Alors? :) @Lucas

Oui je veux bien, mais je vais bientôt y aller. Ce n'est pas trop long?

@AbdullahUYU ?

Non, c'est pas.

D'accord, allons-y

A2: Deux points distincts appartiennents à au moins une ligne.

A1: il y a exactement 4 points.

A3: Deux points distincts appartiennents à au plus une ligne.

A4: tout ligne contient exactement 2 points.

Hello chat

Prouvez <il existe exactement 6 lignes.>

Tu n'y arrives pas ?

Henlo @Astyx!

Henlo to you too

J'ai commencé par dire que <Deux points distincts appartiennents à exactement une ligne.> à partir de A2 et A3

oui

And from a while back, how's it going @Lucas?

Nice and you?

@AbdullahUYU Tu as fait un dessin ?

Doing well, thanks!

non. pourquoi?

Tu devrais

Toujours toujours toujours toujours faire un dessin

Toujours

Tuju

mais attendez, c'est pas finis.

on prend 2 points distincts. donc, ce deux poins appartiennent à exct. une ligne. et par a4 cette ligne contient seulement ce deux points.

si on prends une troisième, on trouve deux lignes distincts qui est également different de la première.

Donc une ligne est caractérisée par la donnée de deux des quatre points

oui, c'est ça.

Donc compter le nombre de lignes revient à compter le nombre de paires de points, sachant qu'il y en a 4 en tout

Tu connais les coefficients binomiaux ?

C'est pas grave sinon.

oui, $\binom{4}{2}$

Cool. Donc à ton avis c'est quoi ?

Voilà, ça fait 6

d'accord, on a le droit de faire ça, n'est-ce pas? @Lucas

Oui bien sûr, pourquoi on aurait pas le droit ?

@Daminark Do you find a nice exercise since yesterday ? :)

Je pensais qu'on devait continuer oralement. Mais j'ai compris maintenant. Merci @Lucas

De rien :)

@AkivaWeinberger hello

please how i can prove that $\chi_{Q}$ is Lebegue integrable but not Riemann integrable ?

It is not piecewise continuous

?

@Lucas vous vous adressez a moi

?

@Vrouvrou What are the lower Riemann sums, what are the upper Riemann sums

lower i think 0 the upper 1 i think

Yup

If you're doing $\int_0^1$, at least

Since they don't converge to the same number, it's not Riemann integrable

ok i understand

it is Lebesgue integrable because Q\cap [a,b] is countable

?

I think so

Thus $\Bbb Q\cap[a,b]$ is measurable

thank you

please do you have an example on function which is integrable in the sens of generalized Riemann but not Riemann integrable ?

@TedShifrin I'm thinking that maybe for my |H|^{d} thing one could perhaps consider the integrand coming from big chern-gauss-bonnet as the motivation for considering the expression (H^{2} - K) comes from the fact that by gauss-bonnet integrating this and minimizing is the same as minimizing willmore

@Lucas: Piecewise continuity is not a requirement for Riemann integrability. The set of discontinuities needs to be a set of measure $0$.

@EricSilva: OK, so you're thinking of Pfaffian of curvature, but for an even-dimensional hypersurface that's no different from the Gaussian curvature (times $dV$), modulo constants.

hmm right

the thing is whatever you consider you would need that minimizing $\int |H|^{d} - whatever$ has to be the same as minimizing $\int |H|^{d}$ (lets suppose that $d = 2n$ just to be safe)

so you want $\int whatever$ to be something intrinsic hopefully

Right, so, assuming you're fixing topological type, Gaussian curvature is probably the only thing.

according to wikipedia in dim 4 you might consider $\int |H|^{4} - |\text{Riem}|^{2} + 4|\text{Ric}|^{2} - R^{2} d\text{vol}$

hi

Hi semi

which is already just awful if you wanna write everything out in terms of the coefficients of $II$

Note that that has $|H|^2$, not $|H|^4$, Eric.

wait whoops

that was my bad

Oh.

So I wonder how those three curvature terms combined relate to Gaussian curvature (or Pfaffian).

the $H^{4}$ actually cancels with part of the scalar curvature term

So $|\text{curvature}|^2$ has the right homogeneity for dim 4. I guess in general, this would be $^n$ for $M^{2n}$.

agreed

i guess using the gauss equation it's not actually too crazy to try to write this all down in terms of the coefficients of II hmm

$R^{2} = H^{4} - 2H^{2}|II|^{2} + |II|^{4}$ is the easy one

I had enough fun writing down explicit Chern form integrals for the Whitney umbrella years ago :)

i think i might try to sit down and try to compute what u get and see if i can imitate the proof in dim 2 in general even dim but not till after finals

Yeah, post finals sounds smart.

wdup homies

You might also ask Neves what's been done.

I got a question for ya'll

I think the homies aren't in at the moment.

what's a better book to use for self-study - Peter Lax's Linear Algebra or Michael Artin's Algebra?

yeah ill ask him after ive had a whack at at least showing the lower bound

@TedShifrin you ma homie, dawg

Well, I'm a retired dawg.

Used to be a UGA faculty dawg for many years. :)

if it's even true, i havent found it anywhere from googling and i suspect it's probably too hard to try to think of the critical points for that dude anyway

I know @TedShifrin :)

you da man, dawg

Hmm, who are you, @D.Hutchinson?

given that classical willmore is wide open after genus 1

I've seen your videos - big thanks to you, dawg :)

Ohhh ...

yeah, dawg

really helpful, dawg :)

So what brings you by today?

two questions: (1) shall I use Lax's linear algebra or Artin's Algebra?

Those aren't comparable.

I know that Artin's book is very modernized abstract algebra, with many linear algebra topics

hmm ...

sort of comparable, no?

anyway ill read up some chern so i have a better understanding of the pfaffian integrand before i come back to that i am off to study

Right, Artin incorporates the linear algebra, using tools from algebra (groups and modules) to make a lot of it more conceptual. I love Artin.

Bye, Eric.

I see

@D.Hutchinson: What is your specific goal? And how advanced are you?

one of Artin's students is our algebra prof. :)

Oh, who?

I learned most of my algebra from Artin (in person).

goal? eh ... was gonna be a PhD but I am doubtful - because of the near-impossible job search afterwards, even in a best-case scenario, I think, dawg :(

really? Oooo

nice!

So who's your prof?

hmm ... if I said it, it would be close to personally-identifying-info, no?

What book are you using in algebra?

I'm scared :(

I can't imagine how, but don't do anything scary.

notes from artin's student's (my prof) collaborator :)

Hmm .... well, that's singularly unhelpful.

hahaha

can I ask you something else @TedShifrin?

If you can handle Artin, it's a great mathematics course.

I see ...

What's the question?

how real is the prejudice towards older PhDs, when they apply for academic jobs?

like, very, very real?

older, meaning, late 30s and older ...

I'm not aware of that, particularly.

I see ...

I know of some classic famous mathematicians who returned to math later in life.

I see ...

It's the caliber of the work and the strength of letters that matter more, I think.

Anyone able to help me with part (d) of this problem: imgur.com/a/Q3c7o

I see ...

If I want to find the radius of convergence do I use the various power series tests like the ratio test, root test , limit comparison test,...?

Limit comparison won't help for radius of convergence. Ratio test and root test are the standard ones. There are some more sophisticated tests that you can find, but those should be all you need.

K thanks. I'll try that when I get home. I'm thinking of what to eat for lunch :p

brb

LOL ... I wish I had a mathy lunch to suggest, @usukidoll.

Shepherd's pie

That's a heavy lunch, Lozansky ... but maybe you meant pi? :P

That'd take forever to finish

My book did an example with the heat equation on a spherical Christmas ham :P

There's a nice gif on the Wikipedia page for "spherical cow"

I did an example when I taught applied math of figuring out the temperature variation in a wine cellar $d$ feet under the earth ... so as to figure the ideal depth.

Hi

Anyone able to help on my problem?

I can't slep

help

Ok. Yeah idk it is on functional analysis so I guess not too many people can help :(

lmao that wasn't actually to you

@DrewBrady: There are only a few people who come by here who think about that stuff. I might have been able to help when I was a grad student, but not now.

I am expressing my intense agony on not being able to sleep

knocks Balarka over the head

Hey guys, I can use your help again with my analysis midterm practice questions :) here are two that I feel are related:

Let $f \in L^1(\mathbb R^n)$,

true or false, $\lim_{r \to +\infty} \displaystyle \int_{\mathbb R^n \setminus B_r} f d \lambda = 0$

I guess I'll listen to Binaural ASMR Pineapple Relaxation now

same hypothesis, true or false:

$\lim_{r \to 0} \displaystyle \int_{B_r} |f| d \lambda = 0$

This is an ASMR-free chat please

I feel like they should both be true, and that I should create a monotone sequence of functions converging to $f$, which is possible by a theorem

I don't think they're related, @GFauxPas.

[is 1 minute in that video] Lord please disclose to me what in Dante am I doing with my life

hmm, okay, so let's work on the first one first

It looks like a Lebesgue point thing

this is already a sequence if we let $r$ run over the naturals

I can’t tell since I’m on mobile but if you put it in mathb.in I might be able to help

what is mathb

It’s a latex thing

@GFauxPas: Did you learn about Lebesgue points? I don't think it's nearly that fancy.

mathb.in/23041

no

i dont think so, it might have been on the day I missed, et me check

OK. So you were thinking about taking balls of radius $n$ and getting a sequence of functions?

no, we did not

yes, then we have a sequence, and since $f \in L^1$ by hypothesis, we can commute the limit inside the integral

it's going to be monotone becausew

Oh never mind

each circle is strictly bigger than the previous one

Be careful, @GFauxPas. What is your explicit sequence of FUNCTIONS?

oh dear, I was making a sequence of integrals. Bad gfauxpas

Your explicit family of functions is going to be f times an indicator on that ball

@Drew! HUSH.

Well, my explicit family of functions is lol

give me 30 seconds to think about that

@Ted Salut !

Sorry! You need to build a monotone pointwise ae cnvgt sequence if you plan to apply MCT

Yes, sure, we can definitely do that

@GabrielRomon: Salut, mon cher!

multiply $f$ by $\mathbf 1_{B_n}$

OK. Do you see how to finish the first one now?

(Hint: what does it mean for f to be L1?)

(When can you subtract something in a limiting sequence without worrying about its affect on convergence?)

(Do the integrals of f times the indicators converge to something? If so, what? If not why not?)

it means it exists as the lim inf of simple functions approximating $f \cdot \mathbb 1_S$ on measurable sets?

It’s easier than that GFauxPas

What is the statement of the monotone convergence theorem?

it means $\int |f| < \infty$

uh, well, lets bring it up

Yeah that is true.....

What can you say then about int of f ?

it's dominated by $\int |f|$?

okay, statement of MCT

Let $(f_k)$ be an increasing monotone sequence of measurable non-negative functions on $\mathbb R^n$

then $\lim_{k \to \infty} \int f_k \, \mathrm d \lambda = \int \left({\lim_{k \to \infty }f_k}\right) \, \mathrm d \lambda$

Ok so what would be a good choice of f_k?

$f \cdot \mathbb 1_{B_k}$

$r=k$?

yes sorry

thanks for your patience guys, you guys are always patient with me here :)

I try to be mean :)

I think Drew's enjoying helping/teaching.

wait, no, the question doesnt say $f$ is nonnegative

but not sure it matters because $f = f^+ - f^-$

Ah, but what do you know about the Lebesgue integral (different from Riemann)?

Also...that may tip you off to use a different limit integral theorem :)

@TedShifrin regarding Lax or Artin, which would you choose to work on for fun?

hmm

there are three main integral limit theorems for non-finite measure spaces: 1 Fatou, 2 MCT, 3 DCT

often you can use any of them but one will be easiest

I forgot Fatou's, let me recall it

I don't remember Lax's notes, @D.Hutchinson. Artin is masterful and shows beautiful mathematics throughout, but parts are quite hard. If you're explicitly trying to learn just linear algebra, then maybe you want more focus. But the right way to understand things like Jordan canonical form and rational canonical form is with modules.

side note, it seems too important to deserve being called a lemma

why don't you take a look at DCT instead GFauxPas

(hint hint)

Rational canonical form? How does that go? (We only did Jordan)

ooph I hated RCF it was very confusing to me when I first learned it

oh, then $f \cdot 1_{B_k} \le f \cdot 1_{\mathbb R^n}$

Rational canonical works when the field isn't algebraically complete, Demonark.

It comes from module decomposition quite analogous to JCF.

GFauxPas ok....and what is f \cdot 1_{\mathbb R^n} ?

@TedShifrin I see

$\int f$

no

obviously not, i was, uh, joking

f \cdot 1_{\mathbb R^n} is f .

oh, right, of course

(since the domain of f is R^n !)

thats obvious now

in hindsight

first of all what do your f1_{B_n} converge to?

in n

I don't know which property Ted was alluding to when he asked what Lebesgue int. has that Riemann int. doesnt

But your inequality only works when $f\ge 0$, @GFauxPas, so maybe you should split $f$ as you suggested.

okay, well, for the time being assume $f \ge 0$,

@GFauxPas: $f$ integrable $\iff |f|$ integrable. Very false for Riemann.

which inequality? I thought we were working on the limit statement

You're not using monotone or dominated?

Oh, you're back to Fatou?

Doesn't Fatou assume nonnegativity?

ugh I get them confused

let me look at Fatou's statmenet

no, we're looking at DCT

Well, dominated needs an inequality :P

Which won't work if $f$ is changing signs.

But Ted. f1_{B_n} -> f pointwise

Let's start by assuming $f \ge 0$ and later on we can write $f = f^+ - f^-$

and f1_{B_n} <= |f|

OK ... Fine.

so the result (I claim) is in some sense immediate once you recognize DCT is the right way to go

Together with my comment about the definition of Lebesgue integrability, yes.

also GFauxPas, I will leave it to you to explain why it is sufficient to consider sequences of the form f1_{B_n}.

(Recall your statement does technically look at limits as r -> +\infty).

Hello

Salut, @Astyx :)

[hint: extract an increasing subsequence, what can you say?]

Comment va ?

Basically @GFauxPas: You want to show that lim_{r \to +\infty} \int f \, d\lambda = 0

Ça va, plus ou moins. J'ai mal au cou :(

Ah bon pourquoi ?

This is equivalent to saying for any sequence r_n \to \infty, \lim_{n \to +\infty} \int_{R \setminus B_{r_n}}} f\, d \lambda = 0.

sure

For any such sequence r_n \to \infty, one may extract a strictly increasing subsequence, r_n(k)

Tout mon corps est en train de dégénerer, @Astyx ...

Now define f_k = f1_{B_n(k)}

Bolzano-Weirstrass

no

Tu vas voir un médecin ?

it is not Bolzano Weierstrass

it is just what it means for a sequence to diverge...

oh i didnt catch that in the latex

(if you could not find such a subsequence then f cannot diverge to +\infty!)

so its dominated from above by 0

Anyways once you find this increasing subsequence you now apply the DCT argument we had to show that \int_{B_r{n(k)}} f d\lambda = \int _{R^n} f d \lambda

now the RHS is finite since f is integrable

by hypothesis

sure

Oui, @Astyx, bien sûr.

oops I mean \lim_{k \to \infty} int_{B_r{n(k)}} f d\lambda = \int _{R^n} f d \lambda

Et alors ?

So now subtract the RHS from the LHS

0 = \int _{R^n} f d \lambda - \lim_{k \to \infty} int_{B_r{n(k)}} f d\lambda

in particular 0 = \lim_{k \to \infty} \int _{R^n} f d \lambda - int_{B_r{n(k)}} f d\lambda

so by linearity

Put the TeX in dollar signs, it'll actually render

@Astyx: Je ne sais pas le mot en français, mais je vais voir un "physical therapist" chaque semaine.

yeah its taking me a bit of time to convert it in my head

to latex

I might have to put @Drew on ignore for that ! :P

$0 = \lim_{k \to \infty} \int _{R^n} f (1 - 1_{B_r{n(k)}}) d\lambda$

Isn't that a tiny bit harsh for something so minor?

Kinésithérapeute ?

Aha @Astyx.

I was joking, Demonark.

but that's in some sense what we knew already

Thus, $0 = \lim_{k \to \infty} \int f (1_{R^n \setminus B_{r_{n(k)}}}) d\lambda$

Oh lmao

En tous cas je te souhaite de te rétablir vite

i forgot how we needed DCT to get this

:(

Merci bien, @Astyx. Tout va bien chez toi?

DCT is love, DCT is life

DCT you need to establish the limit result!

Oui ça va

Je reprends les maths dans un mois

oh, right

Formidable, @Astyx :)

that concludes the proofo

let me see if i can say it back now

J'ai hâte

Je m'imagine ...

Thanks @TedShifrin :)

actually GFauxPas you need not pass to an increasing subsequence

and congrats on your retirement :)

Didn't do much, @D.Hutchinson, but sure.

you can just do it by density on some rational increasing sequence

@Drew: Don't overdo it. Stuff is hard enough ;)

well I think you may need to actually argue this way

because it is not good enough to do an increasing subsequence

We want to find $\lim_{k} \int_{R^n \setminus B_k} = \lim_k \int (f - \mathbf 1_{Bk} f )d\lambda$

If you do $|f|$ you get a monotone function of $r$.

unfortunately convergence of subsequence != convergence of sequence

What do you mean ?

What are we trying to do ?

@Astyx mathb.in/23041

Yeah, you can apply MCT on |f| as well

if that's what you mean, Ted

Then you don't have to worry about $k$ versus $r$. :)

Why not?

Because $\int_{B_r} |f|\,d\lambda$ is a monotone increasing function of $r$.

we have a sequence $f_k 1_{B_k}$ which is dominated by $f$

So if it converges for $r\in\Bbb N$, it converges.

yeah, that's true

yeah, if you don't want to argue by density then I think Ted's approach is easier, GFauxPas!

and also no, unfortunately you said something wrong GFauxPas

:( I may need my hand held again going through this

what was wrong

"we have a sequence $f_k 1_{B_k}$ which is dominated by $f$"

err I didnt define $f_k$, I meant $f$

suppose there is some x in B_k^c such that f(x) < 0....

that's not the issue GFauxPas!

it's dominated in the sense that $|f 1_{B_k}| \le f$ Imeant

@GFauxPas, I think you wanted to assume $f\ge 0$ earlier and just do $f_+$ and $f_-$ separately.

GFauxPas, I recommend you slow down a bit.

takes a deep breath

good boy :)

this is harder than the homework, im trying, ookaay

you have two options DCT and MCT. (Also Fatou, but please let's not think about lim inf right now)