Mathematics - 2017-05-02 (page 1 of 7)

Ted Shifrin

12:05 AM

@DavidRobertJones: Um, no. You need $a[i]-1\le a[i-1]\le a[i]$.

Mike Miller

i'm too lazy to write down a translation

i think i'll just write down what he does

Ted Shifrin

laziness knows no bounds, @MikeM?

PVAL-inactive

Hi @Mike

Hi @Ted

Mike Miller

hi

Ted Shifrin

heya @PVAL

PVAL-inactive

12:15 AM

Some shit went down today here. If you guys have seen the news.

Mike Miller

Yah. Glad you didn't get stabbed.

Ted Shifrin

Yikes. I have been keeping up with the shooting a mile from USCD.

PVAL-inactive

I'm like 70% sure the suspect was in one of my calc classes a few years back.

Ted Shifrin

holy shit, really?

I need to go google what's going on.

PVAL-inactive

I think so I don't have a good recollection of who he is or in what class, but I instantly recognized the face

and I figure he was in some 100 -some person class I was taing for.

Kasmir Khaan

12:18 AM

Hi chat

What is going on ?

Tell me the news

:D

WDUK

Is there any simple ways to recognise what is a suitable substitution for solving an integral? I frequently pick the wrong choice and make it more difficult for myself.

Kasmir Khaan

@WDUK if you see a function and its derivative thee , up to a constant , then you let f(x) = u

Ted Shifrin

well, @PVAL, if everyone were armed with guns, life would have been much simpler.

@WDUK: Weeks of practice.

Daminark

Hey everyone!

WDUK

some times its not got the exact derivative though

so its hard to spot

Daminark

12:21 AM

@PVAL glad you're safe!

PVAL-inactive

@TedShifrin Pretty certain everyone can be here.

Daminark

Ah

UT has campus carry?

Kasmir Khaan

guys what happned?

._.

I did not find anything on the news page

WDUK

@TedShifrin and shorter

Ted Shifrin

@Kasmir: Here you go.

Zee

12:25 AM

@TedShifrin can I ask your opinion on something semi mathematical?

why bother, zee?

I don't understand

omg :(

@BAYMAX can you help me check if my workings for deriving Gauss-Legendre rule on $[0,1]$ using 3points are right?

https://math.stackexchange.com/questions/2261425/derive-the-gauss-legendre-quadrature-rule-on-0-1-using-3-points

Zee

@TedShifrin I know I can be obnoxious sometime, I still respect your credentials though

Ted Shifrin

12:27 AM

I'm tired of your obnoxiousness. I will not engage with you.

Zee

@TedShifrin I understand, sorry

WDUK

hope i dont fail my exam end of this month =/

Ted Shifrin

just practice lots of problems ... that's all I can say, @WDUK. Of course, there are tell-tale signs to look for.

WDUK

do you know a good website with practice problems ?

Ted Shifrin

Nah.

Most calculus books have a selection of 100 integrals at the end of the chapter.

Zee

12:34 AM

@TedShifrin goodbye ted

Ted Shifrin

bubye, zee

Daminark

Seems like Federer has the same reference system as does Rudin

Not sure what I think about this

Ted Shifrin

too arcane a comment for me to respond

hi @TimThe

Daminark

Using (17), we apply 2.4.4-2.4.7 to get ____

Tim The Enchanter

Hello @TedShifrin

Ted Shifrin

12:45 AM

oh right ... well, most of us do that to some extent, but you're right that Federer is particularly overbearing

I made it a point almost never to number equations so as to avoid a lot of that, Demonark.

But Theorems, Propositions get numbered.

Daminark

Good idea

Ted Shifrin

On rare occasions, I would put a star or dagger and then refer to it.

Daminark

Also Neves said that apparently Federer is "the world leader" in defining constants to be determined later

How do you distinguish between referring to a result as a theorem versus a proposition in your text?

Ted Shifrin

Hierarchy of importance/significance.

Daminark

Ah, makes sense

Ted Shifrin

12:50 AM

Some books are notorious for not letting you know what's important and what's not.

Daminark

That's unfortunate. Any come to mind?

Eric

@Daminark have you done besicovitch

Daminark

Yeah, we proved Besicovitch today, and we stated Lebesgue density

Eric

Proved or "proved"

Ted Shifrin

I hope you settled that Fubini's problem by now.

Daminark

12:54 AM

The former, I was rather satisfied

Not quite, I can't think of how to get around $x=0$ so I tabled the problem

Eric

I remember being satisfied with her proof last year, then when studying realized she actually skipped 98% of the proof.

Anyway I mentioned it because last year she defined 19 as a variable when discussing besicovitch lol

Daminark

Beautiful

Eric

In the vein of "constant to be defined later"

Daminark

And I mean, she didn't go into the particular constants or anything but it was all good as far as I'm concerned

Ted Shifrin

@Demonark: Remember two things. One, you're trying to decide $L^1$ on the square. Two, what does Tonelli say?

Daminark

12:58 AM

Lol last class when doing Vitali she let $2\to\infty$, which I found amusing for sure. Though that actually missed a detail once I checked back

So, Tonelli says non-negative and measurable allows you to use Fubini

Eric

Mattilla is a good source for this part of her class

Ted Shifrin

Doesn't it say a bit more about when you can switch orders of integration?

Eric

It's in Federer too but then you need to become an expert in deciphering the book

Daminark

Yeah someone mentioned that for Vitali. The thing is, once we repeat the algorithm on $A\setminus \bigcup_{i=1}^N B_i$, we didn't establish that the new balls we create are actually disjoint from $B_1,\ldots,B_N$

And I mean there are finitely many so I'm pretty sure there's a non-crappy way around it but still..

Uh, we didn't quite say much more if I recall correctly

Eric

Isn't tonelli the same as fubini but replacing the assumption that $|f|$ is integrable with being nonnegative measurable

Mike Miller

1:02 AM

it's different than torelli though

Daminark

That's what I thought

And lol @Mike just looked up Torelli and yeah...

Ted Shifrin

Torelli is neat algebraic geometry.

Mike Miller

and all of these are distinct from archipelagos

Daminark

The collection of islands?

Or yet another theorem?

Ted Shifrin

LOL

Anyhow, I suggest you think more about Fubini-Tonelli.

Eric

1:08 AM

Archipelagos would be a really cool last name

Daminark

Is there more to it than just replacing $L^1$ with non-negative and measurable? Or is this enough to salvage from?

Mike Miller

you're too impatient

you could answer that yourself in another 10 minutes

Ted Shifrin

There's more to figure out. Or you can prove directly that $f(t)\chi_{[x,1]}/t$ is $L^1$ on the square.

Daminark

Hmm, alright, I'll get back to it then

And true @Eric

Anyway, see you guys around!

Mike Miller

speaking of geometric measure theory though today there was a great talk by larry guth from mit on a proof that there are absolute constants $C(m,n)$ s.t. for any null-homotopic Lipschitz map $f: S^m \to S^n$ with Lipschitz constant $L$, there is a null-homotopy with Lipschitz constant at most $C(m,n)L$ if $n$ is odd and $C(m,n)L^2$ if $n$ is even

Ted Shifrin

1:13 AM

I'm heading out for dinner ... that's weird, @MikeM.

Mike Miller

aka, if you have a small map, you can bound how complicated a null-homotopy can be

Ted Shifrin

weird that the exponent comes in depending on parity.

Mike Miller

the proof involves looking at the rational homotopy groups of the codomain

there's either 1 or 2 nonzero ones depending on the parity

Ted Shifrin

ohhhh ... bizarre.

Mike Miller

imagine trying to extend it across the n-skeleton of the ball. picking such an extension we get a "degree" cocycle (integrate the pullback of the volume form on each simplex)

this has some baseline amount of complication you can't get rid of

Ted Shifrin

1:16 AM

I sorta get it ... sorry, need to head out

Mike Miller

nite

Ted Shifrin

see you

Mike Miller

1:32 AM

TIL

PVAL-inactive

@MikeMiller Strange how Audin was born after he died, yet describes him as a young man.

Wikipedia for you.

Mike Miller

da

Semiclassical

Hi chat.

Tim The Enchanter

1:57 AM

@Semiclassical Hey semi

Forever Mozart

3:16 AM

ag

arctic tern

silver?

Zee

anybody down for a problem in basic measure theory?

Ted Shifrin

3:51 AM

baaaaaad, tern :P

Zee

did you miss me @TedShifrin

Anonymous

5:15 AM

How to find extremum values of say $xy+2yz+3zx$ when $x^2+y^2+z^2=1$ ?

Balarka Sen

@blue Lagrange lagrange lagrange

Anonymous

Any other basic way?

Anonymous

(Without lagrange)

Balarka Sen

hm

arctic tern

what's wrong with lagrange?

Anonymous

5:17 AM

Nothing's wrong, just wondering if there is an algebraic method

Anonymous

Anyhow, I think Lagrange is the only option left =P

Anonymous

I was trying to manipulate using AM-GM

Anonymous

Doesn't work

Balarka Sen

I was thinking Cauchy-Schwarz but nah

Anonymous

Umm forget it :P

Anonymous

5:20 AM

I'll make do with Lagrange then

Anonymous

:)

Anonymous

This was actually a problem from my algebra textbook that's why I asked

Anonymous

(No solutions are given unfortunately)

Daminark

5:32 AM

Hey everyone!

Balarka Sen

Hi @Daminark

Daminark

@blue Algebra textbook? What?

Are they doing Lagrange multipliers for the spectral theorem?

How's it going @Balarka?

Balarka Sen

algebra means high school algebra man

Anonymous

@Daminark We don't have lagrange multipliers in syllabus :P

Anonymous

But it is very useful

Anonymous

5:34 AM

I'll use it to shorten problems

Balarka Sen

infinitely so

Anonymous

:D

Balarka Sen

woke up a few minutes ago @Daminark

Daminark

In which case there still ought be a way to solve this without using Lagrange. Still, this type of extremum problem still strikes me as bizarre even in a high school algebra book :P

Fun, I'll probably be heading to bed in about an hour or so

Got manifolds tomorrow

Semiclassical

Lagrange should turn it into a system of linear equations in this case

Balarka Sen

5:35 AM

the mystical manifolds

Semiclassical

since both the constraint and objective functions are quadratic polynomials.

Balarka Sen

Yeah there should be a way to do it without Lagrange but I don't care much

Semiclassical

heh.

Daminark

I mean same, Lagrange always makes your life better

Anonymous

@Daminark Well, it is not exactly a high school maths book. More of a competitive maths book.

Daminark

5:37 AM

And yeah, I think we'll do some transversality, oriented manifolds, and then I think we're gonna generalize degree mod 2 to more general intersection theory

Ah, that makes more sense

Balarka Sen

There's pain and blood involved in doing oriented intersection theory. But it's worth it.

Daminark

Nice

(Also aah)

Is this "pain and blood" just like, time?

Or not "just" time but, will it take a long time?

Balarka Sen

it's just that you have to keep track of more stuff than when you were doing Z/2 intersection

namely, the "orientation number" of the intersection points

Daminark

Ah

Semiclassical

Plugging the Lagrange multiplier conditions into Mathematica and demanding it simplify yields the max value as the positive root of $2\lambda^3-7\lambda-3=0.$

No idea why that particular cubic.

Daminark

5:45 AM

spooked

Hey @EricS!

Balarka Sen

do you know that transversality is stable/generic yet?

Daminark

We got through most of that proof

Balarka Sen

Ah ok

Daminark

Thing is, last class we had our midterm, so he lectured for the first 20 minutes and then had the midterm for the last hour. There was one step missing, but for the life of me I can't remember what

I was kinda freaking out over that midterm so I was not paying as much attention as I should've to the lecture

Balarka Sen

quite understandably

Daminark

5:50 AM

Most of the people in class were significantly better at calculational multi than I, and most had taken point-set last quarter

So I was kinda worried that the midterm would have questions for which this background rift would've caused a problem

Luckily this was not the case

Balarka Sen

between you and i, i don't really remember a lot of point-set topology

and i'm fine

Daminark

I mean yeah, just that we had a couple homework problems about proving things were homeomorphic, and some of it used this quotienting stuff I didn't know about at the time

I've heard of a quotient space, didn't know anything about it

Balarka Sen

I do think multivariable is somewhat essential but I can't calculate either :D

Daminark

Haha, woo, not alone!

Balarka Sen

Ah I see. Quotient spaces are one of the most important part of point-set topology

Daminark

5:54 AM

But yeah at some point I'm gonna go and figure a lot of this stuff out

Balarka Sen

good idea

Daminark

Multi, I mean with regard to calculations I don't often find that I need to get into anything too fancy/messy, I also think I just didn't absorb the theory as well as I should've

Since we basically just reviewed derivatives on $\mathbb{R}$, then defined them on $\mathbb{R}^n$, then did MVT, all in a day. Then on to hypersurfaces and Lagrange multipliers

Balarka Sen

Yeah your courses are rather abstract and I'm not sure if that's too good for health

Daminark

I haven't thought much about it in a health context

I guess I'm a bit too into abstraction but perhaps I can see why. In the case of multi it was mostly my fault because we did have a long pset that week which I didn't start until Friday night, so I dropped a lot of the multi problems in favor of the continuity stuff

Most of the ones I skipped were Lagrange, though when Schlag did the spectral theorem he reviewed those and actually gave a picture, so they made more sense at that point. Result: I tend to default to Lagrange now

Hey @AlessandroCodenotti!

Alessandro Codenotti

Hi @Dami and @Balarka

Daminark

6:06 AM

How's it going?

Balarka Sen

hey @Alessandro

what do i want to do now

Daminark

Besicovitch covering theorem?

Lol jk

Balarka Sen

gmt is a'right

I need to choose between foliations, Riemannian geometry, school stuff

Daminark

The responsible choice is probably school stuff, the dank choice is probably foliations... So do Riemannian geometry!

Balarka Sen

riemannian geometry it is then

Daminark

6:15 AM

What are you using?

DoCarmo?

Balarka Sen

That and Gallot-Hulin-Lafontaine. Mostly the latter

I am bad at Riemannian geometry man

Daminark

I'm sure it's just a matter of time before you'll get much better! ('-')/

Alessandro Codenotti

@Daminark pretty well, thanks. What about you?

Daminark

Doing aight, thanks!

Yo @Eric and @Steamy!

SteamyRoot

Morning

Daminark

6:29 AM

And @ForeverMozart!

Forever Mozart

research=vanity?

that's what I'm wondering now

nobody appears to care about my results, but I'm proud of them

Daminark

@Balarka how would you compare the two books you're using for Riemannian Geo?

Balarka Sen

not sure what that question means

Daminark

As in, what do you think of DoCarmo vs GHL?

You say you're using more GHL, is that a general preference or is it because there's stuff in there that you're specifically into which is not in DoCarmo?

Balarka Sen

doCarmo is lucid, GHL is abstract but has a lot of stuff

general preference

Forever Mozart

6:34 AM

I like how de carmo leads up to the sphere theorem

the end

a whole book leading to one result that uses everything is quite nice

SteamyRoot

I miss Riemannian geometry sometimes... But then I look at the calculations I had to do for some projects and my thesis, and then I'm glad I chose algebra in the end ^^

Forever Mozart

this is quite beautiful youtube.com/watch?v=AmgkSdhK4K8

Daminark

Oh 3Blue1Brown is good, I've seen a couple of their videos

@Steamy Which subfield of algebra are you into more?

SteamyRoot

Group Theory / Algebraic Topology

Daminark

Noice

SteamyRoot

6:41 AM

(in particular: Nielsen-Reidemeister fixed point theory on infra-nilmanifolds and generalisations thereof)

Daminark

Oh, you've narrowed down a good bit, I imagine you're well into your graduate years

SteamyRoot

European system :P

Have to finish a master before you can even apply for a PhD, so yeah... did my master thesis in Riemannian geometry ("Lagrangian submanifolds of complex space forms"), and now I'm 1.5 years into my PhD ("Reidemeister spectrum of infra-nilmanifolds")

Daminark

Ah, I see

Well frenz, I should go to sleep now, so I'll see you around!

Tim The Enchanter

Cya Dami

Anonymous

7:17 AM

How to prove |I-AB|=|I-BA| if A and B are not invertible? (A,B are square matrices of same order)

Anonymous

@BalarkaSen Any idea?

Balarka Sen

@blue There's a trick for this I am forgetting. If I - AB is invertible (so det(I - AB) \neq 0), ones tries to naively write (I - AB)^-1 "=" 1 + AB + (AB)(AB) + ... and manipulate to prove I - BA is invertible too.

That might be the way to go

Anonymous

@BalarkaSen What! How do you apply (I-AB)^{-1} on matrices?

Balarka Sen

You don't. That's why I put the scare quotes.

AH. B(1 - AB)^-1A "=" BA + (BA)(BA) + ... "=" (I - BA)^-1 + 1

I think (I - BA)^-1 = 1 + B(1 - AB)^-1A

Anonymous

Can't understand :P

Anonymous

7:30 AM

@BalarkaSen How do you get this?

Balarka Sen

Ignore the process of getting it for a moment. Assuming I - AB is invertible, what is (I - BA)(I + B(I - AB)^-1A) = ?

meh I am writing 1 and I at the same time, mentally correct those

Anonymous

Does (I + B(I - AB)^-1A) simplify to something?

Balarka Sen

No, but expand the product.

Anonymous

Some huge 4 terms

Anonymous

Can see anything getting simplified

Balarka Sen

7:39 AM

It simplifies. I - BA + B(1 - AB)^-1A - BAB(I - AB)^-1A = (I - BA) + B[(I - AB)^-1 - AB(I - AB)^-1]A

That term in the third bracket is I

so you end up with I - BA + BA = I

Hence, (I - BA)^-1 = I + B(I - AB)^-1A, as postulated.

Astyx

Hi chat

Balarka Sen

hi

Anonymous

@BalarkaSen Ah, right

Anonymous

But I still don't understand how we can write inverse like that'

Balarka Sen

You can't. It's just a dope cheating trick.

Krijn

7:43 AM

Hey @Bala, and chat

Looking at the stars, should we rename this chat Ted Shifrin Fan Club?

Anonymous

@BalarkaSen :P

Anonymous

wth

Balarka Sen

You can actually write (I - A)^-1 = I + A + A^2 + ... under appropriate convergence conditions on A. I think if the terms of A satisfy $|a_{ij}| < 1$, you're ok

This is similar to how (1 - x)^-1 = 1 + x + x^2 + ... only works if |x| < 1

Maybe my first comment is not true though.

Astyx

Is it not $\lt {1\over n}$ ?

You need the triple norm to be less than $1$

Balarka Sen

I am forgetting what the triple norm is. What's $n$?

Astyx

7:47 AM

Dimension of your vector space

Or matrices, same thing

Balarka Sen

Ah, dimension of the matrix. Yeah, that could be it.

Astyx

Triple norm is $\sup_{||x||=1}||A(x)||$, so of course that depends on the first norms you choose

And Chatjax won't load :(

Balarka Sen

I haven't heard of that terminology, but yeah, that's norm of a matrix to me. Not very familiar with it, myself

huh, it seems to work fine on my side

Astyx

Maybe it's french terminology

No it's definitely not working

Balarka Sen

Hi @Krijn, sorry, I missed that ping

Krijn

8:06 AM

@BalarkaSen No problem tho

Balarka Sen

What's up?

Krijn

I got back some data this friday

Balarka Sen

data?

Krijn

Been going through that for the weekend

Yeah, one of the profs is helping me calculate some stuff about morphisms $C \to \mathbb P^ 1$

Balarka Sen

Ah

Krijn

8:08 AM

I can make some pretty okay heuristic arguments, but no formal proof yet

Data is supporting my heuristic arguments, which is nice

Balarka Sen

Cool!

Krijn

I see that I mistyped $\mathbb P$ for $\mathbb O$ there

Can' t edit it now though

Balarka Sen

Yeah I mentally corrected that

Krijn

By the way, have you ever looked at stuff from Szamuely's book?

On fundamental groups and Galois groups

Balarka Sen

I have peeked at it and I know the story a little but nothing seriously

Anonymous

8:18 AM

I saw on some websites that if we can prove that AB and BA have same eigen values then det|I-AB|=det|I-BA| can be proven directly. Any idea how to do that given we know AB and BA have same eigen values? @BalarkaSen

Alessandro Codenotti

If they have the same eigenvalues they have the same characteristic polynomial and det|I-AB| is the characteristic polynomial evaluated in 1

Anonymous

@AlessandroCodenotti Evaluated in $1$ means?

Balarka Sen

what alessandro said

@blue Char poly is $\text{det}(\lambda I - A)$.

At $\lambda = 1$ this is your thing

Anonymous

Oh, but how do we know that $\lambda=1$ ?

Balarka Sen

What do you mean? $\lambda$ is a variable.

Anonymous

8:23 AM

Let me say what I understood till now. AB and BA have the same sets of $\lambda$. Right?

Anonymous

But we don't know if $\lambda=1$ for either of them

Balarka Sen

You are misunderstanding.

Anonymous

maybe

Balarka Sen

The characteristic polynomial is $p(t) = \text{det}(tI - A)$

Alessandro Codenotti

No, $\lambda$ is a variable, $\text{det}(\lambda I-A)$ is a polynomial in $\lambda$

Balarka Sen

8:24 AM

This can be expanded as a product $t - \lambda_i$ where $\lambda_i$ are the eigenvalues of $A$

So if two matrices have the same eigenvalues, they have the same characteristic polynomials.

If they have the same $p(t)$, they have the same $p(1)$

$p(1) = \text{det}(I - A)$ is precisely your expression, so you're done.

Whether or not $1$ is an eigenvalue is immaterial

Anonymous

@BalarkaSen Oh oh! I see now! Thanks

Anonymous

It is just a polynomial like (x-1)(x-2)(x-3)...(say)

Anonymous

Where we can put x=1 to get the desired result

Anonymous

Got it

Anonymous

1,2,3 etc (say) are the eigen values...which are same for AB and BA

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$Mathematics$ Mathematics