residue theorem, but I'm confused on how it relates to the LHS

Sorry if I'm bad with my terminology and notation, I sorta learned this stuff yesterday

DHMO

Would this help? $$\int_0^\infty\frac{v^z}{(1+v)^2}dv = \oint_C\frac{v^z}{(1+v)^2}dv + e^{2\pi iz}\int_0^\infty\frac{v^z}{(1+v)^2}dv$$

Simple Art

Yeah, that reinforces what I was thinking along the lines of. But why the $e^{2\pi iz}?$ What's the $z$ in the exponent for?

Oh

I can move into the integral?

Well, thanks @DHMO :) Cheers

DHMO

@SimpleArt what's the next step?

Simple Art

? What do you mean?

DHMO

have you solved it?

Simple Art

15:24

oh yeah, after that, everything is nice and easy

DHMO

could you teach me?

Simple Art

XD

I can't, gtg

DHMO

alright

Simple Art

Um, residue of the contour integral and complex definition of $\sin$

Balarka Sen

@SimpleArt You mean integration around those paths gives the same result? Yes.

That's because integrating is homotopy-independent.

DHMO

15:27

@SimpleArt You know, $e^{2\pi iz}$ rings a bell of a transformation...

Balarka Sen

(essentially an application of Green's theorem)

DHMO

I don't even know how to find the residue of $\dfrac{v^z}{(1+v)^2}$...

$\displaystyle I = \oint_C\frac{v^z}{(1+v)^2}\ \mathrm dv$
Let $v=e^{itr}-1$, $\mathrm dv = ire^{itr} \mathrm dt$
$\displaystyle I = \int_0^{2\pi} \frac{ir(e^{itr}-1)^z}{e^{itr}}\ \mathrm dt$

Am I supposed to let $r\to0$?

Let $u=1+v$.
$\quad\dfrac{v^z}{(1+v)^2}$
$ = \dfrac{(u-1)^z}{u^2}$
$ = \dfrac{e^{i\pi z}(1-u)^z}{u^2}$
$ = \dfrac{e^{i\pi z}}{u^2}\left(1 - zu + O(u^2)\right)$
Therefore, the residue is $-ze^{i\pi z}$ ?

$\dfrac{\mathrm d}{\mathrm dv} v^z = zv^{z-1}$
Therefore, residue = $z(-1)^{z-1}$ = $-ze^{i\pi z}$.

The theorem to prove becomes: $$\int_0^\infty\frac{v^z}{(1+v)^2}dv = -2i\pi ze^{i\pi z} + e^{2\pi iz}\int_0^\infty\frac{v^z}{(1+v)^2}dv$$

Aaron Abraham

16:14

@DHMO, back to the earlier question; if I were to take the resultant of any two of those vectors (equal magnitude), how do I find the angle between the resultant and the other vectors left over? [If this were a 2D situation, I know I could've used tanα = P.sinθ/(P+Pcosθ) : θ is the angle between the two vectors having the same magnitude-P and α is the angle between either vector and the resultant; but I'm pretty sure that doesn't work here :/ ]

s.harp

17:09

@DHMO note that for $z\notin\mathbb Z$ $\nu^z$ is cannot be defined to be analytic on all of $\mathbb C$. Infact the usual choice of the function gives it a line of non-analyticity on the negative real axis. Your "residue" however is at $(-1)$. So you need to make a choice of which branch of your function you chose, ie what the value of $(-1)^\alpha$ should be. This choice will change the value of the residue!

dsillman2000

Morning guys!

s.harp

To be specific $(-1+u)^z=e^{z\ln(-1+u)}$, there are different choices of the complex logarithm $L_n, n\in\mathbb Z$ at $(-1)$, with $L_n-L_m=2\pi i(n-m)$. So
$$(-1+u)^z_n = e^{z L_0(-1+u)}e^{2\pi i n z}$$
the expansion of $L_0$ at $-1$ is $L_0(-1+u)=i\pi - u+O(u^2)$. So
$$e^{z L_0(-1+u)}=e^{z i\pi} e^{-u+u^2f(u)}=e^{z i\pi}(1-u)+O(u^2)$$
then your function is
$$e^{i \pi (1+2n) z}\frac{1-u+O(u^2)}{u^2}=e^{i\pi(1+2n)z}(\frac1{u^2}-\frac1u)+\mathrm{analytic}$$

ah, damn i forgot to multiply $z$ onto the expansion of $L_0$, so the result actually is$$ze^{i\pi(1+2n)z}(\frac1{u^2}-\frac1u)+\mathrm{analytic}$$

it is interesting that doing a numerical calculation gives a different result, because my software has chooses a version of the function that is not analytic on the negative real axis :)

here are the real and imaginary parts of the integrand $e^{-i t}(1-e^{i t})^z$ for $z=1/2$, you can see that it chooses a discontinuous continuation at $t=\pi$

Mike Miller

17:38

Hi @s.harp.

s.harp

Hello @MikeMiller ^

@MikeMiller do you use spectral sequences in your work? I got subscribed into a seminar about it and the prof sold it like as if these were the most amazing tools in all of topology

Mike Miller

I use them at a pretty cursory level. Sometimes things are easier to calculate when you have a spectral sequence. (Well, that's pretty much the whole point, but I don't use the really fancy ones.) Sometimes it's easier to show a map between chain complexes is a quasi-iso by showing that it's an iso on the first page of a spectral sequence or something.

Somehow they're just really basic fundamental tools. About as fundamental as exact sequences. They're just harder.

(Not to imply I really understood them before like six months to a year ago.)

s.harp

ok, I'll try to follow the seminar, but the actual derivation of the specific sequences seems like a huge pile of work

Mike Miller

black box the construction of spectral sequences from exact couples or filtrations.

the general fact is that if you have a filtered chain complex (an increasing sequence of subcomplexes $\mathcal F_p$, with the union over all of them being everything) there is a spectral sequence going from the homology of the associated graded complex $\mathcal F_p/\mathcal F_{p+1}$ to the associated graded of the homology

and that's sometimes enough info to reassemble the homology; the homology of the associated graded is usually pretty easy to compute

once you have that, spectral sequences just come from filtrations

s.harp

17:54

the example in the introduction was that for a fibre bundle $X\to B$ a CW structure on $B$ gives a filtration of $X$, and the prof computed I think $E^1$ and $E^2$ from this, said "you can continue this, but its complicated", explained about the sequence converging to an $E^\infty$ sequence and how that relates to the homology of $X$

Danu

@MikeMiller That fundamental? Damn

Mike Miller

@s.harp Yup, that's right. Calculating past the $E^2$ page tends to be pretty hard. A lot of the time you work so that you can set things up so that it stops doing stuff after $E^2$.

In fact, this paper's main contribution... is computing one differential on the $E^3$ page of a spectral sequence.

@Danu you gotta compute

Semiclassical

hi chat

Andrew Thompson

@Danu One of my biggest regrets so far is that I haven't taken the time to learn spectral sequences properly. If something suggests that you should know how to use them I encourage you to learn that.

Mike Miller

learn by using

Andrew Thompson

18:03

Yup.

Semiclassical

every so often i hear that phrase, but I've no comprehension of what they are

Mike Miller

you take the differential in a chain complex and write it as a sequence of better and better approximations to the real differential

sort of like a taylor series

then the sepctral sequence details how you go from the homology of the first approximation, to the homology of the second approximation, to ...

Semiclassical

mmmkay

why 'spectral'?

Mike Miller

idk

Andrew Thompson

@MikeMiller Is this paper still being peer reviewed or did it already appear somewhere? (Google suggests the former, but not sure.)

@Semiclassical Because mathematics. Everything is spectral.

Semiclassical

18:05

lol

homological recursion would be a fun name for it, though probably bad in its own way

Mike Miller

@Semiclassical probably in the same sense that if you sum over all the eigenspaces with $|\lambda| < N$ of an elliptic operator, that provides a finite-dimensional approximation to the actual operator

Semiclassical

hmm

Mike Miller

@AndrewThompson presumably the former, but these are good grad students at good places, and it's rather unlikely that it would be on the arXiv without their advisors looking over it in detail

Semiclassical

can't say I understand why one would care about it, but then I'd probably need a much more refined sense of what interesting problems in homology are

Mike Miller

@Semiclassical you don't understand why people would care about a tool that makes computations easier?

slash possible

Semiclassical

18:09

I can't if I don't know what kinds of computations we're talking about

I can imagine that there are problems where it's crucial, but I'm ignorant of what those would be

Danu

@AndrewThompson Not for now, luckily :P

Andrew Thompson

(I've heard) they're not that scary once you get used to them, as most things in math.

Studentmath

Concepts in math are usually either terrifying or fundamental, depends on whether you know them well already or not.

s.harp

I would be very interested in something that is still scary once you get used to it

Mike Miller

ghosts

Studentmath

18:14

war

s.harp

in math X(

Mike Miller

better answer

s.harp

unless you mean ghosts in QFT

Mike Miller

bubbling

bubbling terrifies me

s.harp

i guess the path integral is scary to most mathematicians that got used to it :)

Studentmath

18:14

I've got an abstract algebra. I am sure it is stupid but I am really rusty and I don't trust my judgement.

s.harp

whats bubbling?

Studentmath

I recall that for every isomorphism, $f(q)=q$ for every $q\in Q$;

So if I have a field extension $Q(a)$, $Q(b)$ $a,b\notin Q$ and they are isomorphic, then the isomorphism $f$ between them necesserily holds $f(x)=x$ if $x\in Q$, and $f(a)=b$. right?

I don't trust my judgement that I can jump to the conclusion that $f(a)=b$

Andrew Thompson

Yes, you have made no argument as to why that should be true.

s.harp

im not entirely sure how field extensions are defined, but consider extending $\mathbb Q$ by $i$ and by $3i$, these two fields are both $Q\oplus i Q$ but the isomorphism wont map the two generatros onto each other

Studentmath

@s.harp I don't exactly recall what the latter sign means, but field extension $Q(a)$ is defined as the smallest field (i.e. the intersection of all the fields) containing $Q$ and $a$.
@AndrewThompson if I try to follow that up: assuming there is an isomorphism $f$, and knowing that for every $q\in Q, f(q)=q$, because isomorphism is one-to-one and unto $f(a)$ - ohhh I see. It's not necessarily true.

But it is true to say that $f(a)=q*b$ for some $q\in Q$, due to the one-to-one and unto of the isomorphism.

Mike Miller

18:31

No. Take a= root 2, b = 1+root 2

Studentmath

Okay, they are isomorphic?

Ah

Studentmath

18:45

So, it's not true that if $Q(a)$ and $Q(b)$ are isomorphic and $a,b \notin Q$ then $a,b$ are the roots of the same irreducible polynimial. This example is a rather obvious contradiction

Balarka Sen

4 stars on Ted's message. should I interpret that to mean they are happy for me for not getting smacked by Ted in a while, or that they want me to get smacked by Ted soon?

Mike Miller

19:03

Whatever makes you happy.

s.harp

Cobra Verde or Aguirre, Wrath of God?

Studentmath

Thanks @mike @s.harp @AndrewThompson.. I contradicted $Q(\sqrt{i})$ isomorphic to $Q(\sqrt{2})$ and also contradicted $Q(\sqrt{2})$ isomorphic to $Q(\sqrt{3})$, so I had something in me telling me that statement oughta be true. Guess it was wrong after all :)

Danu

@s.harp The latter (haven't actually seen either, though)

s.harp

19:27

I already saw Aguirre, but it was a long time ago and I'm sort of itching to see it again, so I'll go with that

Mike Miller

@Danu You figure things out?

Mary Star

Hello!!

Let $E/F$ be an algebraic extension and $C$ the algebraic closure of $E$. I want to show that the field $C$ is the algebraic closuree also for $F$.

We have that $C=\{c\in E\mid c \text{ algebraic over } E\}$, i.e., every polynomial $f(x)\in E[x]$ splits completely in $C$.

Since $F\leq E$ we have that so every polynomial of $F[x]$ is also a polynomial of $E[x]$. Therefore, these polynomials have also their roots in $C$, and so $C$ is also the algebraic closure for $F$.

Is this correct?

syzygy

it's trivial what you are asking

$C/F$ is algebraic, let $f\in F[X]$ be a nonzero polynomial

it can be viewed as being in $E[X]$, so it has a root in $C$ by definition

Harsha G.

19:48

I'm new to this website, and curious to know the background or the kind of people that are part of this awesome community. I am a high schooler, what about you guys or girls?

Danu

20:05

@MikeMiller Regarding what? :P

@HarshaG. Hi! I'm a physics-student-turned-mathematics-student doing my master's degree :)

Mike Miller

whatever

Danu

@MikeMiller I'm doing okay-ish right now.

The bits about connections are similar to some stuff I already know.

arctic tern

@HarshaG. most frequent-posting regulars are probably math majors (undergrads and grads), math teachers, or math enthusiasts. other people come in for help or outside perspective too - math students trying to understand math or do their coursework, laypeople curious about things, or people in other areas (of academia or industry) like computer science, physics, engineering, etc.

Ted Shifrin

@Studentmath!! Glad to see evidence you're alive :)

Hi Tern, @Danu

@Studentmath: What if $\alpha = \sqrt3$ and $\beta=2-\sqrt3$. Does $f(\alpha)=\beta$ work?

Danu

@TedShifrin Hey there

Mike Miller

20:17

@TedShifrin We already did that one.

Semiclassical

hi chat

Ted Shifrin

Oh, I didn't read everything ... Okey dokey.

Hi @semiclassic

Ah, I see. Well, not identical, but yes :P

Danu

Toots and the Maytals - Take Me Home Country Roads

►

Ted Shifrin

Interesting play on John Denver's song/lyrics.

Appalachia->Jamaica

Danu

This version is so great

So different from the original

Ted Shifrin

20:21

folk -> reggae?

Danu

Early reggae is really really great

Ted Shifrin

I see you're distracting yourself again :)

Danu

No, I'm typing! :D

Connections on $\operatorname{Hom}(E_1,E_2)$ and all that linear algebra stuff

Studentmath

@TedShifrin haha, glad to see glad of that, also glad to see you're alive considering how pissed you probably are from the elections

Danu

I should think about that as arising from some kind of product rule right?

$\nabla(f(s_1))=\nabla(f)s_1+f\nabla s_1$

Is there a formal name for this kind of idea (requiring "product rule")?

Ted Shifrin

20:23

You mean thinking of it as $E_1^*\otimes E_2$? Yes.

Danu

@TedShifrin Well, what you can do is define the thing on Hom before you define it on the dual

Then you get the dual from there

Ted Shifrin

I'd rather think about dual first and then do the general case from what I said. To each his own.

Danu

But yeah, I guess I should just think of this as tensor products giving product rules

Mike Miller

They're all effectively equivalent.

Danu

:)

Ted Shifrin

20:24

@Studentmath: Nothing about these days overjoys me. Let's put it that way.

I may actually be visiting Israel. A friend of mine and I are thinking of doing a tour — probably March.

Danu

It'll soon be over, @Ted :P And I guess the election is kind of run by now.

Ted Shifrin

The toxic damage will last a long time, @Danu.

I expect nontrivial violence.

Danu

Perhaps most supporters will be kinda like "dafuq" based on the last few weeks' meltdown

Or so one might hope

anyways, let's not get bogged down in this topic again :P Sorry for getting into it.

Bob Marley - Trenchtown Rock

►

Ted Shifrin

Yup, back to your getting tensor and tensor.

Tobias Kildetoft

@TedShifrin Hi

Ted Shifrin

20:28

Heya @Tobias

Danu

@TedShifrin I learned about the Chern connection today

Ted Shifrin

And hi @Fargle. Haven't seen you in ages.

Mike Miller

@Danu One could have hoped thism onths ago, too.

Ted Shifrin

Oddly enough, @Danu, I only heard it called that on MSE a few years ago. In all my life, I had never heard the term.

Danu

@TedShifrin Hehe.

What'd you call it?

Ted Shifrin

20:29

Canonical connection ...

Danu

I guess you learned about it from the man himself so he probably didn't call it Chern connection ;D

Ted Shifrin

Chern certainly never named it after himself. It took him almost 40 years before he would say Chern classes.

Danu

...or maybe you already knew

@TedShifrin Hahaha :D That must be so freaking weird

Ted Shifrin

I wonder who actually named it. Griffiths/Harris don't use the term; nor does Wells.

Danu

If you're a big shot like him... so much stuff is named after you

Ted Shifrin

20:30

The biggest are probably Chern-Gauss-Bonnet and Chern classes.

Adeek

hi @TedShifrin

Ted Shifrin

Probably the most modest mathematician I ever met.

Hi Karim.

Is @Balarka hiding to eschew smacks? Cuz he hasn't found me a tube around a curve yet?

Ah, Google is wonderful.

Mike Miller

I've no idea what a Chern connection is. Am I supposed to>

Danu

Unique connection on holomorphic vector bundles that is compatible with both holomorphic and Hermitian structure

Ted Shifrin

Oh, no, Kikkawa named a canonical connection in the setting of web geometry the Chern connection. Still hunting.

Mike Miller

20:32

oh

Danu

It's such a shame that this stuff wasn't around before.

Would've been so cool to see Chern there.

The interview with Kirby is very interesting. In fact I think most of the interviews are.

Ted Shifrin

Well, Simons was one of Chern's early students, of course.

There's a video made at MSRI with interviews about Chern, too. Quite excellent.

I bought it a few years ago when it appeared.

Googling for the Chern connection turned up this unanswered question on MSE.

Danu

@TedShifrin !! nice. Oh, it's not online for free?

Ted Shifrin

I don't know, @Danu. Is it now online for free? They were charging for it 3 years ago.

Danu

@TedShifrin Do you have a title or something else to go by?

Ted Shifrin

20:39

"Taking the Long View: The Life of Shiing-shen Chern"

Danu

Taking the Long View: The Life of Shiing-shen Chern 山长水远：陈省身的一生

►

:D

Ted Shifrin

Is that a legal version?

Danu

I don't know. It's on youtube.

Ted Shifrin

Yup. MSRI has posted it free now. And they say if downloading fails, you can buy a DVD for $20. I don't begrudge the $20 I spent. :)

Danu

I have to say that I'm mostly happy that it's online for me to watch it... :)

Studentmath

20:41

@TedShifrin Let me know if you do! I may have a few advices in that case :)

Ted Shifrin

Danu, look at the unanswered question I linked up there ^^^ and tell me what you'd say.

Danu

@TedShifrin It's a good thing to spend money on such valuable content.

Ted Shifrin

@Studentmath: If I can find you ... you're in hiding these days.

Studentmath

@TedShifrin Also, I finished something with excellence so now I might be able to start research M.Sc in Math next semester during weekends. Got the confirmation from one end, the university asks that I prove weekends are enough for me on two courses I like less... so getting the rust off.

Of course I am in hiding, everything is a huge mess around here

Danu

@TedShifrin ...gauge transformations?!

Ted Shifrin

20:43

Yeah, @Studentmath ... I'm glad to see you!

Studentmath

Didn't have a second to breath, 2 years done and 3 to go doesn't make me that happy either :P

Ted Shifrin

What's the problem with that, @Danu? Physics talk for change of frame :)

No, I'm sure not, @Studentmath.

Danu

This site also has some great content (more to sift through though!).

Mike Miller

Math talk, too, @Ted...

Ted Shifrin

I still say change of frame, @MikeM :P

Mike Miller

20:44

The group of changes of frame, uh-huh

Ted Shifrin

Do you have any comment on that question, @MikeM? I have one off the top of my head.

Danu

I like reserving gauge transformation for principal bundles.

@TedShifrin The physics terminology is reserved for gauge theory, though

Murray Gell-Mann - Another visit to The Institute for Advanced study. Shiing-Shen Chern (71/200)

►

Pretty funny

Ted Shifrin

You'd have to take an associated bundle to get the principal bundle — starting with a complex vector bundle with a hermitian structure. Equivalent, of course.

Mike Miller

He's working with principal bundles.

Danu

@TedShifrin You can get a principal one from a vector bundle?

Mike Miller

20:47

@TedShifrin I'm working right now, so nmot paying too close attention.

@Danu Write down some charts and transition functions and use those exact same transition functions for the principal bundle.

More geometrically, literally take the space of frames.

Ted Shifrin

Sure, @Danu, in general the $GL$ frame bundle. But if you have a Riemannian or Hermitian structure, then $O$- or $U$-frame bundle.

Danu

Oh, I know all of those words

But I haven't thought about it all that much :P

Mike Miller

sounds like a great time to start

Danu

Not really, since the semester just started.

Mike Miller

whatever

Danu

20:49

Now now, don't be offended

Tobias Kildetoft

semester just started there @Danu? I am already marking exams here.

Ted Shifrin

Ah, there's the sleep offender.

Danu

@TobiasKildetoft Germany's holiday times are all messed up.

Mike Miller

@Danu Condescension is a good way to get someone to not want to pay attention to you. So I won't.

Balarka Sen

@TedShifrin I haven't done any math the past few days. Very pressed with schoolwork.

Ted Shifrin

20:51

I find the dismissive use of "whatever" one of the worst pieces of vocabulary to arise in the last decade or so. I do not like it

Danu

@MikeMiller Whaaa? What I said was not at all meant to be condescending. I'm just saying that I currently have a lot of stuff to do, so I'm not spending time on this particular thing.

Ted Shifrin

Just don't forget to sleep, @Balarka.

Studentmath

Or to eat @BalarkaSen, can happen too.

Danu

I have no idea how that could've sounded condescending (I actually looked up the definition again because I wasn't sure if I misunderstood what that word meant). "an attitude of patronizing superiority"...

Ted Shifrin

I don't know what was supposed to be condescending, @Danu. Personally, I find "whatever" exactly that. But let's all move on.

Balarka Sen

20:54

@TedShifrin Trying not to. I have to get up like 6 in the morning consecutively for the next few days.

Danu

@TedShifrin Yes, let's move on.

Ted Shifrin

Good night, then, @Balarka.

pingOfDoom

hello all!

Ted Shifrin

hi doomedPing

Mussulini

@TedShifrin I'm told someone who wishes to remain anonymous is quite fond of your bald patch

Danu

20:54

^lmao

pingOfDoom

my ping is fucked man!

Ted Shifrin

LOL, @Mussulini. WTH are you talking about?

pingOfDoom

Can someone tell me why in discrete time, (-1)^(n^2) is not a periodic signal?

Mussulini

because pi is irrational

pingOfDoom

lolwut?

Mussulini

20:55

$(-1)^x=\cis(\pi)^x=\cis(\pi x)$

Ted Shifrin

Huh?

Wait, are these the standard Fourier coefficients in a Fourier expansion? What are we doing?

pingOfDoom

like if you graph it then its the exact same as (-1)^n

Tobias Kildetoft

I think the lecturer did quite a good job with the exam this time. Several exercises that can be solved by anyone who knows the definitions, but which become much easier for someone who has an understanding for how everything interconnects in the subject.

Balarka Sen

@TedShifrin Maybe I'll quit the chat, yeah, but still have to get some work done before sleeping.

I owe apologies to you and Mike for not getting work done as promised

Ted Shifrin

Balarka: No apologies needed. Get rest and do a good job on your work.

Tobias Kildetoft

20:57

Like one where first they are asked for the rank of a matrix, then whether the columns are linearly independent, then for the rank of the transposed matrix.

Semiclassical

if it's exactly the same as (-1)^n it would be periodic, but evidently it's not

Danu

@TobiasKildetoft Hmm,

In my experience (as a student, purely!) it's very hard to make mathematical exams.

Ted Shifrin

@Tobias: Interestingly, a question I answered in 10 milliseconds this morning was due to someone's not remembering that if the columns of a matrix form an orthonormal basis, then so do its rows.

pingOfDoom

i thought that it would be some math thing where x[n+N] != x[n] but if you do the expansion then it does work

Semiclassical

What?

Ted Shifrin

20:58

I give up on this, since no one will explain what we're talking about.

Mussulini

@pingOfDoom I'm sorry, don't listen to me I made a mistake.

Tobias Kildetoft

@TedShifrin well, that is also less obvious than it sounds once you are used to it (and of course requires the orthonormality to be in the standard basis)

Ted Shifrin

@Alessandro !!

Alessandro

Hi @Ted!

pingOfDoom

like for a signal to be periodic, then you need some N such that x[n] = x[n+N]

Studentmath

20:59

If anyone has some probability/graph theory question I can try to get some rust off there too, feel free to ping me.

Ted Shifrin

@Tobias: Not necessarily, but some orthonormal ground basis, yes.

Transcript for

$Mathematics$ Mathematics