Mathematics - 2015-02-27 (page 3 of 4)

ys wong

6:00 PM

Im currently at Linear Algebra now

Balarka Sen

The intersection of our open subset is htpy equiv to $S^1$

teadawg1337

I'm studying real and complex analysis independently @yswong

Sayan Chattopadhyay

Hey guys can we prove that there are infinitely many constants.......

Balarka Sen

and we need to see what happens to the induced maps of $S^1 \hookrightarrow U$ and $S^1 \hookrightarrow V$

ys wong

Btw guys do u all learn Quantum Mechanics

G-man

6:02 PM

@sayan how do you define a constant?

Balarka Sen

$\pi_1(i_U)$ clearly maps a loop of $S^1$ to $S^1 \vee S^1$

G-man

i know some elementary relativistic mechanic, yeah @ys

ys wong

Is Quantum mechanics releated to maths?

teadawg1337

Physics is a very mathematical science

G-man

^agree

ys wong

6:04 PM

Im intending to go towards that field rather than become a mathmatician

Julian Rachman

@Ted aw man. Maybe you can be my higher mathematics tutor.

ys wong

Btw do u all know what skill set must i have in order to learn Quantum properly?

Im intending to become a Quantum Mechanist?

Julian Rachman

@Sayan I haven't yet. I am half way through the problems

user130018

@yswong Linear algebra

G-man

is mechanist a word? im not sure..

ys wong

6:08 PM

User130018: Which is what im learning now

Mike Miller

@AlexWertheim: still here?

user130018

@MikeMiller What are your favorite topology books

ys wong

User130018: Thats all ? Then my life as a Quantum Mechanist would be very easy

User130018: Easier than that of a mathmatician

Mike Miller

Like, point set stuff? I never read any of those. The one we used for the course I took was Kelley's, I think. I wasn't a big fan, since it got the definition of a topology wrong.

user130018

@MikeMiller No, like algebraic topology, homotopy theory, etc.

Mike Miller

6:11 PM

Oh. Uh, Hatcher? Bredon's book is good too, if you know smooth manifolds already.

Bott-Tu is another "good if you know smooth manifolds and de Rham cohomology" book.

ys wong

User130018:But seriously, how far do i have to go in mathmatics if i want to learn Quantum Mechanics. Im already at linear Algebra and ODE. How much more maths do i need.

Balarka Sen

@MikeMiller Really, though, the cell decomposition is a way easier way to go about this.

user130018

@MikeMiller Do you have any reference recommendations for learning smooth manifolds and cohomologies

Mike Miller

Lee's book is good. Lots of exposition.

user130018

Is his topological manifolds book any good

Mike Miller

6:14 PM

No idea. Probably.

ys wong

User130018: Could u answer me?

user130018

@yswong Try looking at a quantum mechanics book to check

Mike Miller

I doubt he knows precisely what mathematics you need to know for wuantum mechanics. Anyway, a good undergraduate math education can do nothing but help you if you want to be a physicist.

Daniel Fischer

@LeGrandDODOM That's one way, the one I was evidently thinking of then. There are of course many other ways.

Axoren

@robjohn Thanks for looking into it, but there's still something missing.

I corrected the mistake you found.

ys wong

6:18 PM

User130018: An interesting observation i made though. I realize that the more maths i know the easier Quantum becomes. Even though i did not really specifically study Quantum Mechnaics.

Stan Shunpike

@TedShifrin is it true curves on a manifold are invariant under coordinate transformation?

Balarka Sen

@MikeMiller I think $\pi_1(i_A)$ is the homotopy class of the map that sends a circle to wedge of two circles by pasting boundary of the square according to $aba^{-1}b^{-1}$ and $\pi_1(i_B^{-1})$ is the homotopy class of the map that sends a circle to RP^2 via $x \mapsto x^2$.

no, wait.

evinda

@DanielFischer I found the following: The statement is always true for $f(n)=\Omega(1)$. But how do we deduce it?

Balarka Sen

OK, this is getting complicated.

Kaj Hansen

@BalarkaSen @PedroTamaroff @JulianRachman Pete uploaded more course notes for what we've been looking at the past few weeks (and more). math.uga.edu/~pete/4200_metric_spaces.pdf

Daniel Fischer

6:23 PM

@evinda $f(n) = O(f(n)^2)$ doesn't make sense, since the one is a function, the other a class of functions. Should it be $f(n) \in O(f(n)^2)$?

ys wong

Can i ask u guys another thing

evinda

@DanielFischer This is meant.. we just write it like that..

ys wong

How does a knowledge in Mathematics help u in Economics

Daniel Fischer

@evinda Well, don't write it like that. It's wrong. Regarding the matter: What does it mean that $f(n) \in O(g(n))$, by the definition?

user130018

@yswong Well sometimes there are physical phenomena (that require math) related to economics...like the equation for the diffusion of superneutrinos can be used to model bonds and stuff

evinda

6:26 PM

@DanielFischer $f(n) \in O(g(n))$ means that $\exists c>0, n_0 \in \mathbb{N}$ such that: $f(n) \leq c g(n), \forall n \geq n_0$.

robjohn

@MikeMiller Oh, so I've passed by it on the way home from the Ahmanson many times.

ys wong

User130018: U see im interested in learning mathematics more as an applied tool rather than learning mathematics for its own sake.

evinda

So $f(n) \in O(f(n)^2)$ iff $\exists c>0, n_0 \in \mathbb{N}$ such that $f(n) \leq c f^2(n) \overset{f(n) \neq 0}{\Rightarrow} f(n) c \geq 1 , \forall n \geq n_0$

right? @DanielFischer

robjohn

@MikeMiller I just never knew it was called that.

Daniel Fischer

@evinda Right. You may need to take the possibility that $f(n) = 0$ into account or not.

abcd1234

6:29 PM

Hi guys, I have asked a question here:math.stackexchange.com/questions/1167173/…

For #2, I need some help

evinda

@DanielFischer For $f(n)=0$: $f(n) \leq cf^2(n)$ holds.
But how can we now find all the f(n) for which $f(n) c \geq 1$ holds?

ys wong

User130018: At the same time, i also wanted to train my critical and logical thinking skills and especially problem solving skills im using maths as my training grounds. Is this a good way to do so.

User 130018: Can maths help me accomplish all this

abcd1234

I can understand maybe it is surjective, but is this because $h(a) = a+bi$ where $a+bi \in \mathbb{c}$

Daniel Fischer

@evinda You can rewrite $f(n)c\geqslant 1$ a little.

evinda

@DanielFischer $f(n) \geq \frac{1}{c} \Rightarrow f(n) \geq C$, where $C=\frac{1}{c}$

Balarka Sen

6:33 PM

@MikeMiller I give up. It's maddeningly hard to do it the way I was trying to do.

Alex Wertheim

What's up, @Mike?

ys wong

Hey everyone:i wanted to train my critical and logical thinking skills and especially problem solving skills im using maths as my training grounds. Is this a good way to do so.

Balarka Sen

Cell decomposition is the best way.

Daniel Fischer

@evinda Right. So the condition is that there exists a ...?

evinda

a $C>0$ such that $f(n) \geq C$? @DanielFischer

Balarka Sen

6:36 PM

The next problem wants me to find the universal cover of RP^2 wedge RP^2. Interesting.

Mike Miller

@AlexWertheim: Ever gonna send that email? (The thing I actually wanted to say was: I bet if you started studying soon, like after you accepted the offer, you could probably pass the algebra qual, especially given your number theory background.)

Balarka Sen

I guess all we have to do is to find a simply connected cover with deck transformation Z/2 * Z/2.

Daniel Fischer

@evinda Or $f(n) = 0$, right.

Balarka Sen

oh. heh.

just draw the cayley complex of Z/2 * Z/2 i guess.

iwriteonbananas

@BalarkaSen why are u looking for that?

Balarka Sen

6:38 PM

i am solving a few qual altop problems.

for fun.

iwriteonbananas

whats qual alg top?

Balarka Sen

dis

Alex Wertheim

@Mike: hahaha, I've gradually been asking you most of the questions here, I realized. But yes, probably.

Mike Miller

At this rate we might as well wait until you're here :)

iwriteonbananas

those problems look hard

Balarka Sen

6:40 PM

actually the space is just infinite wedge of spheres

Alex Wertheim

It may well work out that way, haha. :) Even with your nice assurances, I'm still dreadfully slow.

iwriteonbananas

@BalarkaSen what is?

Balarka Sen

the universal cover of RP^2 wedge RP^2

Alex Wertheim

That's nice of you to say. I really had better get studying ASAP...

Balarka Sen

at least i think it is. @Mike?

iwriteonbananas

6:41 PM

it's not simply connected though

Balarka Sen

uh. yes it is, @iwriteonbananas

iwriteonbananas

why?

Balarka Sen

wedge of simply connected spaces is simply connected.

Alex Wertheim

I'm weak on a lot important fronts, I'm afraid. The algebra qual is the only one that's really possible, I think.

evinda

@DanielFischer Nice.. I am looking at this exercise: Prove or disapprove the sentence $f(n) \in O(f(n)^2)$.
Could I answer as followed?
False. Let $f(n)=\frac{1}{n}$, $\forall n \in \mathbb{N}$. Suppose that $f(n)=O(f(n)^2)$. That means that there are $c>0$ and $n_0 \in \mathbb{N}$ such that $ \frac{1}{n}=f(n) \leq c f(n)^2=c \frac{1}{n^2}, \forall n \geq n_0$.
$\frac{1}{n} \leq c \frac{1}{n^2} \Rightarrow \frac{n^2}{n} \leq c \Rightarrow n \geq c$, contradiction.
We see that $f(n) \in O(f(n)^2)$ iff $f(n)=\Omega(1)$.

iwriteonbananas

6:43 PM

oh i was thinking wedge sum of $S^1$'s

Balarka Sen

that's stupid, since there is no interaction of S^2 at all, and S^2 is the universal cover of RP^2.

iwriteonbananas

i know.....

Mike Miller

@AlexWertheim: I just know I learned a lot in the span between when I started studying and when I took the quals. I think it's about 5-6 months out? That's a long time.

Yes.

Balarka Sen

yes, but we are looking at wedge

iwriteonbananas

i just realized

Alex Wertheim

6:45 PM

@MikeMiller: that's true, I suppose. Before you began your independent work, what was your topology/analysis background like?

(By independent work, I mean your studying for the quals)

Mike Miller

Remember that you only have to take two. I'd taken a general topology course and knew about the fundamental group. Starting in Jan I took an algebraic topology course that lasted the last two quarters of school, and included me learning (on the side) smooth manifold stuff.

Analysis, I 'knew' measure theory (good god I'm bad at it)/Lebesgue stuff, and various topics in complex analysis (basic stuff, riemann surfaces, some harmonic analysis)... and that's about it. I didn't study for that test.

Daniel Fischer

@evinda You have a typo in the penultimate line, $n \geq c$ instead of $n\leq c$. If you are only considering functions that never attain the value $0$, it is correct, if $f(n) = 0$ is a possibility, you need to account for that too. That would of course destroy the nice $\Omega(1)$ characterisation.

Balarka Sen

this one looks easier, except that i don't know what cohomology is.

iwriteonbananas

neither do i

but ive heard someone say that word before...

Alex Wertheim

@Mike: Hrmm. Interesting, thanks for sharing. Did you mainly just read and do problems?

evinda

6:52 PM

@DanielFischer So could we say this?
We see that $f(n) \in O(f(n)^2)$ iff $f(n) \in \Omega(1) \cup \{0\}$.

Axoren

Cohomology? Homology together?

Balarka Sen

who the hell would want to compute fundamental group of $SL_2(\Bbb C)$? shudders

ys wong

Btw guys, how hard is calculus releative to topics like Real Analysis, Complex Analysis, Fuctional Analyiss, etc?

Axoren

Sounds like an interesting partnership.

@BalarkaSen Someone out there in a cabin in the woods is trying.

Daniel Fischer

@evinda No, it's not as easy. you can have $f(n) = 0$ for some $n$ but not all, consider $f(n) = 1 + (-1)^n$.

Axoren

6:53 PM

There's always someone

Mike Miller

Yeah, but problems also includes the problems associated with the texts I was reading. Generally, the quals themselves were harder, because the book I mainly used was Hungerford, and it's not a hard book.

Alex Wertheim

The quals look like they've increased substantially in difficulty. I didn't find the 2001 Fall algebra exam too bad, but for the 2014 exam (the one you took, lol), I felt like I could answer very few questions satisfactory.

Mike Miller

@BalarkaSen: It's simply connected.

evinda

@DanielFischer In our case $f(n)$ is an asymptotically positive fuction. So can it take the value 0?
How else could we find all the f(n)?

Mike Miller

@AlexWertheim: Only look at the ones 2010ish and later. There was a huge jump across the board around then.

iwriteonbananas

6:55 PM

lolol

Balarka Sen

@MikeMiller Not obvious.

At least, not to me.

It's abelian, and that's as far as I would go.

Mike Miller

Oh, I'm sorry. It's $\Bbb Z$.

Daniel Fischer

@evinda If asymptotically positive means that $f$ can have only finitely many zeros, then $\Omega(1)$ is right. Only when you need to take functions like the above example into account things become complicated.

Alex Wertheim

@Mike: Makes sense, that seemed to be the case. Unfortunately, I know almost no representation theory, and don't know much commutative algebra. The exam seems to have shifted towards these topics in some sense. I was thinking about picking up Serre's beautiful book on linear representations of finite groups.

Mike Miller

@AlexWertheim That's where I learned the rep theory I used for it. I only read the first chapter and a little bit of the second - huge difficulty jump.

There's very little you need to learn of it.

evinda

6:57 PM

@DanielFischer Could I also justify it only as follows? :/
False. Let $f(n)=\frac{1}{n}$, $\forall n \in \mathbb{N}$. Suppose that $f(n)=O(f(n)^2)$. That means that there are $c>0$ and $n_0 \in \mathbb{N}$ such that $ \frac{1}{n}=f(n) \leq c f(n)^2=c \frac{1}{n^2}, \forall n \geq n_0$.
$\frac{1}{n} \leq c \frac{1}{n^2} \Rightarrow \frac{n^2}{n} \leq c \Rightarrow n \geq c$, contradiction.

Mike Miller

I think Hungerford (and definitely Lang) covers pretty much all of the commutative algebra you need.

Balarka Sen

@MikeMiller OK. How'd you do that?

Alex Wertheim

@Mike: awesome, good to know. I wonder if I'm being mainly intimidated by not knowing some of the language used here, as opposed to the problems being intrinsically difficult. Of course, I'm sure many of them are quite tough.

Mike Miller

Oh, I should tell you. Passing scores are generally 60-70/100.

And on the most recent one, it was 39.

Daniel Fischer

@evinda Same typo as above, you should have $n \leq c$ at the end. As I said, if you don't need to care about functions with $f(n) = 0$ for infinitely many $n$, it's right.

Mike Miller

6:58 PM

It was a notoriously hard test.

Alex Wertheim

Phew. God, looking at it I was just floored.

Mike Miller

Yeah, me too. I think I scored a 52? That or a 59. I don't remember.

Alex Wertheim

I was under the impression that one was supposed to be able to answer most questions. That is a big relief. Not that I intend on answering few, but still.

Mike Miller

(Also: nobody knows if you fail a test except you and the secretaries.)

evinda

@DanielFischer You mean that it is false when $n \to +\infty$ ?

Alex Wertheim

7:00 PM

2013 looks significantly easier, although not easy, of course.

Daniel Fischer

@evinda I don't understand your question.

ys wong

Btw guys: How important is calculus for a Mathematics Major

Mike Miller

@BalarkaSen The maximal compact subgroup of $SL_2(\Bbb C)$ is $U(2)$ (and a Lie group splits - not as a group, but as a smooth manifold - as a product $M \times \Bbb R^k$, where $M$ is the maximal compact subgroup and $k$ is an appropriate integer.) So it suffices to compute $\pi_1(U(2))$. But we have a fiber bundle $S^1 \to U(2) \to SU(2)$, and recalling that $SU(2) \cong S^3$, you can use the long exact sequence in homotopy groups to finish the job.

iwriteonbananas

T___T

wow

Balarka Sen

That looks subtle.

evinda

7:03 PM

@DanielFischer So if we also take into consideration the functions with $f(n)=0$ for infinitely many $n$, what could we say?
At the counterexample $f(n)=\frac{1}{n}$ the function goes to $0$ only if $n \to +\infty$. So do we have to take this limit into consideration?

Balarka Sen

I didn't know that $\pi_1(G) = \pi_1(H)$ if $H$ is maximal compact subgroup of $G$.

Mike Miller

It's easier to show that it deformation retracts onto $U(2)$; that it splits as a product is harder. And the former thing is essentially Graham-Schmidt.

Balarka Sen

Guess I'd have to know a bit Lie theory to understand that.

Alex Wertheim

I think my main worry is that I got started on serious mathematics late, and now that I have the chance to build a much more complete background, I don't want to get ahead of myself.

iwriteonbananas

gram*

Mike Miller

7:05 PM

You need $G$ to be a Lie group. Topological groups don't necessarily have a maxcompact.

Daniel Fischer

@evinda You mentioned "asymptotically positive" before. What does that mean? On that hinges (almost) everything.

Mike Miller

@AlexWertheim I understand that. But at the very least, studying will mean you learn a lot on the way.

evinda

According to my book a function that is positive from a value of its argument and further is called asymptotically positive. @DanielFischer

Alex Wertheim

@Mike: oh absolutely. I'm not opposed to studying, so much as I'm nervous about skipping the courses if I pass the qual! But I suppose that is much too far ahead of myself. :)

ys wong

Btw guys: Is Calculus the pinnacle of a Mathmatics

Of Mathmatics

Mike Miller

7:07 PM

You shouldn't worry about that. If you can pass the qual, you wouldn't learn anything particularly substantial from the courses.

ys wong

I oned used to think that after i mastered calculus, thats means i have already conquered all of Mathmatics

Alex Wertheim

Really? That's indeed surprising, just because there's SO much material listed on the qual syllabi.

Mike Miller

Right, and to pass the quals you should try to learn it :)

Also keep in mind that your education in those fields doesn't stop there; you pass the algebra qual and you go ahead and take commutative algebra, which reinforces your knowledge of the stuff you learned before...

Alex Wertheim

That's true. I guess it only expands the opportunity to take interesting classes. :)

Mike Miller

That was my perspective.

Daniel Fischer

7:11 PM

@evinda Okay. (I would rather call that "eventually positive", but no big deal.) Then your example $f(n) = 1/n$ is admissible, and hence proves that we don't generally have $f(n) \in O(f(n)^2)$. And, as you have shown, the condition for $f(n) \in O(f(n)^2)$ for asymptotically positive functions $f$ is that $f\in \Omega(1)$.

Mike Miller

and the algebra class conflicted with the Lie groups/algebras class I took, so I'm glad that didn't turn out to be a problem :)

Alex Wertheim

229? I was looking at that. It looks like an awesome class.

Balarka Sen

I guess I'd have to rebrush my fundamental group and covering spaces knowledge by going through the exercises in Hatcher thoroughly (as I learned that stuff from Munkres, I left some of those out) after studying homology, seeing that I am not entirely comfortable with Van Kampen.

For once, I want to study something very thoroughly.

Mike Miller

I didn't enjoy mine so much, @AlexWertheim, but it probably depends very heavily on who teaches it.

Jack D'Aurizio

Is someone willing to review this answer: math.stackexchange.com/questions/1168209/… - I think it is a pity it goes unnoticed.

Alex Wertheim

7:15 PM

Aw, that's a shame. Which classes have you really enjoyed, out of curiousity?

(I'll brb)

Mike Miller

Anything Paul Balmer teaches is going to be a gem. His commutative algebra class was really outstanding. I'm enjoying gauge theory, I enjoyed learning for the topology seminar, I'm having a great time in PDEs... it depends pretty closely on the instructor (and the topics, ofc) I think.

I didn't so much enjoy the instructor of 229, but it was also more focused on the algebras; I wanted to learn about the Lie groups.

There was a project at the end, though, that I had a lot of fun with.

Stan Shunpike

is it true curves on a manifold are invariant under coordinate transformation?

if so, why? I thought nothing would be invariant under coordinate transformation. But when I heard the above remark, suddenly a whole bunch of things made sense. Yet I still don't understand why curves would be invariant under coordinate transformation

Alex Wertheim

Sweet. I figured, yeah. I was thinking about that particular instructor as a potential advisor, so interesting to know.

evinda

@DanielFischer So after saying that it is not true, could I say the following?
We notice that $f(n) \in O(f(n)^2)$ iff $\exists c>0, n_0 \in \mathbb{N}$ such that $\forall n \geq n_0: f(n) \leq c f(n)^2 \overset{f(n) \neq 0}{ \Rightarrow } 1 \leq c f(n) \Rightarrow f(n) \geq \frac{1}{c}:=C \Rightarrow f(n) \in \Omega(1)$.

Mike Miller

What about his work interests you? I don't really have a good handle on what he does.

I've heard he does lots of... categorifying invariants. Nobody really gave me more specifics.

Alex Wertheim

7:32 PM

@Mike: I don't have an excellent idea either. But it seems like he does a lot of interesting work on Hecke algebras and modular representation theory, which piqued my interests.

Mike Miller

Ah, I see. Modular representation theory is, to my knowledge, very categorical nowadays. Balmer gave a friend of mine a book on it as an introduction to applications of triangulated categories...

Alex Wertheim

Hrm. I enjoy category theory as a useful language for talking about things, but I've generally been turned off to it when it enters the research realm. Then again, I don't really know anything about anything. :)

Mike Miller

I used to feel that way, but my opinions have changed. (If you want to be convinced about $\infty$-categories, read the first few bits of Jacob Lurie's paper on the cobordism hypothesis.)

Alex Wertheim

Really? I'll definitely give it a look. I've heard so much praise about Lurie's work.

Mike Miller

It's not all to my taste, but I'm convinced he does what he does because he thinks it's the right way to prove things. He's very down to earth. (If you look at the courses he's taught, only a handful are straight up category theory!)

Alex Wertheim

7:39 PM

I need to stop my brain from shutting off when I see research about monoidal such-and-such $E_{\infinity}$ ring spectra and so on. ;)

Mike Miller

Eh, I haven't trained mine not to yet.

But I also don't know what they are yet.

Alex Wertheim

That's a relief. I've found other descriptions much too overwhelming.

Mike Miller

Who did you sign up to meet with?

Alex Wertheim

Well, I didn't want to bombard them with too many names. I listed Hida and Khare officially, since if I do number theory, it will almost certainly be with one of them.

Mike Miller

I did something similar, but it turns out pretty much everybody signed up for a lot (like, 4 or so - I signed up for three?)

You shouldn't feel so bad about signing up to meet with lots of people if you can still edit it.

Alex Wertheim

7:47 PM

I don't think I can, sadly. Still, I'm not sure I could have a fruitful conversation with many faculty members. I don't know much number theory, but I know even less about other things. =P

Mike Miller

That's the case with everybody coming in :)

When I talked to Haesemeyer he just gave a brief (10 minute) outline of the idea behind what he does.

Alex Wertheim

Funny, he probably would have been the third person I would have asked.

Mike Miller

But certainly some are hard to talk to... I met with Totaro and he was basically just silent.

I met with four, now that I think of it, and that was me trying not to cause undue pressure. :P

Alex Wertheim

I get that sense with Totaro. Very serious about algebraic geometry, but maybe difficult to approach.

Mike Miller

Yes, he's not very talkative. His students apparently have the same experience with him. :P

Well, it's probably best to keep gossip to email.

Alex Wertheim

7:52 PM

Yes, of course. Sorry, I'll remove those.

Mike Miller

Haha, no worries.

Alex Wertheim

:) Did you feel like you got an accurate overall impression of the feel of the department during the visit weekend?

Mike Miller

Fairly, yes.

The free food they give you is NOT similar to the standard amount.

Alex Wertheim

Hehehe. Does UCLA have a Friday tea time sort of thing? I wasn't sure.

Mike Miller

Thursday post-colloquium snacks and coffee.

I was mortified that I missed it yesterday.

Alex Wertheim

7:59 PM

Man. I'm so psyched. :)

Mike Miller

When do you fly home on Thursday?

Alex Wertheim

A little before 4, so I'll probably have to leave around 1:15 or so.

Unfortunately, there are 0 direct flights, and with the time difference, I don't get back in until 11pm. Crazy.

Mike Miller

Ah, drat. I don't know if there's a colloquium but it would be far too late for you

Alex Wertheim

I figured. :( Are they usually more in the evening?

Mike Miller

Colloquium's Thurs, 3PM. But seminars in general never start before around 2.

There'll be an algebra seminar on wednesday...

Alex Wertheim

8:11 PM

Darn, I'll probably miss that, since Wednesday is basically full up.

Andrew Thompson

How would you read $u(\cdot ) v$ in the setting that there is a bilinear map $U^* \times V \to \text{Hom}_F (U, V)$ sending $(u, v) \mapsto u(\cdot) v$ ?

Mike Miller

It should say "$(u, v)$ maps to the linear map, $w \mapsto u(w)v$."

Alex Wertheim

I gotta run. Nice talking as always, @Mike.

Mike Miller

So $u(\cdot \;) v$ should be read as "the linear map $w \mapsto u(w)v$.

See ya!

Andrew Thompson

Ah, that makes sense. Thank you!

Mike Miller

8:14 PM

No worries.

infinitesimal

Does anyone remember how to derive the formula $2^{64}-1$ for the total number of grains of wheat on a chessboard?

Mike Miller

$1 + x + \dots + x^n = \frac{x^{n-1} - 1}{x-1}$. plug in $x=2$

infinitesimal

Thanks :-)

Axoren

Does anyone know if there's a definition for the largest possible ordinal?

Daniel Fischer

@Axoren There is no largest possible ordinal. There's always $\alpha + 1$.

infinitesimal

8:29 PM

in The Heap™ - Consultancy ©®, 3 hours ago, by Paul White

RIP Leonard Nimoy.

Tweeted by TheTweetOfGod on February 27, 2015 at 5:28 PM

:(

Axoren

@DanielFischer I guess I'm asking more for an upper bound on the size of ordinals, in the same way that infinity upper bounds the natural numbers.

Mike Miller

There isn't one.

Axoren

We can always take $n+1$, then why do we have infinity?

evinda

@DanielFischer Are these implicatios right? Or should we have $\Leftrightarrow$ instaed of $\Rightarrow$?
We notice that $f(n)=O(f(n)^2)$ iff $\exists c'>0, n_0' \in \mathbb{N} $ such that $\forall n \geq n_0'$: $f(n) \leq c f(n)^2 \overset{f(n) \neq 0}{ \Rightarrow } 1 \leq c f(n) \Rightarrow f(n) \geq \frac{1}{c}:=C \Rightarrow f(n) \in \Omega(1)$.

infinitesimal

Why do we have the infinitesimal @Axoren?

Daniel Fischer

8:32 PM

@infinitesimal For posting random remarks in chat, duh.

Axoren

@infinitesimal Because your mother and father fell into the throes of passion one fateful night.

infinitesimal

-_-

Axoren

But honestly, we can always say "Let $\alpha$ be a quantity greater than every ordinal."

But I was hoping there was a more meaningful definition.

infinitesimal

How do you define "quantity"?

Daniel Fischer

@Axoren We can say that, but it has no meaning. The ordinals are unbounded. If $\alpha$ is an ordinal, take a well-ordering of $2^\alpha$, and you have an ordinal with a greater cardinality than $\alpha$.

Axoren

8:37 PM

Consider the ordinal which is the limit of the following sequence: $\alpha, 2^{\alpha}, 2^{2^{\alpha}}, 2^{2^{2^{\alpha}}}, ...$

Suddenly, we can't do that anymore to get a larger ordinal

Daniel Fischer

@Axoren Call it $\beta$. Then look at $\beta, 2^\beta, 2^{2^\beta},\dotsc$

Repeat.

Axoren

Now you're starting to see where I'm going with this.

We take the limit of such a process

Mike Miller

And there's an ordinal bigger than it.

Axoren

$\varepsilon_0$ is the infinite tetration of $\omega$.

I need to look up the latex for the notation I'm about to use

But I think there's a process that will produce an ordinal for which we can't do that anymore.

Let's consider $\alpha = \lim_{n \to \infty} \omega \mathbin{\uparrow}^n \omega$

Actually, could we just say $\omega \mathbin{\uparrow}^\omega \omega$?

Julian Rachman

@Kaj I decided to use Pete's notes as a supplement to my learning of Topology. :)

Axoren

8:45 PM

Raising $\omega$ to that power suddenly makes it a smaller ordinal.

Actually, I think we still need to consider $\alpha = \lim \{ \omega \mathbin{\uparrow}^1 \omega, \omega \mathbin{\uparrow}^2 \omega, \omega \mathbin{\uparrow}^3 \omega, ... \}$

Mike Miller

If you have any ordinal, $2^\alpha$ is bigger. This is a theorem, it doesn't matter if you pick a wacky ordinal. If you have a limit that exists, then it's an ordinal, and every ordinal has another bigger than it.

Axoren

@MikeMiller However, we can clearly see that $2^\alpha$ is not larger. With our definition of $\alpha$, $\alpha = \alpha^\alpha > 2^\alpha$.

Daniel Fischer

9:04 PM

@Axoren No, it's not. The bleeping up-arrow notation means $\omega \uparrow^1 \omega = \omega^\omega$, and then $\omega \uparrow^{n+1} \omega = \omega^{\omega \uparrow^n \omega}$, right?

A Beautiful Mind

It seems that Chris is not in chat again, hmm.

Axoren

Ugh, then I'm lacking the proper notation for the case where it also changes the base.

For example: $x, x^x, {(x^x)^{(x^x)}}$

Daniel Fischer

@Axoren I'm asking what it means.

Mike Miller

@DanielFischer I have a question that I can claim is about distributions.

Daniel Fischer

@MikeMiller Would that claim be accurate?

Mike Miller

9:06 PM

Essentially.

Axoren

You called it bleeping (as I assumed an expletive) so I take it you're right.

I just checked, you're right.

Let's say there's some function $f(x) = x^x$. We take the limit of this sequence instead:

$\{ x, f(x), f(f(x)), f(f(f(x))), ...\}$

Mike Miller

I've got a differential operator $D$, and a compactly supported distribution $E$ with $DE = \delta - \omega$, $\omega$ Schwartz. Then I want to show that for all $k$, there exists an $\alpha$ such that $\mathcal F(D^\beta x^\alpha E)$ is $L^1$ for all $|\beta| \leq k$. Any ideas? I fiddled with it for some time to no avail. Surely the fact that $E$ is a parametrix is important, I just don't see how to abuse it.

Daniel Fischer

@Axoren Okay. Then you think that for the limit - call it $\lambda$ for a change - you have $f(\lambda) = \lambda$?

Axoren

Yes, that's what I'm claiming.

Daniel Fischer

@Axoren But if you start with $x = 2$, then all your steps are finite, hence your limit is $\omega$. However, $\omega^\omega > \omega$.

Axoren

9:11 PM

We wouldn't start with $2$, we'd start with $\omega$

Can we not do that?

Daniel Fischer

@Axoren That's just a small example to show that the reasoning doesn't go through. Just because $\lambda$ is the limit of $f^n(x)$ doesn't mean it's a fixed point of $f$.

"$f$ is not continuous"

Axoren

Let me think about this for a little longer.

Gato

Hi, how can I prove that $p(U\cap (E\times \{u'\})$ is open in $(E,d)$where $p$ is the natural projection and $U$ is open in $E\times E'$ and $u'\in E'$ ?

Daniel Fischer

@MikeMiller What kind of differential operator is $D$? Linear with constant coefficients?

Gato

I know that $U\cap (E\times \{u'\})$ is open in $E\times \{u'\}$

Mike Miller

9:16 PM

Not linear, but polynomial with constant coefficients, yes. Changing my notation, if $D_k = -i\partial_k$, we can write our operator $P(D) = \sum_{\alpha} a_\alpha D^\alpha$, where $a_\alpha \in \Bbb C$.

(This notation is convenience because $\widehat{P(D)} = P(\xi)$.)

It's elliptic, but I think that's only relevant as far as proving that there is a parametrix.

Daniel Fischer

@MikeMiller Still a linear operator. Which is nice, because of the interaction with the Fourier transform (among other things).

Mike Miller

Ah, misunderstood what you meant by linear.

Gato

@DanielFischer Can you look at my question please ?

Daniel Fischer

@Gato The shortest way is to observe that $e \mapsto (e,u')$ is an embedding of $E$ into $E\times E'$. But I guess you're expected to use the (concrete) definition of the product topology. So: What is "the canonical" basis of the product topology on $E\times E'$?

Gato

9:34 PM

@DanielFischer In fact I don't have any course on product topology, just a multivariable calculus course and I am reading some topology result to make sure I understood what I am doing. The canonical basis ? Perhaps it's the canonical projection $p(u,u')=u$ and $p'(u,u')=u' ?$

Daniel Fischer

@Gato Ah, not a topology course, I see. Do you know what a basis of a topology is?

Gato

No.. but I can prove that $e\mapsto (e,u')$ is an embedding

@DanielFischer

Daniel Fischer

@Gato How do you prove that?

Axoren

@DanielFischer So, I just need to show that $f^n$ is continuous, right?

Daniel Fischer

@MikeMiller By the way, I don't see anything either so far.

Axoren

9:38 PM

If I do, then it's limit exists and it's a fixed point.

Gato

If I denote $j_{u'}(v)=(v,u')$ is clearly an isometry witch is always an injection. @DanielFischer

Axoren

Well, not $f^n$, I guess I need the limit of $f^n$ first.

Mike Miller

@DanielFischer Here were the ideas I tried but got nothing with: consider instead $\mathcal F(D^\beta x^\alpha E)\mathcal F(P(D)E)$, or maybe approximating the $D^\beta$ by $P(D)$.

To no avail, I guess.

Daniel Fischer

@Gato Ah, metric spaces. Okay, nice, you have an isometry. Now you should show that (bijective) isometries map open sets to open sets, then you're done.

user159870

@DanielFischer I think that you could help me. math.stackexchange.com/questions/1168439/…

Gato

9:44 PM

@DanielFischer Oh okay $j_{u'}: E\rightarrow E\times \{u'\}$ is a surjection with inverse $p$ restrained to $E\times \{u'\}$. I will do the rest :-). Thanks

Daniel Fischer

@Axoren Since the class of ordinals is a proper class and not a set, we can't really speak of continuity directly. We can fudge that, however by pretending we restrict the domain to some ordinal and then the codomain to an appropriately large other ordinal. Then we can look at the continuity of these restrictions. And then you find out that $f(x) = x^x$ doesn't give you a continuous function.

Axoren

Are you already sure that it doesn't?

Daniel Fischer

@Axoren Yes. Because for $\alpha > 1$ we have $\alpha^\alpha \geqslant 2^\alpha > \alpha$. So $f$ has no fixed point other than $1$.

And if it were continuous, then it would have infinite fixed points.

Axoren

Even if we picked a fixed point, we couldn't pick the largest fixed point either.

So that doesn't get anywhere even if it worked.

Daniel Fischer

@user159870 The order relation is transitive, so when you sort the functions, you only need to check where in the sorted initial part the next function fits in. You need not compare each function to all others.

user159870

9:55 PM

You mean that we use $\log^a n< n^b < c^n, \forall a,b>0$ and $\forall c>1$?

Daniel Fischer

@user159870 Something like that. If you already have e.g. $1 < \log n < n^3 < e^n < n!$, and the next thing is $n\log n$, you don't need to compare that to $e^n$ and $n!$ after you compared it to $n^3$ and found $n\log n < n^3$.

Mike Miller

10:12 PM

hello karl

evinda

@DanielFischer I want to prove or disprove the statement $f(n)+o(f(n))=\Theta(f(n))$.
Do I have to set $g(n)=o(f(n))$ and use the definition, that is that $\forall c>0, \exists n_0(c) \in \mathbb{N}$ such that $\forall n \geq n_0: g(n)<cf(n)$ ?

Karl Kronenfeld

@MikeMiller sup

user159870

I saw that $1$ and $n^{\frac{1}{\lg n}}$ are at the same equivalence class, so $1$ is in $\Theta (n^{\frac{1}{\lg n}})$. How do we conclude to that? Is it obvious that it is so? $n^{\frac{1}{\lg n}}$ doesn't belong to the class of constants. Des it belong to the class of logarithms? Or is there an other way to think to check these two specific functions? @DanielFischer

Daniel Fischer

10:28 PM

@user159870 $\operatorname{lg}$ is the base-$2$ logarithm? Write $n$ as a power of $2$.

evinda

Hey @Axoren!!! Do you maybe have an idea about the following?
I want to prove or disprove the statement $f(n)+o(f(n))=\Theta(f(n))$.
Do I have to set $g(n)=o(f(n))$ and use the definition, that is that $\forall c>0, \exists n_0(c) \in \mathbb{N}$ such that $\forall n \geq n_0: g(n)<cf(n)$ ?

user159870

$\operatorname{lg}$ is the base-$2$ logarithm? : Yes.

$n=2^{\lg n}$ Correct? How does this help? @DanielFischer

Daniel Fischer

@user159870 Correct. So $$n^{\frac{1}{\operatorname{lg} n}} = \Bigl(2^{\operatorname{lg} n}\Bigr)^{\frac{1}{\operatorname{lg} n}}.$$

user159870

So it is an exponential? @DanielFischer

At the order $\log^a n< n^b < c^n$ the constans are the smallest and the exponentials the largest.

I am mixed up.

Daniel Fischer

@user159870 No, not an exponential. Look and simplify.

user159870

10:37 PM

OOhh. It is $2$. Correct?

@DanielFischer

Daniel Fischer

Yes.

user159870

10:48 PM

At the beginning of the exercise, what do I have to do? Do I have to clasify the functions in constants, logarithms, polynomials, exponentials?

@DanielFischer

Axoren

Hey, @evinda Yeah, that's actually fairly simple. Your solution will work.

evinda

@Axoren So don't we have to pick a specific c?

Axoren

But you can use the fact that they're the same function to show that $f(n) + cf(n) = (c + 1)f(n)$ for $n > n_0$

At that point, it should be simple to see that it's tightly bounded.

evinda

@Axoren Since at little-o the equality doesn't hold, don't we show it like that for O(f(n)) and not for o(f(n)) ? Or am I wrong?

Axoren

Wait, I thought that was a typo of Big-O

Daniel Fischer

10:52 PM

@user159870 You can do that, and then sort each group separately. Or just take each function from the supplied list, and insert it into the sorted part at the right place.

Axoren

What is little-o?

evinda

If $g(n)=o(f(n))$ then $\exists c>0, n_0(c) \in \mathbb{N}$ such that $\forall n \geq n_0$: $0 \leq g(n)<cf(n)$ @Axoren

user159870

When some functions contain for example both logarithms and exponentials? What do we do in this case? @DanielFischer

Daniel Fischer

@user159870 in the coarse classification, they'd belong among the exponentials. You have for example $$2^n < \frac{e^n}{\log n} < e^n < e^n\log n < 3^n.$$

user159870

What can I say at the beginning about these functions when I want to make the classification? @DanielFischer

Transcript for

$Mathematics$ Mathematics