Mathematics - 2016-10-21 (page 1 of 4)

Adeek

12:03 AM

oh I see how I can do it.

first we have to show that $\{z_n\}$ is cauchy sequence in X.

then we can use that some how

0celo7

12:16 AM

yes

but that's clear, no?

we have $||z_n-z_m||_X\le (1+C)||z_n-z_m||_X<\epsilon$

where $||T||_\mathrm{op}=C$.

oh wait

that's not $<\epsilon$

But that still works!

$||z_n-z_m||_X\le ||z_n-z_m||_X+||T(z_n-z_m)||_Y<\epsilon$

@Adeek Agreed?

Similarly $(T(z_n))$ is Cauchy in $Y$.

@BalarkaSen Shouldn't you be asleep?

Balarka Sen

woke up

0celo7

@BalarkaSen Is the projection a closed map in the product topology?

I know it's open.

Mike Miller

@0celo7 $xy = 1$

Balarka Sen

^

It is for compact factors however

0celo7

@MikeMiller Rats. We talked about this topology and that was the example he used

PVAL-inactive

12:27 AM

@MikeMiller What do I do if I've had to prove something I think that someone else already knew, but hasn't written down ( I think)? Do I just email them saying "you knew this already, right?"?

Mike Miller

@PVAL-inactive That's what I would do. This is just something they know and not something they're trying to write a paper on, yes?

PVAL-inactive

The someone else is the top person in my field.

This is something I assume they've known for 30 years.

Mike Miller

It's not your advisor?

PVAL-inactive

Its YE.

Mike Miller

Oh. Obviously.

PVAL-inactive

12:29 AM

I emailed him once before about something, and he responded.

0celo7

@PVAL-inactive Why didn't they write a paper?

PVAL-inactive

Everyone says hes really nice, but he's one of the few mathematicians whose profile is really intimidating.

Mike Miller

Yeah probably just ask him. I guess you would state it in the paper as "this was first observed by YE" assuming it was.

@0celo7 They had better things to write papers about.

PVAL-inactive

The result is that if $X_1,X_2$ are compact exact symplectic manifolds with convex boundary and $X_1$ symplectically embeds inside $X_2$, then $X_2-X_1$ is an exact symplectic manifold with the same form.

He's written arguments that are close to word for word to thing I proved, but he proved different things with them.

Mike Miller

I liked his note on extending exact cobordisms.

PVAL-inactive

12:33 AM

which note is that?

Mike Miller

The one used in holomorphic disks etc.

PVAL-inactive

@MikeMiller those cobordisms aren't necessairly exact.

Mike Miller

My bad.

You see how well I know symplectic stuff.

PVAL-inactive

Well I emailed my advisor about it so theres a chance he knows a reference for it. I sort of feel like he'll probably say "ya I knew about this, and YE probably knew about this before me" or something.

Mike Miller

It's like Thurston. I feel like YE knows half the things that get published.

0celo7

12:40 AM

Why the acronym?

Balarka Sen

Anonymity.

PVAL-inactive

Because I do not want to reveal publicly who I am talking about contacting.

Mike Miller

There's the thing where he knew a calculation of $\pi_0 \text{Diff}(S^3)$ and wrote that you can do it by filling discs in a paper and then someone wrote a 50 page paper where they carried it out later.

0celo7

It's not hard to figure out who the "top person" in your field is, tbh.

PVAL-inactive

I am aware, but I'd like plausible deniability etc.

0celo7

12:41 AM

Hm, ok.

@PVAL-inactive My advisor recommended his book.

PVAL-inactive

CE or EM

Balarka Sen

I have heard of him a lot but no clue about any of his works.

user777

Hello, I have a question about non-isomorphism tree

0celo7

@PVAL-inactive The GSM one.

@MikeMiller I mean, does that really count as "knowing"? Is it possible to know that a proof doesn't hit a snafu unless you write it out? Especially when it takes 50 pages to do so?

Mike Miller

He knew.

0celo7

12:45 AM

Or did he have it in a notebook somewhere?

PVAL-inactive

probably that.

I think he had probably computed that the moduli spaces of bishop disks were nice enough for the proof to work somewhere for other reasons.

0celo7

@PVAL-inactive Do you know the book I'm talking about?

PVAL-inactive

Ya thats the EM book I said

0celo7

Oh, what does EM mean?

PVAL-inactive

The other author's name begins with an M.

0celo7

12:47 AM

Oh, right.

Mike Miller

Nice book.

PVAL-inactive

I own it. Haven't really read it.

0celo7

What does one need to read it? My advisor was too vague.

PVAL-inactive

I've spent quite a lot of time on CE.

user777

if you asked to count all trees that are non-isomorphism with 6 vertices, then the answer is 6 according to this site math.stackexchange.com/questions/413792/trees-on-six-vertices , but I don't know why the count two trees with the same degree?

Mike Miller

12:48 AM

Don't you mean ET.

PVAL-inactive

Probably just 1st year level diff top.

0celo7

Hmm, ok. That's what he said.

PVAL-inactive

No i mean CE.

user777

According to my understading is that you are asked this kind of question, you start counting from degree 2 (we cannot start with degree 1 since there is no tree never with degree 1 and connected at the same time)

0celo7

Off to homework. Byes

PVAL-inactive

12:49 AM

I've read a good deal of ET too I guess, but I've commited my life to CE.

user777

so with degree 2, you have like this one (. --- . --- . --- . --- . --- .)

assume that . is vertex and --- is an edge

Mike Miller

Oh right.

I've spent no time on it. But again, I know nothing about this.

user777

Sometimes I feel I'm speaking with myself

PVAL-inactive

There's an interesting disparaging review of ET on Amazon.

by this guy mathoverflow.net/users/57930/roger-bagula

Adeek

hm @0celo7 chehcking it

yeah I agree

Mike Miller

12:53 AM

@PVAL-inactive I've seen him around. He's extremely bizarre.

Adeek

I agree completely with you

0celo7

@Adeek Ok I really need to get work done but maybe I've given you an idea

Adeek

yeah I think I got it so we have X and Y complete right

so $z_n \rightarrow z$

and also $T(z_n) \rightarrow T(z)$

I guess $T(z_n) \rightarrow y$

hm

0celo7

@Adeek you need bounded for that

closed -> bounded -> continuous

Adeek

yeah bounded --> continous as we have linear operator

we know that T is bounded

Ted Shifrin

12:57 AM

Ugh. So frustrated. Someone I know from UGA posted a crap solution, then wrote a remark that everything he did was backwards. And told me that sufficed because he'd written the remark. Ugh.

Adeek

so it is continous

so we must have $T(z_n) \rightarrow T(z)$

Balarka Sen

Yikes

Ted Shifrin

@Balarka: What the **** are you doing awake?

Balarka Sen

i woke up!

Ted Shifrin

Oh great. 3 hours of sleep.

I think this place is like a drug for you ... :D

Balarka Sen

12:59 AM

shrug. tell that to my biological clock.

Adeek

I don't know how you function if I don't sleep for 8 hrs then I can't focus

Ted Shifrin

No wonder you're always sickly.

0celo7

He's probably not a human...

Ted Shifrin

Good, Karim. Stay healthy.

Good point, @0celo.

But you're perhaps only semi-human yourself.

0celo7

Oh?

PVAL-inactive

1:00 AM

@TedShifrin How does this possibly affect your life?

Youre not teaching again are you?

Ted Shifrin

Spending significant time here is teaching, yes, @PVAL. And I guess sloppy arrogance in people I like is distressing.

Balarka Sen

I am proud of this particular compliment: chat.stackexchange.com/transcript/36?m=28820219#28820219 (probably that's not what Mike intended it as though)

PVAL-inactive

oh on this site?

Ted Shifrin

Yes, on this site.

Thanks for fixing the spelling, @Balarka :D

Balarka Sen

grr

Ted Shifrin

1:02 AM

@PVAL: I do not pretend to be perfect. But if someone comments that things are all wrong, I try to fix them or I delete. I don't say "I've done enough as it is."

Adeek

oh ok it is trivial afterwards @0celo7

Ted Shifrin

I like the fact that this person is trying to learn stuff better by answering questions, but if his answers screw up other people, that's not helpful.

0celo7

@Adeek Can you post the proof of that Banach thing?

Ted Shifrin

trying to think of EM books

PVAL-inactive

EM are the names of the authors

Ted Shifrin

1:03 AM

Yes, so I deduced. Not E&M. :P

PVAL-inactive

If you scroll up it should obvious who the E is ( I think).

0celo7

@TedShifrin Jackson

Ted Shifrin

LOL, 0celo, yes, but neither an E nor an M.

Adeek

so we have $\{z_n\}$ which goes to z in the X norm. Also, we have $T{z_n}$ goes to T(z) in the Y norm so this means that $||z - z_n||_X + ||T(z_n) - T(z)||$ goes to zero as n goes to infinity.

That completes the proof.

sure @0celo7

I could send you the notes

0celo7

Oh, please do.

Adeek

1:05 AM

can I have your email

PVAL-inactive

@TedShifrin The book has a lower case h in the title.

Ted Shifrin

ROFL, gee, thanks, @PVAL.

Adeek

oke got it

I will send it from my personal uni email

just a sec

Ted Shifrin

OK, Milnor. But there wasn't a coauthor, was there? Oh, Epstein?

PVAL-inactive

@TedShifrin no

0celo7

1:05 AM

Huh?

Ted Shifrin

Sigh.

0celo7

Milnor?

Balarka Sen

I suspect that h is special enough for it to distinguish between books

0celo7

Not Russian enough

Ted Shifrin

I thought the "h" was a solo author.

Mike Miller

1:06 AM

Nope.

0celo7

Who has a lower case h as a first letter of a last name :P

inb4 some Dutch person

Ted Shifrin

last name? I was thinking h-cobordism.

I give up.

close

o

@0celo7 done

1:06 AM

but not quite

Ted Shifrin

goes back to martinis

PVAL-inactive

That's actually quite close.

Adeek

let me know I could actually all notes that I have

the guy explains everything from scratch if you want

PVAL-inactive

It's written by someone Arnold doesn't think very highly of as an expositor (I guess that really doesn't narrow it down).

0celo7

@Adeek the proofs are posponed of the first two things!?

Ted Shifrin

1:08 AM

LOL, @PVAL. I co-wrote a paper that in large part was making rigorous something A'rnold said in a sentence.

Adeek

@0celo7 I have it in later notes if you want it

Ted Shifrin

One of my better papers, actually :P

Adeek

do you want me to share all of the notes ?

it is not that many actually

0celo7

sure

Balarka Sen

He shares advisor with someone whose self-portrait is a baboon (or was it something else in his site page? I don't remember)

PVAL-inactive

1:10 AM

@TedShifrin Do you have an online copy?

Ted Shifrin

Well, in fairness to my friend, I have to report he deleted (rather than spend 5 minutes making it a decent answer).

Yes, @PVAL.

0celo7

@BalarkaSen lol what?

PVAL-inactive

Can you link it?

Ted Shifrin

Balarka, go back to sleep if you're going to be that helpful.

No, @PVAL, not easily, but I can email it.

Balarka Sen

Sorry about that :3

I'll stop.

Adeek

1:11 AM

@TedShifrin I got a scholarship btw

2000 $

Ted Shifrin

Good for you, Karim!

PVAL-inactive

Can you email it to me?

I'm interested.

0celo7

Can you email me $2000 too?

Adeek

sure @PVAL-inactive

yeah I am happy about that pays for rent and food haha

Ted Shifrin

Sure, @PVAL, not exactly down your alley, but this ended up being the beginning of a whole saga for me and coworkers. Let me see if I can resurrect your email.

Adeek

1:12 AM

@PVAL-inactive can I have your email ?

0celo7

@Adeek what are you sending him?

Ted Shifrin

You're going to email him cash?

Adeek

noo lol

0celo7

Does that even work?

Adeek

functional analysis notes ?

0celo7

1:13 AM

He was asking for Ted's paper

Adeek

he asked for it

oh ok

lol

Fractional Inquirer

Ted's papper on what?

Tell me. Now.

0celo7

Arnold

Adeek

okay guys I am gonna go grab the bus

I will be here soon maybe hour

Ted Shifrin

That's a lot of heavy lifting, Karim.

0celo7

1:14 AM

I grabbed the bus today

Adeek

haha @TedShifrin

btw @TedShifrin I found this great algebra book called allufi chapter 0

I am reading it while I am on the bus

Ted Shifrin

Oh, I know about Paolo's grad algebra text. Well respected.

Balarka Sen

It's nice, but has an abstract point of view.

PVAL-inactive

It's a weird book.

Fractional Inquirer

Ted... Tell about what is your paper. Now.

Adeek

1:15 AM

I like that @BalarkaSen

Ted Shifrin

I don't mind that so much at the grad level.

Adeek

I like the book introduces category theory from scratch

Balarka Sen

I don't object to his treatment of classification of groups of a certain order.

Ted Shifrin

@Fractional: This was about cusps of the Gauss map of a generic surface in $\Bbb CP^3$.

Fractional Inquirer

Let's talk about less known facts.

Did you know that there is a very simple algebraic expressions to compute the lesser angle of a right triangle in terms of its sides?

That's to say, it gives a very accurate approximation.

0celo7

1:16 AM

trig is too hard

Fractional Inquirer

Ask what is the expression.

0celo7

I don't dare

@Adeek got it

good lord

PVAL-inactive

@Ted Did you figure out who E is yet?

Fractional Inquirer

Ocelo, this is a MATH CHAT, not transvestite chat.

Prostituta.

Ted Shifrin

No, @PVAL, but I finally found your email address.

0celo7

1:19 AM

@FractionalInquirer What is wrong with you?

PVAL-inactive

He proved one of Arnold's most famous conjecture in the case of (real) surfaces. I don't know if an english translation of that proof ever came to be.

Ted Shifrin

Just give it up. I sent the email.

0celo7

@TedShifrin Even I knew who it was!

PVAL-inactive

Eliashberg

Ted Shifrin

Ohhh ... You could have told me Stanford.

0celo7

1:21 AM

@TedShifrin How do you remember where all these people are?

My advisor does the same thing

Balarka Sen

I was referring to Gromov when I said that baboon thing.

0celo7

lol

Ted Shifrin

Well, cuz I know a lot of them, even if not personally. I actually know a fair number of people at Stanford. I'll be there in a couple of weeks.

Balarka Sen

ihes.fr/~gromov

Ted Shifrin

Well, Balarka, he sorta looks like one.

Not that any math geeks should poke fun for looks.

0celo7

1:22 AM

He looks homeless...

Balarka Sen

I think he looks pretty nice.

Ted Shifrin

Not commenting.

0celo7

Uhhhhh

Ted Shifrin

OK, I'm disappearing. G'night, @Balarka.

Balarka Sen

Night

PVAL-inactive

1:25 AM

@Ted Why is this titled Brieskorn paper?

There's no interesting topology around the singularities is there?

Ted Shifrin

1:59 AM

@PVAL: No, it's because I replied to the last email you sent me. Sorry about that. I was so confuzled.

Adeek

2:34 AM

back

0celo7

3:00 AM

guess who's back

Adeek

@0celo7 how do you like the notes ?

0celo7

Haven't looked at them. Currently freaking out about QM.

He should let us have an equation sheet

Adeek

It is lucky your doing math and physics

0celo7

I'm an engineer

Adeek

your not doing double major in math and physics ?

0celo7

3:05 AM

double major in engineering and math.

QM is relevant to my engineering work

the homework was easy, then it got super hard

so I hope he puts questions like those on the first homeworks

Adeek

your not gonna continue onto math ?

0celo7

I'm not particularly good at math, so no.

oh god what if I have to compute something like $\int_\Bbb Rx^4 e^{-x^2}dx$

oh that would kill me

Better memorize Gaussian integrals.

Ted Shifrin

3:37 AM

@0celo: Honestly, I hate it when you and @Balarka say stuff like "I'm not particularly good at math." Quit that. The fact that you can assimilate (granted, not like one of the stars in the field) sophisticated graduate level math at your age(s) disqualifies both of you from talking like this. Seriously, quit it.

Mike Miller

h

arctic tern

message.rar.zip.tar.gz

Ted Shifrin

Regoodnight @MikeM ... and good evening to tern!

arctic tern

hello

DHMO

6:31 AM

@FractionalInquirer arctan(b/a)?

@0celo7 use the fact that $\int_\Bbb Re^{-x^2}\ \mathrm dx=2\pi$? (standard distribution)

Balarka Sen

@TedShifrin I just felt at times that I'm only quick at assimilating ideas, but not at proving things with them. I've grown to understand that as useless, hence the comment. But I have slowly stopped caring about being or not being good at math; the important thing is I like doing it, and it's part of my life (I guess I'm not very good at living either!)

DHMO

I mean $\sqrt{2\pi}$

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

$\quad\int_\Bbb R x^4 e^{-x^2}\ \mathrm dx$
$= - \dfrac12 \int_\Bbb R x^3\ \mathrm de^{-x^2}$
$= - \dfrac12 (x^3 e^{-x^2})_{-\infty}^\infty + \dfrac12 \int_\Bbb R e^{-x^2}\ \mathrm d(x^3)$
$= - \dfrac12 (x^3 e^{-x^2})_{-\infty}^\infty + \dfrac32 \int_\Bbb R x^2 e^{-x^2}\ \mathrm dx$
$= - \dfrac12 (x^3 e^{-x^2})_{-\infty}^\infty - \dfrac34 \int_\Bbb R x\ \mathrm de^{-x^2}$
$= - \dfrac12 (x^3 e^{-x^2})_{-\infty}^\infty - \dfrac34 (x e^{-x^2})_{-\infty}^\infty + \dfrac34 \int_\Bbb R e^{-x^2}\ \mathrm dx$

Balarka Sen

I'm sure 0celo7 can do the relevant computations

DHMO

I just want to practise

Balarka Sen

ah, ok

DHMO

6:46 AM

oops, it is $\sqrt\pi$ not $\sqrt{2\pi}$

but why?

Oh, I transformed $r\ \mathrm dr$ erroneously to $\mathrm d(r^2)$

DHMO

7:29 AM

Is there any straightforward proof of $\sum i^2=\dfrac16(n)(n+1)(2n+1)$? I have seen two proofs (1. consider $\sum(i+1)^3-i^3$, 2. mathematical induction) but I always believe in the beauty and elegance of mathematics...

user243301

10:37 AM

@user51189 this message is only to do feedback about one of your questions, thus is not required a response. Good luck. My idea as companion of the answers, is that number theory study integers (whose factorisation in terms of prime numbers) and arithmetic functions. An important class of such are multiplicative functions, if you declare that $2$ is not a prime then this is a complication for the theorems involving multiplicative functions.

s.harp

@DHMO if compare $x^2$ on $[i,i+1]$ with $i^2 \chi_{[i,i+1]}(x)$ you find that the difference of the integrals will be $\frac13 [(i+1)^3-i^3] - i^2 =\frac13 (i^3+3i^2+3i+1-i^3)-i^2= i+\frac13$.

DHMO

@s.harp what is $\chi$?

s.harp

so the difference of $\int_0^n x^2 dx$ and $\sum_i^n i^2$ is $\sum_i^n (i+\frac13)=\frac{n(n-1)}2+\frac n3$

$\chi_A(x)=\begin{cases} 0 &x\notin A\\ 1 & x\in A\end{cases}$

DHMO

nice

s.harp

i dont know if this proof is elegant, but it is a simple analysis calculation of the same result without much moving of summands

DHMO

10:44 AM

thanks