Mathematics

I just discovered an old friend from almost 20 years ago, @Mike, who now lives up in the hills overlooking Stanford -- he's a nurse in San Mateo County

Mike Miller

12:51 AM

How'd you discover him?

Ted Shifrin

I ran across some things that reminded me, googled, had no luck on Facebook, and finally Facebooked a friend of his in Atlanta (whom I found by googling) and asked for his email :P

I'm a better stalker than I'd realized.

Anyhow, I'll go visit when I drive up to the Bay Area ...

Mike Miller

Well, if you wanted to be a really good stalker, you'd visit without emailing

ʙᴀᴅ ᴀᴛ ᴍᴀᴛʜ

lol

Ted Shifrin

no, I wouldn't know the address

Mike Miller

I did say you'd have to be really good :)

Ted Shifrin

12:53 AM

I'm not that good.

Alex Wertheim

1:09 AM

Hello friends.

ʙᴀᴅ ᴀᴛ ᴍᴀᴛʜ

Hi @AlexWertheim

Alex Wertheim

Hello @ʙᴀᴅᴀᴛᴍᴀᴛʜ. I don't think we've met before, unless you have another name.

I almost mistook you for Mike. =P

ʙᴀᴅ ᴀᴛ ᴍᴀᴛʜ

@AlexWertheim Sorry my avatar keeps changing on its own randomly to different colors/designs, don't know why

Alex Wertheim

No need to apologize. I more thought it was funny. :)

ʙᴀᴅ ᴀᴛ ᴍᴀᴛʜ

I wish I could be like Mike, though

David Wheeler

1:21 AM

I like my avatar because it has $D_4$ symmetry

ʙᴀᴅ ᴀᴛ ᴍᴀᴛʜ

1:32 AM

Don't all of the default avatars have that property?

user112495

I'm considering the set $\mathbb{R}$ with the zariski topology. How do I determine the limit points of the sequence $x_n=n$ under this topology?

@DavidWheeler Would you be able to help me with this?

David Wheeler

What is your thinking on it?

Pick a real number, say, $y$, what does a deleted neighborhood of $y$ look like?

user112495

1:50 AM

@DavidWheeler My current thinking is that any number in $\mathbb{R}$ is a limit point. This is because the closure of $ \{1, 2, \ldots \}-\{x\}$ is $ \mathbb{R}$

David Wheeler

What is your definition of "limit point"?

What I am trying to get to is this: the number of points NOT in any proper (not all of $\Bbb R$) ZT-open set $U$ is finite. So the number of points has a maximum. Pick an integer $k$ such that $k > u$ for all $u \in \Bbb R - U$. Then $U$ contains all integers greater than $k$.

user112495

@DavidWheeler A point $x \in X$ is a limit point of $S$ if every neighbourhood of x contains at least one point of $S$ not equal to $x$

David Wheeler

Now the $U$ I mentioned above contains INFINITELY many $x_n$ (all the ones greater than $k$), and at most, only ONE of those could be $y$, even if $y$ happened to be an integer.

So $U -\{y\}$ contains not only "some" points of $\{x_n\}$, but infinitely many.

Pedro Tamaroff

@user112495 That's wrong.

The closure of that space is that space.

Perhaps you mean that the closure of $\Bbb R-\{x\}$ is $\{x\}$.

user112495

@PedroTamaroff Why is the closure of that space that space? I thought only finite subsets and the set ($\mathbb{R}$) itself were closed, whereas the set above is countably infinite?

Pedro Tamaroff

2:06 AM

@user112495 Ah, I didn't know that you had changed the topology on $\Bbb R$. =)

You're taking the cofinite topology.

user112495

@PedroTamaroff Is that the same as the zariski topology?

David Wheeler

@PedroTamaroff ya, actually the ZT on $\Bbb R$, same diff.

Pedro Tamaroff

Hold up.

David Wheeler

it's the co-finite, as you can easily see, each open set is the complement of the set of zeroes of a polynomial.

Committing to a challenge

What is $\operatorname{Arg } z$?

Pedro Tamaroff

2:08 AM

@DavidWheeler You're thinking about $\Bbb R[X]$, then.

David Wheeler

@Committingtoachallenge "argument" function (the "theta" in $z = re^{i\theta}$)

Committing to a challenge

Thank you.

Pedro Tamaroff

That space's prime spectrum doesn't correspond to the elements of $\Bbb R$, though.

David Wheeler

personal.psu.edu/axk29/TOPOLOGY/Chapter1.pdf page 9

Pedro Tamaroff

@DavidWheeler Oh, I had never encountered the cofinite topology being named Zariski topology.

David Wheeler

2:12 AM

see my above remark

it's the "classical" definition, not the more modern approach.

yes

Pedro Tamaroff

Oops.

I wanted to edit.

Committing to a challenge

How many named(or otherwise well known) conformal mappings(??) are there? Such as the Joukowsky transform?

A very large amount?

Bib

2:34 AM

any ideas on how to find a nontrivial irreducible representation of $\mathbb{Z}$ which factors through a finite quotient and weakly contains the trivial representation?

David Wheeler

@PedroTamaroff This may be a duplicate: math.stackexchange.com/questions/1172701/…

Committing to a challenge

This is a trivial question(for most of you surely, but not for me), but what should I expect to be looking at with something like bounding $\{\frac{\pi}{2}\lt \operatorname{Arg } z \leq \frac{3\pi}{4}\}$

Pedro Tamaroff

@Committingtoachallenge Do you know about the representation of a complex number as $re^{i\theta}$, $0\leqslant \theta<2\pi $?

Then $\arg (re^{i\theta})=\theta$.

Your set is a wedge (like an infinite pizza slice) bounded by two halflines from the origin, with the respective angles.

Committing to a challenge

@PedroTamaroff I actually haven't seen this before, what is the name of this so I can read the wiki page

Pedro Tamaroff

@Committingtoachallenge Polar coordinates.

Committing to a challenge

2:41 AM

Ahhh one of those things I neglected in calculus :\

Polar/cylindrical/spherical were all surface learnt as an engineering student, and now I suffer

Can all complex numbers be converted to such a form?

Bib

Yes :)

David Wheeler

The important thing about identifying a point in the plane is it takes two pieces of information to do so-"cartesian coordinates" use up/down and left/right, whereas polar coordinates use distance and direction.

Bib

$r$ is the modulus and $\theta$ is the angle from the principal axis. Use trigonometry from high school to get $\theta$.

David Wheeler

One often sees "cis" form: $z = r(\cos\theta + i\sin\theta)$

Committing to a challenge

Fair enough, so we want $d(p,0)$ and angle, is there any connection with metric spaces here?

David Wheeler

2:48 AM

So, for example $1+i = \sqrt{2}(\cos(\pi/4) + i\sin(\pi/4))$

Bib

Well, $d$ is a metric...

$\mathbb{C}$ is definitely a metric space, no?

David Wheeler

so $|1+i| = \sqrt{2}$ and $\text{Arg}(1+i) = \pi/4$

there's some "tricky rules" for Arg, but basically you want with $z = x + iy$, $\arctan(y/x)$ with some rules to get the proper sign, and the special cases where $x = 0$

Committing to a challenge

@Bib True true

David Wheeler

One has to be careful, arg isn't continuous (for hopefully obvious reasons)

Committing to a challenge

@DavidWheeler Thank you, that all makes sense

David Wheeler

2:56 AM

Complex analysis is very "pretty" there's a certain sense of it being the pinnacle of analysis.

Bib

@SanathDevalapurkar how does one end up knowing as much as you do in high school?

I hope you don't mind me asking

David Wheeler

@Committingtoachallenge And yes, $\Bbb C$ is most definitely a metric space, and the usual topology is the metric topology. In fact, $d(z,w) = |z - w|$, so $d(z,0) = |z|$.

David Wheeler

3:12 AM

@PedroTamaroff Thanks

user105491

@Bib Hi @Bib I wouldn't say I know too much math.

Bib

Perhaps not, but I'm still super curious as to how someone works up to it without at least a university degree. many things at the higher levels such as things you do require a lot of background to process with.

user105491

I guess I just got really interested in math at a young age, so I had time (namely, that I wasn't bogged down with school) to study math.

user105491

It's also coupled with some sense of imagination, I guess.

user105491

Let me try to be more precise.

user105491

3:25 AM

In ordinary university schooling, I think that students are given the conventional picture of concepts. When I learn things by myself, I create a simpler (but more restricted) picture to help understand things better.

Bib

I think I understand. Surely you had some guidance before high school though?

user105491

But all in all, I just learnt it for fun, because as a kid with honestly nothing to do (school was, as I said, barely any work), I though math would be fun.

user105491

@Bib Not exactly. I only got my mentor at UCLA last year. But of course, my way of learning has a lot of disadvantages.

Bib

That's pretty fascinating...

user105491

For example, I found that some (thankfully not many) of the pictures I painted for myself were incorrect. I ended up having to relearn a few concepts.

user105491

3:28 AM

I also learnt that I had skipped a few essential things (I didn't really know too much complex analysis, for example, and I only studied that last year).

user105491

So while I may know a few things in math, there still may be a lot of places in which I could be deficient.

Bib

Nonetheless, I find it remarkable how you've been able to wade through some very complicated mathematics, no matter the nature of your understanding

user105491

Thanks; of course, I'm understanding concepts in stable homotopy a lot more after I met my advisor. :-)

user105491

@Bib Could I ask what year of undergrad you're in?

Bib

I'm in third year right now.

user105491

3:35 AM

Ah, OK. As you might have deduced, I'm not too familiar with what topics are covered when in the course of an undergrad study. Could you outline/briefly sketch the topics covered in each year?

Bib

I come from a Canadian university, so things might be different in the states, especially considering that my university is not as renowned as U of T

in first year, we see calculus and an intro to analysis, as well as the basic linear algebra

user105491

OK

Bib

in second year, we see abstract algebra, more calculus, some analysis in $\mathbb{R}$, differential equations, discrete mathematics, probability and stats, linear algebra, and logic

user105491

OK. Can I ask a question?

Bib

third year is where we start seeing analysis on metric spaces, linear algebra, and other topics

go ahead

user105491

3:38 AM

What topics in logic are covered?

user105491

I know stuff only up to forcing @Bib.

user105491

Elementary forcing, as introduced by Cohen.

Bib

in second year logic, we see set theory, predicate logic, truth tables... basically it acquaints us with ZFC

user105491

Oh, ok.

user105491

Thanks.

user105491

3:43 AM

What do you learn in the third year @Bib?

Bib

third year is where you get to start picking courses you're interested in

so real analysis, complex analysis, linear algebra, ring theory are necessary to take in third year

but you get to see topology, dynamical systems, etc.

junior and senior year seem blurred together in a sense: you're expected to take many third year courses in senior year

user105491

@Bib That's cool. It's kind of like the UK education system (I have a friend there).

Bib

do you mean the UK system in universities?

user105491

Yes.

Bib

I have no idea how the Tripos works for example

user105491

3:49 AM

This is a good introduction, as of course, is the Cambridge website.

user105491

Apparently the http:// is needed for getting the link to work.

user105491

Hi @PedroTamaroff.

Pedro Tamaroff

Hello.

user105491

How are the moderator duties coming along?

Bib

it does seem somewhat similar

user105491

3:51 AM

@Bib Yeah, it does. My friend's studying at a different university, though. I think the only difference is that the Tripos is faster-paced than regular courses.

user105491

Anyway, I have to leave.

user105491

Nice talking to you, @Bib!

Bib

Take care!

Bib

4:10 AM

Is anyone here versed in representation theory?

user105491

@Bib I guess I know some representation theory.

Bib

Perhaps you can help with my question chat.stackexchange.com/transcript/message/20338388#20338388

user105491

@Bib You're working with complex representations, right?

Bib

exactly

user105491

All the irreducible representations of $\mathbf{Z}$ are one-dimensional, right?

Bib

4:19 AM

correct

user105491

I'll talk to you later, my parents are calling me for dinner.

user105491

Bye

Bib

Bye

Committing to a challenge

So are modules over a field just a fancy way of saying vector space over a field?

David Wheeler

yes

Committing to a challenge

4:31 AM

Is there any difference between the two? I imagine modules can be taken away from the field or something(based on otherwise language redundancy)

Or perhaps vector spaces are just nicer to learn than modules

Alex Wertheim

Modules are in some sense a "generalized" notion of a vector space.

They're structures over rings. Vector spaces are explicitly over fields.

David Wheeler

Some rings are "bad" in that they have undesirable algebraic features

this limits how "easy" it is to solve problems in that structure

Alex Wertheim

So modules encapsulate the notion of a vector space. Without some of the niceties, but the added generality allows you to solve some other problems.

David Wheeler

but certain basic things are true, even with "bad" rings, and algebraists like to be as general as possible until forced to be specific

Committing to a challenge

Fair enough

David Wheeler

4:36 AM

with certain kinds of structures, mappings from an example of that to structure to itself can often be given "more structure"

Alex Wertheim

In general, linear algebra is very nice. Not as nice when you get away from fields, which endow vector spaces with some remarkable properties. But nice nevertheless. One thing modules allow us to do is to make linear algebraic computations over more more generalized algebraic objects.

David Wheeler

for example, given a set, you can have the monoid of transformations, or the group of bijections

this allows us to create an action of the "more structure" on the "less structure"

with modules, the "less structure" is an abelian group, which has a natural ring associated with it, its ring of endomorphisms

Committing to a challenge

I don't understand anything you have said after "with certain kinds of structures...", so give me a minute to work this out

Mike Miller

the point being that you can do linear algebra over, say, $\Bbb Z$, and lots of results will still carry over

they'll be a tad more technical, in general, but many things still just work

Committing to a challenge

@MikeMiller $\Bbb Z$? Using modules(since this doesn't work on V.S.?)

Mike Miller

4:40 AM

yes

David Wheeler

linear algebra over $\Bbb Z$ is essentially just studying an abelian group

Committing to a challenge

For addition

Fair enough

David Wheeler

no, even considering the scalar multiplication

Every abelian group is a $\Bbb Z$-module, and vice versa

Alex Wertheim

@David: probably a bit much, a bit fast for someone who is just learning what a module is.

Maybe not, though. :)

Committing to a challenge

Not 'maybe not' it would seem :P

I think the way you are talking about it is a little too casual for my level

David Wheeler

4:43 AM

The "scalar multiplication" on an abelian group is simple, though: $k\cdot a \stackrel{\text{def}}{=} a + a +\cdots + a\ (k\ \text{times})$

Committing to a challenge

I have never seen things phrased like " given a set, you can have the monoid of transformations, or the group of bijections"

David Wheeler

Ok, take ANY set, $X$

Alex Wertheim

Btw, howdy @Mike.

Mike Miller

hello!

David Wheeler

the set of bijections on $X$, forms a group under composition

In a certain sense, ALL groups arise this way

Committing to a challenge

4:45 AM

I will think about what you have said. I have to leave now(due to a lecture in 15 minutes across the uni), thanks for your help and information

David Wheeler

It's a theorem, called "Cayley's Theorem"

2

Committing to a challenge

I'll look it up when I finish thanks

Thanks @Alex and @Mike aswell

Alex Wertheim

Certainly @committing, hope what I said was useful at all. Really admire what you're doing, btw. Keep at it!

badskjapanses

6:45 AM

Let $f$ be a continous nonnegative function on $[0,1]$, show that: $$\int_{0}^{1}f^3(x)dx\ge 4\left(\int_{0}^{1}x^2f(x)dx\right)\left(\int_{0}^{1}xf^2(x)dx\right)$$

Transcript for

$Mathematics$ Mathematics