Abstract algebra (more geometric than most), linear algebra, and integrated linear algebra/multivariable calculus/multivariable analysis, @FernandoMartin. All undergrad.

Nice!

Every time I teach grad diff geo I can't find a book I like to follow, but I've never tried to write at that level. Too much better competition.

Thank you for the compliment, @Twink.

@TedShifrin Yes, why?

ROFL ... Because I keep doing it to you and you lose track of which problem you're thinking about :P @Peter

I think it's part of your charm, though :)

OK, @Peter, I'll expect complete reports tomorrow :) At least I'm done grading. Now I only have to spend a day preparing lecture notes ...

@TedShifrin I agree re DG books... I jumped around between doCarmo, O'Neill, Petersen... they all have pros and cons

@TedShifrin Aw, thanks. But how I am losing track?

can I write $\infty \neq 0$ or does it look awkward?

@Twink To what end?

Petersen has a bunch of sophisticated stuff, but it's very hard if you don't think just like he does (and I don't). And I prefer to teach with primarily a differential forms approach (being a descendant of Cartan->Chern). Not many books do that. Of course, I've defined connections on vector fields and will eventually define connections on vector bundles in general (won't have time for principal bundles).

@TedShifrin What if I look at angles on the triangle? I am thinking about tension now, since you mentioned soap.

LOL@Peter ... You were almost at the end of the four lines tunnel :P

YOU mentioned soap, not I.

@TedShifrin Anthony did.

Yes, you can give a physics proof too.

@TedShifrin What do you mean?

@PeterTamaroff to conclude my proof that the sequence does not converge uniformly

"Four lines tunnel".

Oops, it was @Anthony. My apologies to you both.

More physics, @Peter. You know, tunneling (whatever that is).

@Twink I don't follow.

I would write a sentence, @Twink, saying it in words, rather than in symbols. But, yes, that's the point.

@TedShifrin OK, allow me to be naïve now.

Damn, I'll be asleep for tennis tomorrow. :D

Take two points in the plane.

@TedShifrin Five minutes.

starts stopwatch

Then $X$ is just the middle point in the line joining them.

Yup.

ok @TedShifrin

@TedShifrin Why not a proof by infinite iteration? Take the middle sides, mark them, get a new triangle, repeat, repeat, repeat?

Whoa. You've lost me. What's the next step?

I can't see your whiteboard :P

two minutes left :)

@TedShifrin OK, almost done.

Oh, no, stupid.

Nevermind.

is it normal at the undergraduate level to make alot of mistakes on many of these proofs?

Oh, no wait.

FUUUUUUUUUUUU

@user60887 practice makes perfect

You were over time anyhow, @Peter :P

hey everyone i have a doubt

It's very important to get practice writing and get it criticized. Sadly, most faculty can't be bothered grading homework, as it is, frankly, not fun. So at some universities, TAs do it. Some do it well, others don't read/criticize carefully. But get feedback and try to improve, @user60887.

@TedShifrin OK, I still think this is stupid, let me upload the image now.

What I do is I attempt the problems I get wrong a few days later and do them over till I understand them.

Excellent, @user60887. Also study the proofs in the book/class to see how mathematicians write things and think.

its related to the question here: math.stackexchange.com/questions/485913/…

@Peter: That's converging to the centroid, but it's useless for this problem.

@TedShifrin Yes, that was my point.

I like the picture, though. I'm glad you're warping into a geometric person.

OK, g'night, all. May you all have mathematical nightmares :D

Cheers @TedShifrin

bye

and thanks

hello

good night @TedShifrin

Night, @FernandoMartin, @user60887, @PeterTamaroff, @AnthonyCarapetis, and @Twink. Whew.

@PeterTamaroff you removed a comment

night :)

@TedShifrin Byes.

NO, he removed a picture.

@Twink It was a picture.

Night :)

ok :p

anyone?

@Twink: It's been so nice doing math with you. Please stop being argumentative. Although I realize that was done in humor.

of course :D

thanks for your help

Night :)

You're welcome.

@Shahab: sorry, no clue about that.

ay nanita

I still have one function to prove that it converges pointwise but not uniformly

its $f_k(x) = \tan^{-1}(kx)$

@FernandoMartin oh...no problem. actually i am not interested in the solution though, but i want to understand a particular coloring.

converges to $f(x)=\pi/2$ if $0<x \leq 1$

and to $0$ if $x = 0$

I'm gonna post it

@Twink: well, why doesn't it converge uniformly?

I have one theorem

to prove it

What does it say?

$f_k$ converges uniformly to $f$ iff $\lim_{k \to \infty} \sup_{x \in I}|f_k(x)-f(x)|=0 $

That's the definition.

no

@Twink Is the limti continuous? Are the functions continuous?

the definition is with $\epsilon$

The uniform limit of continuous functions is continuous.

@Twink: those definitions are obviously equivalent! go along what @PeterTamaroff suggests.

Pointwise convergence is easy.

the limit is no continous

@Twink There you go.

and the functions are continuous

but the convergence is not uniform

I have to prove it

I can't use that theorem

@Twink: "therefore", not "but"

Why not?

that the uniform limit of continuos functions is continuos

because it's the next theorem

Well, prove that first and then use it.

@Twink Well, can you guess what $\lim\lVert f-f_k\rVert$ is? After guessing it, prove it.

no I can't

use it

$|\tan^{1}(kx)-\pi/2|$?

first I have to find the sup

over all the $x$'s in $[0,1]$

do I do it with the making $f'_k(x)=0$?

@Twink: each $f_k$ is continuous and vanishes at $0$

Use some intermediate value theorem near $0$ and you're done.

no entiendo

:'(

maybe I won't use the theorem

and do it with the definition

Each $f_k$ is continuous and vanishes at $0$

finding the $\epsilon$

@Twink What for?=

I know but how will I find the sup?

Hence there's some neighborhood of $0$ where $f_k(x) < \varepsilon$

for all $x$ in that neighborhood

so $\vert f-f_k\vert \geq \frac{\pi}2-\varepsilon$ at least, and this holds for each $k$.

vanishes at $0$ means $f_k(0)=0$?

Yes.

But $|f(x)-f_k(x)|=|\pi/2-f_k(x)| \leq \pi/2 + |f_k(x)|<\pi/2 + \epsilon$

:S

That does not contradict what I said

oh you're right

@Twink Make a drawing, think, then make more questions.

:-(

You won't learn if you keep asking.

@Twink: Notice that that holds only on some neighborhood of $0$.

(is it 'neighborhood' or 'neighbourhood'?)

@FernandoMartin Depends: are you american or a brit?

=D

Oh, I'm definitely a brit so that should be 'neighbourhood'.

@FernandoMartin Colour, favour, flavour, &c.

@Twink: see how there's always a neighbourhood of $0$ where $f_k$ is really far from $f$?

yes

I understand that but I have to prove it analytically

The analytical proof is exactly what we outlined before!

why $\vert f-f_k\vert \geq \frac{\pi}2-\varepsilon$ ?

I'm gonna post a question

Read what I said, it follows immediately.

mejor hay que hablar en español aprovechando que no estan los gringos

haha

uy no ya llego el anon

math.stackexchange.com/questions/500919/…

that answer is more like a question

@Twink Yes, those are the good answers.

I think I'm gonna take another example

@Twink: stop for a second and read.

to show that there is a function like that

with those propierties

I didn't want to take $f_k(x)=x^k$ because it was very simple

hey anyone knows what c(n) stands for in the answer to this post: mathoverflow.net/questions/142777/…

the colour of $n$

yes, but how is it defined?

I don't know I'm only a not very smart student

okay...thanks for thinking about it

maybe @PeterTamaroff knows

One just has to read carefully: " c(n) is the colour of n in the worst-case colouring for van der Waerden's theorem"

@PeterTamaroff: you were reading Ted's book on diff geo right?

@PeterTamaroff: by worst case coloring for vdw theorem what is meant? I do not understand.

@Shahab Me neither.

@FernandoMartin Not diff geo, multiv. calc.

Ahh, I see.

che boludo

@Twink sos argentino?

no, de españa

deberia haber una sala de chat en español

ya te fuiste @PeterTamaroff?

@robjohn apparently in the complex plane, $1=-1$ :D

@DanielRust Heh... I just commented on that one :-)

@DanielRust I see you voted to close it, too :-)

@robjohn Yeah it wasn't the best posed question and there's been similar ones asked before, and better written.

@DanielRust I agree.

@DanielRust I can't vote on such things since my votes are binding and it is better to have a consensus of users.

unless my vote is the last one, so I have to time it right

@robjohn Yeah I guess being a mod has it's downsides

@robjohn But hey, sweet-ass diamond right?

@DanielRust yeah, membership has its benefits :-)

more like a job really

@Kasper: Try $\|a+b\|_\infty$

How does one prove that the Erdos-Copeland constant, $0.23571113171923...$ is a normal number?

Oh never new that haha $\|$ compare with $||$

first one looks better indeed :), thanks @robjohn

@Kasper They group better... $\|a\|\|b\|$ is better than $||a||||b||$

There's so many LaTeX codes that people don't realise makes formatting look better.

like \colon instead of :

and \mid instead of |

@Alyosha: have you tried reading the original paper? ams.org/journals/bull/1946-52-10/S0002-9904-1946-08657-7/…

@AnthonyCarapetis I couldn't find it: thanks!

@DanielRust However`\mid` does not work in the \left\middle\right construct

no true, it's used in setbuilder notation like $\{x\in\mathbb{C}\mid \|x\|=1\}$

it makes spacing better instead of $\{x\in\mathbb{C}|||x||=1\}$

@DanielRust You can simulate \mid with \,|\,

wonder if you can simular \| using |\! |

$|\!|$

ah that looks crap :P

$$\left.\frac ab\,\middle|\,a\right.$$

The network engineers are trying to divide by 0.

Greetings

Greetings

@robjohn I managed to compute this in one line math.stackexchange.com/questions/366266/finding-the-limits/… (referring mainly to the 2nd series since the first one is done as a consequence of the first one)

@cyberskull greeting the great one!

@Chris'ssis greetings greatest one :D

@cyberskull :D

@Chris'ssis Here just for you.

@cyberskull :))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))

@cyberskull how are you today?

@Chris'ssis Fine thanks, how are you?

hi

hi

@cyberskull excellent! I'm in a good mood since I managed to brilliantly compute some questions. :D

how is it going ?

@what'sup hi

@what'sup alright, how about you?

@what'sup thanks! Excellent today! How about you?

not so bad :-)

:D

what is that ? @cyberskull

The network engineering chat room @what'sup

i can't see any chat there

It just opened.

now a default chatroom.

you mean the name

??

In the title: The /0

ok

brb . hi @DanielRust

@what'sup hey

how are things ? @DanielRust

@robjohn I'd work some more but unfortunately there is no paper any more in my house. For some problems I need some paper ... (it's bad since the stores here are closed today)

:-(

I might do it in LaTeX but I feel better when writing things down on paper ...

@Chris'ssis just write on codecogs.com/latex/eqneditor.php if you're good in latex it will be easy

@what'sup yeah, thanks, but I prefer some paper ...

@Chris'ssis as you like

@what'sup I'm good what'sup just trying to fully work out a problem i gave a hint to yesterday and then realised I don't think I can actually use the hint :P.

:-)

Oh yay, just got a gold badge

for?

steward

1000 close votes

Did you get reps ?

Nah, badges are just cosmetic

:-(

Well, I guess rep is mostly cosmetic too

I guess it gives you access to a few more tools on the site, and you can award it to other people, but that's not much

i thought that when you get many reputations you'll be a mod

at 10k you get access to mod tools

but that doesn't make you an elected community mod

wow , you're lucky @Chris'ssis

@what'sup why? :-)

at 10k you get access to mod tools

@what'sup someone entered once here and said I don't deserve these points. Maybe that one was right.

@what'sup maybe I know too less math. (then it's good I'm at least lucky)

@Chris'ssis we all know too less math :-)

:-)

(-:

@DanielRust requirements to be a mod ??!!

@Chris'ssis You use paper? how quaint...

@what'sup Yes, you have to be mean and ruthless...

@robjohn :-)))))))) (sometimes)

@robjohn o_O

@robjohn I'm "building up" a new question. I still need to check some things around it and it's done.

@Chris'ssis Just as people get downvotes they don't deserve, many get upvotes they don't deserve. However, it is also the case that many don't get the votes they do deserve.

@robjohn are you a mod ?

i think so

@what'sup that is why my name is in blue and there is a diamond after my name :-)

@robjohn that's true.

Good morning America!

@robjohn Once Bill Gates stated "11 Rules You Will Never Learn In School", and the first one in this video is related to what you said above youtube.com/watch?v=tGzbAqcxPcs

@Twink hi

i actually dunno who made this wonderful site ?

!

hi @what'sup :)

@cyberskull CIDR Notation, dear

how are you ? @Twink

it is the equivalent of ANDing a number with zero

not dividing

fine, waking up very early to work in my math hw, and you?

fine too , waking up early and feeling tired :-(

oh i just knew ! :-)

Darn, I leave to let the dog outside, and post my answer too late.

I hate it when I do that.

we all hate that :-(

Why is that linear algebra?

I've had an answer, get up to take care of something, and come back to finish it and submit and it has an answer. Bah

weird o_O

@robjohn you should solve the hard questions only .

;)

@what'sup You get very few votes for them, generally.

i disagree .

@robjohn isn't 77.9k enough ? :-) we are just helping people as @DanielRust said rep is mostly cosmetic too .

and votes

@what'sup Then why do we need reputation at all?

@what'sup For sure, I answered many questions on sci.math, but I was also upset at being scooped there, too.

@what'sup There was no rep on sci.math

oh ! ok :-)

@what'sup who says it is anything to do with rep? Did I say anything about rep?

@robjohn here is an amazingly beautiful question Let $a_0>0$, $a_{n+1}=\sqrt[\large2013]{2013 a_n+1}-1, \forall n\in \mathbb{N}$. Compute $\lim_{n\to\infty} \displaystyle \frac{1}{\log(n)}\sum_{k=0}^{n} a_n$.

so why do you need votes ?

@what'sup I wrote up an answer and someone beat me to it. Thus , my effort went to waste. Besides, if rep is worthless, then why have it?

@what'sup and why do I answer Chris'sSis's questions? There is no rep there :-)

:-)

I'd give up my rep points anytime if that could make me do better at math. :D

:[D

@Chris'ssis trading something ethereal for something real? Is there even a question there?

i hate when someone asks me the reason of interchanging the sum and integral i don't know why ? :-)

@robjohn I only wanted to emphasize the fact that for me it's important to do good at math and enjoy the beauty of things, not the points. To be honest, it's not bad to have some points since there are some helpful tools you may use on this site. I'm sure many people don't believe my attraction to math is the way I tell it since it may seem unreal.

:-)

@what'sup you mean finite or infinite sums?

infinite

o_O

@what'sup It is important because sometimes it is valid and others it is not.

i know i make sure it's valid

but i just hate it . o_O

brb

@Chris'ssis 1006 looks to be the limit...

heya

I haven't checked numerically yet

the way ?

hi @N3buchadnezzar

Hmm

$$\| A\|_p = \left( \sum_i \sum_j \left|a_{ij}\right|^p\right)^{1/p}$$ ?

@robjohn no need to check numerically . let @Chris'ssis say is it the right answer or not ?

@what'sup why? I have worked it to a point where I believe that is the answer.

@what'sup why ?:-) You're funny.

@robjohn If my work is correct then the answer is different. The answer has a different looking for a limit. :-)

if it's right he'll put the answer .

@Chris'ssis you mean that you don't know if your answer is correct . o_O

@Chris'ssis then I will check my work

@what'sup not really. Well, if I consider that my work is correct then the answer is different. So far I have no reason to believe my answer is wrong. (I always admit there might be some mistakes but unless I find something wrong I cannot say that my answer is wrong)

ok let me check too

@Chris'ssis The binomial theorem says that $a_{n+1}-a_n=-1006a_n^2+O\left(a_n^3\right)$

@robjohn I see

so who wins ?

@what'sup those who enjoy doing math. :-)

how -----> who

@what'sup I mixed the letters. :-)

ok

@what'sup I did a circular permutation ... :D

:-)

@Chris'ssis That seems to indicate that $a_n\sim\frac{1006}{n}$

wait a second . @Chris'ssis do you mean $$ \sum_{k=0}^n a_k $$

@what'sup there was a small typo, it was $n$ instead of $k$. Thanks.

ok

@robjohn hmmm, let me check that.

you're welcome

if it was n $ \to $ so easy

@robjohn yeah, it was a mistake in my work.Your answer is correct.

@robjohn all gets reduced to computing this limit $$\lim_{x\to0}\frac{1}{ \displaystyle \frac{1}{ \sqrt[\large2013]{2013x+1}-1} - \frac{1}{x}}$$

Hello I have a question: if I have an $m \times n$ matrix, how do I show that $\dim(\Ker(A))$ isat leaf equal to $n-m$ ?

are you allowed to use the rank nullity theorem?

@DanielRust Yes

@DanielRust Yes

@Chris'ssis yes. That looks like what I was using :-)

Well you clearly have $\dim(Im(A))\in\{0,\ldots,m\}$ and so $\dim(Im(A))\leq m$. Add $\dim(\ker(A))$ to both sides and we get $$\dim(Im(A))+\dim(\ker(A))\leq m+\dim(\ker(A)).$$ Hence $n-m\leq\dim(\ker(A))$.

@robjohn Mathematica says that limit is $1/1006$.

@Jean-FrancoisRossignol The last line uses the rank nullity theorem.

@Chris'ssis which is good :-)

@robjohn right. This is the answer of our initial limit. At first I thought it's $1/H_{2012}$.

Off to the park... BBL

@DanielRust Thanks

@robjohn ok. Have a nice time there.

@DanielRust Just one specification How do I conclude that $n-m\leq\dim(\ker(A))$.

Could you just explain the transition to the last line ?

@Jean-FrancoisRossignol the LHS is equal to $n$ (by rank nullity), then just subtract $m$ from both sides.

@DanielRust Ah! okay thank you very much

Hi

Can I have help here

Hi @DanielRust and @Jean-FrancoisRossignol. Having a good day over on the other side of the pond?

Help on what, @Faheem?

@TedShifrin I have post a question here math.stackexchange.com/questions/501214/…

Sorry. I know nothing about Matlab. :(

@TedShifrin actually I am having trouble in implementing cubic relationship

You need someone who knows some numerical analysis. That's not me, sorry :(

@TedShifrin You see the mathematical expression, which is equivalent of matlab code

Hey @TedShifrin. Doing well. Just enjoying my last sunday before term starts. I start teaching next week.

@TedShifrin Can you suggest anyone who might help as how to implement cubic relationship