Mathematics - 2013-10-03 (page 1 of 3)

Twink

00:02

thanks @MikaelÖhman I'm gonna check

I get this

! Package babel Error: You haven't loaded the option spanish yet.

See the babel package documentation for explanation.
Type H <return> for immediate help.
...

l.6 \select@language{spanish}

?

Danny

twink are y spanish

Twink

yes

Danny

hola

Twink

hola:D

Danny

como estas

Twink

00:10

bien y tú?

Danny

perfecto

Twink

qué bueno!

Danny

Sí, estás estudiando

Twink

sí

Danny

muy bebo

muy beno

Twink

00:22

:D

Pedro Tamaroff

@FernandoMartin Helloes.

Twink

todos aquí hablan español

user71494

Hi, I forgot how the number that doesn't change another number in two-input operation is called. Could anyone tell me? For example x + 0 = x, x*1=x

Danny

No sé

anon

@Runemoro "neutral element" or "identity element" (or simply "identity")

user71494

00:33

thank you

Twink

sí Danny

Pedro habla español

es argentino

Ted Shifrin

Hi @anon

anon

yo

Twink

@Danny anon doesn't like me

Danny

lol why

Twink

00:36

he says I'm a troll

do you think I am a troll?

Danny

No creo que eres guapo

Tengo que ir

Twink

gracias

adiós!

Danny

¡Hasta luego!

Twink

thanks for saying I'm handsome

but you haven't seen any picture of me

or are you talking about my happy face? :D

Danny

we leave that unsaid

if have to go it is 2:42 AM

iam in sweden

Twink

00:43

why do we leave that unsaid?

Danny

because i havent seen

Twink

so why did you say that I'm handsome?

Danny

but surely you are

Twink

in spanish?

guapo = handsome

Danny

yeah

Twink

00:45

I don't understand...

Fernando Martin

Hey there @PedroTamaroff

Danny

twink

Twink

yes?

Danny

i see you tomorrow or some other day i really got to sleep

study well!

Twink

ok good night

Pedro Tamaroff

00:47

@FernandoMartin How did Topo go?

Twink

sleep well

Danny

you must get good grades, aim only expecting A from you

iam

Twink

ok :D

Fernando Martin

Good. The test is on Monday though!

There was a questions-only class today

Twink

hola Fernando

Fernando Martin

00:48

Hi Twink

Pedro Tamaroff

@FernandoMartin Right.

Fernando Martin

@PedroTamaroff: Did you have any tests yet?

Linear algebra?

Pedro Tamaroff

@FernandoMartin Next Tuesday.

Twink

Pedro can you help me with something please?

Pedro Tamaroff

@FernandoMartin Do you mind if I ask why did the guy "know" me?

@Twink I can always try.

Fernando Martin

00:51

Haha, not really. He's a friend of mine and Guillermo, we mentioned you when he asked about Rosario I think

Twink

it's Lemma 1 on page 6 whitman.edu/mathematics/SeniorProjectArchive/2007/huh.pdf

Pedro Tamaroff

@FernandoMartin Ah.

Twink

the definition of Baire one function is Def. 2

Fernando Martin

@PedroTamaroff: Do you know how to prove that $GL(n,\mathbb{R})$ has exactly two connected components?

Pedro Tamaroff

@FernandoMartin Use the $\det$.

Fernando Martin

00:56

That proves that it has at least two

Twink

@FernandoMartin I asked for Pedros' help first

Pedro Tamaroff

@Twink OK, what is it? =)

Fernando Martin

Haha, that's ok. It's not an important problem anyway.

Twink

did you see it Pedro?

Pedro Tamaroff

@Twink I have it open, I don't know what you want.

Twink

00:57

I'm trying to understand why $f_k$ converges to $f$ pointwise

they just say it with words

this is what I did:

but I don't what to do with the case $|f(x)|=M$

Pedro Tamaroff

@FernandoMartin Right.

@Twink Hold on a sec.

Twink

ok :) thanks

Fernando Martin

@PedroTamaroff: Nevermind, I'll look it up. I just asked to know if you knew the problem

thanks!

Pedro Tamaroff

@FernandoMartin Wait.

Ah.

Fernando Martin

We don't know if it has two components.

There may be more

I'm pretty sure there's exactly two

Pedro Tamaroff

01:01

@FernandoMartin Ah, you have to prove the set of positive and negative dets are connected.

Fernando Martin

Yep

Pedro Tamaroff

Aren't they actually path connected or something?

Fernando Martin

I think they're path connected

Yeah

But I couldn't prove it

here it is

Pedro Tamaroff

@Twink

@FernandoMartin Cawl.

Fernando Martin

The argument is a bit more involved than what I thought, haha

Twink

01:03

yes?

maybe it's not necessary to separate in cases

Pedro Tamaroff

@Twink What is your problem?

Twink

but that's all that came to my mind

in my proof

I haven'ts handled the case $|f(x)|=M$

Pedro Tamaroff

@Twink What part of the pdf are you looking at?

Twink

page 6

Lemma1

and the definition of baire one function is Def. 2

Ted Shifrin

@Fernando, yes, they're path connected. What linear algebra do you know?

Pedro Tamaroff

01:07

@Twink The function they define, if I am not crazy, is $\max(\min(g_n,M),-M)$.

Thus $-M\leqslant f_n\leqslant M$ by definition.

Twink

ye

s

$f_n$ is bounded by $M$

what I want is to prove that $f_n$ converges pointwise to $f$

Pedro Tamaroff

@Twink Well, $\max,\min$ are defined using $|\cdot |$.

Twink

so I took an $x$ in the interval $I=[a,b]$

Pedro Tamaroff

And $|\cdot |$ is continuous.

Twink

and I want to prove that $f_k(x) \to f(x)$

Fernando Martin

01:09

@TedShifrin: I read the argument there. Thanks!

Ted Shifrin

There? OK.

Fernando Martin

It used polar decompositions.

Pedro Tamaroff

Thus $f_n\to\max( \min(f,M),-M)$, @Twink

@TedShifrin What is your argument?

Fernando Martin

@Ted: I posted a link to the answer above. This is it

Ted Shifrin

Oh, it is easier than that, @Fernando. Gram-Schmidt is all you need.

Pedro Tamaroff

01:11

@TedShifrin ORLY?

Fernando Martin

Hmmm, using G-S I think I can prove that any invertible matrix is path connected to an orthogonal one.

Twink

hmm

Pedro Tamaroff

@Twink But you had that $-M\leqslant f\leqslant M$ right?

Twink

I don't understand that : max min

yes

ok

Pedro Tamaroff

@Twink $$\max(f,g)=\begin{cases} f&;f\geqslant g\\ g&;g>f\end{cases}$$

Twink

01:12

thank you :)

Pedro Tamaroff

@Twink ;)

Ted Shifrin

Right, @Fernando. Then it takes some linear algebra to prove $SO(n)$ is connected. If you know the spectral theorem for normal matrices, it's easy.

I retract my "all you need". Sorry.

There's a much more elementary inductive argument, too.

Pedro Tamaroff

@TedShifrin You're kinda in favour of "simple", right?

Ted Shifrin

Not necessarily. But I prefer conceptually elegant to symbol-pushing :)

Understanding an element of $SO(n)$ as a product of an even number of reflections and a bunch of 2-dim rotations is prima ;)

Pedro Tamaroff

@TedShifrin ${\rm SO}(n)$ is $\det A=1$?

Ted Shifrin

01:22

Don't even need reflections, since $-I_2$ is a rotation. Yes, @Pedro.

Plus orthogonal!

$SL(n)$ is $\det=1$ only.

Pedro Tamaroff

@TedShifrin What is $\rm SO$?

$(\det A)^2=1$?

anon

$\det=1$ is special, $A^TA=I$ is orthogonal, SO is special & orthogonal

Pedro Tamaroff

Euclidean motions?

Ted Shifrin

Orientation-preserving motions fixing $0$

Pedro Tamaroff

@TedShifrin Ah. What is it called?

Fernando Martin

01:25

Special ortogonal group

$SO(n) = SL(n)\cap O(n)$

@PedroTamaroff: How do you get your SOs and SLs straight in TeX?

Ted Shifrin

Then ... We complex geometers come along and do unitary, hermitian, etc. to complicate your life.

Pedro Tamaroff

@FernandoMartin I use \rm

Fernando Martin

Thanks!

I'll be back in a while

Pedro Tamaroff

@FernandoMartin So the $\rm O$ are the orthogonal ones, and $\rm SL$ are the ones with $\det = 1$?

Ted Shifrin

Or don't put it in math mode :) I prefer ital.

Pedro Tamaroff

01:27

@TedShifrin NOOOOOOOOOOOOOOOOOOOOOOOOO

Ted Shifrin

Yes @Pedro ... On both. I also don't like roman d$x$ in integrals.

Pedro Tamaroff

@TedShifrin Oh, I don't either.

Fernando Martin

Yes Pedro.

Ted Shifrin

Well, you have some redeeming qualities.

Fernando Martin

I like to think of $\rm{SL}$ as the kernel of the determinant.

Ted Shifrin

01:29

Yup.

Fernando Martin

Well, I'm off now.

Ted Shifrin

Bubye.

Pedro Tamaroff

@TedShifrin Ted.

Daniel Fischer

01:46

Is anybody here knowledgeable in set theory?

Pedro Tamaroff

@DanielFischer Heh, not the higher set theory, sorry.

What's the problem?

(Nonetheless, if a problem eludes you it probably eludes me.)

Daniel Fischer

I wonder whether the existence of non-zero linear functionals on every non-zero vector space is equivalent to the existence of bases.

(And hence the axiom of choice.)

Pedro Tamaroff

@DanielFischer Asaf is your guy here! Heh!

Daniel Fischer

Well, he's not here right now, is he?

Pedro Tamaroff

@DanielFischer He doesn't chat here, nope.

You can "summon" him on main and use IRC.

Daniel Fischer

01:50

Even if, he's probably asleep now. I should be too.

Ted Shifrin

Yes, @Pedro?

Pedro Tamaroff

@DanielFischer You can ask Andres Caicedo.

He's online.

@TedShifrin I am trying to find a new topic that interests me.

Daniel Fischer

Not in this room, it seems. How do I reach him? Comment under one of his answers?

Ted Shifrin

@Daniel: Note the discussion on Wiki under Hahn-Banach re axiom of choice.

Pedro Tamaroff

@DanielFischer Yeah.

@TedShifrin I started reading "Introduction to Real Analysis" by Kolmogorov and Fomin.

Ted Shifrin

01:55

@Pedro: what about projective geometry if you're tired of analysis?

Pedro Tamaroff

@TedShifrin What are the prereqs for that?

Ted Shifrin

Not much. Look at the chapter I sent you. And Pedoe has a classic book on geometry that is quite sophisticated.

Pedro Tamaroff

@TedShifrin OK. Melooks.

Ted Shifrin

amazon.com/Geometry-Comprehensive-Course-Dover-Mathematics/dp/…

Pedro Tamaroff

@TedShifrin Oh, and Dover. Cool. Do you know the guy?

Ted Shifrin

01:59

Of course, then you'll ask me questions I can't do :)

Pedro Tamaroff

@TedShifrin ?

Ted Shifrin

No, he may be dead or else 90. He was a student of the famous Scottish geometer Hodge.

Pedro Tamaroff

@TedShifrin Oh. Hodge.

Ted Shifrin

You've heard of him?

Daniel Fischer

Isn't he a star?

Pedro Tamaroff

02:02

@DanielFischer Hehe.

@TedShifrin Your book has the "Hodge star".

Ted Shifrin

A dead one, yes, very much. Hodge Theory, Hodge-Riemann bilinear relations ...

cyberskull

Does anyone still use Euclid's Elements for courses?

Ted Shifrin

Ah, yes, that's the warm-up for theory of harmonic diff forms.

Occasionally, cyber.

cyberskull

hmm..

Ted Shifrin

Sorry, @Daniel ... Too long a day. I totally missed your pun.

Daniel Fischer

02:05

Well, you haven't missed too much there :/

Ted Shifrin

So @Pedro, hodge and pedoe have a 3-vol treatise on alg geometry.

Pedro Tamaroff

@TedShifrin Pedoe is so close to Pedro. Daydreams.

Ted Shifrin

LOLZ

cyberskull

dreamer

;-)

Pedro Tamaroff

@TedShifrin I'm reading Chap 8 of yours.

Ted Shifrin

02:21

OK. There are some good exercises :) Very different flavor from all this analysis stuff.

Twink

@PedroTamaroff I'm a little confused about the max min function

$$f_k(x)=\max\big\{\min\{g_k(x), M\}, -M\big\} = \left\{\begin{array}{rcl} -M & \mbox{if} &g_k(x)<-M, \\ g_k(x) & \mbox{if} & -M \leq g_k(x) \leq M, \\ M & \mbox{if} &g_k(x)>M. \end{array} \right.$$

Pedro Tamaroff

@Twink ?

Twink

because first I take tha min

between g_k and M

Pedro Tamaroff

@Twink OK.

Twink

but then I take the max

between g_k and -M

not between the min and -M

what if the min is M?

Ted Shifrin

02:26

@Pedro: A high school kid I'm doing Spivak Calculus with had a creative idea. He asked about non-integer derivatives and wanted to graph $(x,s,f^{(s)}(x))$. Intriguing ...

Pedro Tamaroff

@TedShifrin IIRC one can do fractional derivatives, but it is ugly.

@Twink Then you get $\max(M,-M)=M$.

Ted Shifrin

Oh, I told him the idea of how they're defined. Fourier transforms :)

But his graphing idea is cool.

Pedro Tamaroff

@TedShifrin Why?

Ted Shifrin

I don't know how this behaves with $s$, but I should think about it in a few months.

Twink

$$f_k(x)=\max\big\{\min\{g_k(x), M\}, -M\big\} = \left\{\begin{array}{rcl} -M & \mbox{if} &\min\{g_k(x), M\}<-M, \\ g_k(x) & \mbox{if} & -M \geq \min\{g_k(x), M\} \text{ and } g_k(x) \leq M, \\ M & \mbox{if} &g_k(x)>M. \end{array} \right.$$

Ted Shifrin

02:30

It should be continuous in $s$, methinks, based on Fourier/inverse Fourier ...

Twink

wouldn't that be the correct definition?

no wait

that?

Pedro Tamaroff

@Twink I would have to check. Maybe I fucked up =)

Twink

$$f_k(x)=\max\big\{\min\{g_k(x), M\}, -M\big\} = \left\{\begin{array}{rcl} -M & \mbox{if} &\min\{g_k(x), M\}<-M, \\ g_k(x) & \mbox{if} & -M \leq \min\{g_k(x), M\} \text{ and } g_k(x) \leq M, \\ M & \mbox{if} &g_k(x)>M. \end{array} \right.$$

what do you think @TedShifrin ?

is that equality correct?

and this one

11 mins ago, by Twink

$$f_k(x)=\max\big\{\min\{g_k(x), M\}, -M\big\} = \left\{\begin{array}{rcl} -M & \mbox{if} &g_k(x)<-M, \\ g_k(x) & \mbox{if} & -M \leq g_k(x) \leq M, \\ M & \mbox{if} &g_k(x)>M. \end{array} \right.$$

they're the same? :S

Ted Shifrin

Looks like you can simplify it more, since you know $-M<M$.

Pedro Tamaroff

@Twink Just make a drawing =)

Ted Shifrin

02:38

Damn, now I've convinced @Pedro to draw :D

Pedro Tamaroff

@TedShifrin Well, I did like drawing before Ted-era!

Ted Shifrin

Yes, @Twink, your earlier version wins, I think

Good, @Pedro :)

Twink

I already made a drawing

but it doesn't help

I guess I'll try to continue doing it analytically

2 hours ago, by Twink

I have that

but I don't know how to handle the case $|f(x)|=M$

wait, @TedShifrin which version did you say was correct?

this one?

8 mins ago, by Twink

11 mins ago, by Twink

$$f_k(x)=\max\big\{\min\{g_k(x), M\}, -M\big\} = \left\{\begin{array}{rcl} -M & \mbox{if} &g_k(x)<-M, \\ g_k(x) & \mbox{if} & -M \leq g_k(x) \leq M, \\ M & \mbox{if} &g_k(x)>M. \end{array} \right.$$

Pedro Tamaroff

@Twink Just remember we have a pointwise limit Twink!

Twink

I think I almost finish

with my proof

but there's something I don't see

Pedro Tamaroff

02:45

Since $g_k\to f$ and $f\leqslant M$, $g_k\leqslant M$ eventually.

Twink

in the case $|f(x)|=M$

yes

that's what I proved

but if $f=M$

it's not necessirly true that $g_k \leq M$

eventually

left and right

that's my problem when I handle the case $|f(x)|=M$

as you can see in my proof

like a sequence that converges to 0 by both sides

Pedro Tamaroff

@Twink If $g_k(x_0)>M$ then $f_k(x_0)=M$, so no worries.

Twink

yes but it's not for a fixed $k$

it's for the limit

the proof is for the limit

I want to prove that for a fixed $x$, $f_k(x) \to f(x)$

in the case that $|f(x)|=M$

Ted Shifrin

They're both correct, but your earlier one is more informative, @Twink.

Twink

which one is the earlier one?

so they're correct?

both?

if both are correct then I there's no problem

11 mins ago, by Twink

8 mins ago, by Twink

11 mins ago, by Twink

$$f_k(x)=\max\big\{\min\{g_k(x), M\}, -M\big\} = \left\{\begin{array}{rcl} -M & \mbox{if} &g_k(x)<-M, \\ g_k(x) & \mbox{if} & -M \leq g_k(x) \leq M, \\ M & \mbox{if} &g_k(x)>M. \end{array} \right.$$

if this one is correct the problem is over

Ted Shifrin

02:57

Yes, I believe it's correct, but my brain is half-dead :)

2

Twink

I'm trying to understand it seeing it like a composition of two functions max and min

but I can't :S

for me it makes more sense this one

29 mins ago, by Twink

$$f_k(x)=\max\big\{\min\{g_k(x), M\}, -M\big\} = \left\{\begin{array}{rcl} -M & \mbox{if} &\min\{g_k(x), M\}<-M, \\ g_k(x) & \mbox{if} & -M \leq \min\{g_k(x), M\} \text{ and } g_k(x) \leq M, \\ M & \mbox{if} &g_k(x)>M. \end{array} \right.$$

but I don't understand why they're equivalent

I think I already understood :D

even if I don't know how to go from the second one to the first one xD

I'm sorry for my LOOONG proof

that took me some hours :(

Pedro Tamaroff

03:17

@Twink Note that $\min\{g_k,M\}<M$. So you may drop the $M$ in there, that is, you can just state the "middle" part as $-M\leqslant g_k\leqslant M$.

Similarily, if $\min\{g_k,M\}=M$, it cannot be less than $-M$.

Thus only the option $g_k<-M$ remains.

Twink

so, $\min\{g_k,M\}$ cannot be $M$?

Pedro Tamaroff

@Twink If you got to that case, no. Follow the possible outcomes!

Twink

I see

thank you so much Pedro :D

you're awesome

Kevin Driscoll

03:40

@PedroTamaroff Congrats on your recent discovery of the Tamaroff-Loy-Farin Theorem..... top billing is no laughing matter

Twink

:O!!

@KevinDriscoll which theorem is that!?

Kevin Driscoll

check the starred comments ------------------------------->

Twink

LOOOOOOOOL ¬¬

Pedro Tamaroff

@AndresCaicedo Are you there?

Adriano

04:02

Hey, quick question. I'm a bit rusty with sigma notation, and I was wondering if the following equation holds true:
$$
\left[\sum_{i=0}^m a_ix^i \right]\left[\sum_{j=0}^n b_jx^j \right] = \sum_{i=0}^m \left[\sum_{j=0}^n \left( a_ib_jx^ix^j \right)\right]
$$

Pedro Tamaroff

@Adriano Try it for some special cases and see!

Adriano

@PedroTamaroff It seems to work out. =]

Twink

when they're finite sums you can do almost whatever you want

like interchanging the sigmas

associating however you want... etc

Pedro Tamaroff

@Adriano Just use distributivity.

Gustavo Bandeira

@PedroTamaroff Can you take a look at something?

Pedro Tamaroff

04:13

@GustavoBandeira Yeah.

Gustavo Bandeira

@PedroTamaroff This.

Any ideas?

Pedro Tamaroff

@GustavoBandeira What is troubling you?

Gustavo Bandeira

@PedroTamaroff I'm not sure if all subsets of the mapping $\mathbb{N} \mapsto \mathbb{N}$ are symmetric.

Pedro Tamaroff

@GustavoBandeira What is the mapping? And what do you mean by a subset being symmetric?

Gustavo Bandeira

@PedroTamaroff @PedroTamaroff Subsets like $(n,n)$.

Pedro Tamaroff

04:17

@GustavoBandeira Those are elements, not subsets.

Gustavo Bandeira

Aren't they also small sets?

Pedro Tamaroff

Well, $(a,b)=\{\{a\},\{a,b\}\}$, but that is not what you're thinking, are you?

@GustavoBandeira "Small"?

Gustavo Bandeira

Yes. I mean, they have only two elements.

I know the usage os small is meaningless.

Pedro Tamaroff

@GustavoBandeira Well, using my definition above an order pair is a set having two elements.

But they are sets that are elements of a bigger set.

They are not subsets of the set in question.

Gustavo Bandeira

Got it.

Twink

04:19

@PedroTamaroff do you want to be a teacher?

Gustavo Bandeira

I'll have to got out in a hurry.

BBL

Pedro Tamaroff

@Twink You mean professionally?

Twink

yes

Pedro Tamaroff

@Twink I wouldn't want to dedicate my career to teaching, not really.

But I do like "teaching" if you can call it that. =)

Twink

so, you want to dedicate to research?

Pedro Tamaroff

04:33

@Twink Yes.

Twink

but almos all researches from universities must teach some courses

generally 2 courses when they're titular professors

Pedro Tamaroff

@Twink Yes, sure.

I just mean I don't want to devote my math-life to teaching.

Twink

I think you're going to be a very nice teacher

hi pal @cyberskull

why didn't you say hello?

cyberskull

@Twink hi pal

:-)

Twink

:D

cyberskull

04:46

:D

Twink

how are you pal?

I hadn't seen you for a long time

cyberskull

@Twink Fine thanks. How are you pal?

Twink

fine too, thanks :)

cyberskull

:)

Twink

I missed you :$

cyberskull

04:49

Aww thanks...that's very nice of you to say.

Twink

:$ :)

cyberskull

:D

Transcript for

$Mathematics$ Mathematics