As long as it's not the b****ing exercise I think you're fine

@Mike What word is that?

Bitching.

I guess.

Mike seems to be careful about matching asterisks.

@Mike

Dude.

I've been pawndering.

Pawndering?

Pawndering.

Consider the category of pointed sets.

Mhm?

The point of the coproduct is to create another (relatively small?) pointed set with unique copies of the given sets, while the product produces a large pointed set with many copies of the sets inside.

For example, the coproduct of two pointed circles is a figure eight with the point at the middle, while the product is a torus with a point $(x_0,y_0)$ on its surface given by the location of the original points in the circles?

Yep.

@Mike Yet say for abelian groups, coproducts and products are the same? For the finite case.

Yep.

For the infinite case the coproduct is the direct sum, and the product is the direct product yes?

Yeah

But you can perhaps think of that as still being "The smallest abelian group that contains both the others"

You're monosyllabous today.

Mhm had two syllables.

So the coproduct in pointed spaces is "The smallest pointed space that contains the others"

Yes.

This is a naive way of doing it since this probably won't work for every category, especially not bizarre unfamiliar ones. But it works for ones I can think of.

Dude.

Go on.

Let me explain the solution to that counting problem.

The sine one? Bah.

Yeah.

Don't wuss out.

It reads like one of Chris's Sis's things.

Heart is broken.

I'm a heartbreaker.

Well, I'll shut up now.

No, feel free to explain it.

I'll listen.

Hmm....

OK.

So, recall we want to count integer solutions to $-n\leqslant x,y,z \leqslant n$ and $|x+y+z|\leqslant s$.

The observation is that this is the same as counting solutions to $k_1+k_2+k_3=i$, with $|k_j|\leqslant n$ and $|j|\leqslant s$.

Agreed?

Well, now remember $$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{ikt}dt=[k=0]$$

Thus, $$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{(k_1+k_2+k_3+j)it}dt=[k_1+k_2+k_3+j=0]$$

Now we sum through $|k_l|\leqslant n$, $l=1,2,3$ and $|j|\leqslant s$.

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\sum_{|k|\leqslant n}e^{kit}\right)^3\sum_{|j|\leqslant s}e^{ijt}dt=\#\text{ solutions to }k_1+k_2+k_3+j=0\text{ with } |k_l|\leqslant n,|j|\leqslant s$$

@Mike Pretty neat, right?

That's a lot more elegant than I thoguht it would be.

=)

Wait, $i$ here isn't the imaginary unit?

Yes, that's why I changed to $j$.

Oh, ok.

Got it.

Well, and $$D_n(t)=\sum_{|k|\leqslant n}e^{kit}=\frac{\sin \frac{2n+1}2 t}{\sin \frac t 2}$$

I could guess that there was some transformation from the summations to sines

Well, that is the famous Dirichlet kernel.

I'll admit that it's not famous to me.

So finally. $$\frac{1}{{2\pi }}\int_{ - \pi }^\pi {{{\left( {\frac{{\sin \left( {n + \frac{1}{2}} \right)t}}{{\sin \frac{t}{2}}}} \right)}^3}} \frac{{\sin \left( {s + \frac{1}{2}} \right)t}}{{\sin \frac{t}{2}}}dt = \left\{ {\left( {x,y,z} \right) \in {{\left[ { - n,n} \right]}^3} \cap {{\Bbb Z}^3}:\left| {x + y + z} \right| \leqslant s} \right\}$$

@Mike Well, slightly famous then.

The point is $S_n(f)=D_n\star f$

Where the LHS is the $n$-th Fourier sum for $f$.

Yah.

So it allows us to write a Fouirer partial sum as a singular integral.

@Mike I sent you an email.

I responded in kind.

Did you do the effing exercise?

Yes, but it is after the section I was studying.

So I was going to finish it and then look at it.

Doesn't look to crazy, though.

Or is it?

@AlexYoucis I liked your examples of morphisms post.

@Mike Hey, thanks haha

They're all standard.

@AlexYoucis Hey dude.

Couldn't comment on that.

I wanted to give you +250 on an answer you posted. 'twas spectacular.

@PedroTamaroff Thanks man. It really wasn't that great. Georges should have, and does, get more points or whatever. His answer is really nice.

I think Pedro's talking about something different.

@AlexYoucis I was talking about this.

You've been posting long answers lately!

@Ethan How did it go=?

@PedroTamaroff Oh, right, haha. Yeah, that took a loooooong time to post. The server gets really buggy when your post starts to get long since it recompiles every like second or so

it was fine lol

It's bad.

@AlexYoucis Yes, careful.

Sometimes you might lose and edit.

model theory... fun

@PedroTamaroff Yeah, I had it saved in notepad while I was typing it haha

That's how I type pretty much everything... even when they're not long, it gets obnoxious.

yea it takes a long time to scroll up and make sure things are formatted correctly

after you type it in

if your answer is long

@AlexYoucis So, I am back to studying algebra.

@PedroTamaroff Good. It's important for everything (kind of) haha

@Mike has convinced me of reading Hungerford, which has a little on categories.

I am at the moment reading about groups... studying hopefully more advanced stuff, getting to know things better.

@PedroTamaroff Hungerford is good--can be a little dry at sometimes, but most algebra books can.

At the moment I was wondering why abelianeness is important for coproducts to exist.

@PedroTamaroff Coproducts exist in Grp.

They're just not what you're thinking of.

OK?

@PedroTamaroff He's right, they are what's called the "free product

I mean why the coproduct in $\bf Ab$ doesn't extend nicely.

@AlexYoucis Ah, I know what a free product is! =)

@PedroTamaroff Well then, there you go. That is the coproduct in Grp.

@Pedro And there's a counterexample to the 'the coproduct picks the smallest object that contains both' thing. The free product is, of course, gigantic.

It's also a nice example of how large of a discrepancy there can be between colimits between a category and a (full) subcategory.

@Mike What's your background? :)

Thinner than I'd like

@Mike Hehe, I see.

Haha, I meant what are you? A student, etc.

Oh, yeah. Student.

In my last year of undergrad.

@Mike Nice! Where at, if you don't mind me asking.

Not far from you, Santa Clara.

@AlexYoucis A brief problem from Polya and Szego solved some minues ago. Quick, it's still fresh!

Haha, I have no idea where Santa Clara is. I'm not from CA.

Y'know where San Jose is?

@Mike Rougly. It's between Berkeley and Palo Alto, no?

Or is it a bit farther south...

This is embarrassing :)

Little bit further south. Santa Clara is basically at San Jose

Haha, I won't push then :P

Haha, no problem! What's in Santa Clara? Do I know your school?

You don't. Santa Clara... University. :P

@Mike Very cool. What are you interested in? Thinking about going to grad school?

Number theory, algebraic geometry, their intersection. And yeah, I'm waiting on results from applications.

@Mike Apply to Berkeley? :)

PS, we have the same interests

This is how I feel when I try to solve @Chris'ssis problems

I did, but you guys haven't gotten back to me, so I don't think that's where I'll end up :P

@Mike Hear back from anyone?

Also, that's cool. I thought you were a straight geometry guy.

Yeah, I got an offer from LA. That's probably where I'll end up, barring an unexpected result elsewhere.

@Mike No, no. Well, kind of. I like algebraic geometry on arithmetic things. I don't care about primes.

@Alizter looool

@Mike My first acceptance was also LA ;)

@Mike Were you thinking about Hida?

@Alex And Khare

@Mike Khare spends a lot of his time in India, form what I've heard.

Have you visited yet?

My visit day is in mid-March.

Nice, nice. I know some people who go to UCLA, good guys.

(So no.)

You defintely have to talk to Merkeurjev

@Chris'ssis You know that Gamma one

I can never remember if I spell that right..

@Alex Who do you hope to work with at Berkeley, if I might ask?

@Mike Right now the most likely choice is Martin Olsson. Although, we made some recent hires who are very appealing.

I am also doing a reading course with Xinyi Yuan.

$\lim_{n\to\infty}1/n (\Gamma(1/1) \Gamma(1/2)...\Gamma(1/n))^{1/n}$

@Alizter aha. Are you done with it?

You must explain how this even qulifies as doable

I've heard he's a wonderful adviser (Olsson).

@Mike Yeah, Olsson is awesome. I love him. I'm doing a reading course with him right now and it's great.

he's one of them.

It might not be, I nuked the message.

We also just had Bhatt here. I don't know if he'll come though.

@Mike How did you know then?

I emailed Ribet a while ago about people doing arithmetic at Berkeley, since a friend of mine said he's very willing to talk to students.

@Alizter I remember I said that one is pretty straightforward. Did I say something else?

@Mike Yeah, Ribet is the best.

@Mike He like lends me books and stuff

That's awesome, haha

@Mike Yeah, very, very cool dude.

Do you want to see something great?

Always.

"I'm still sparkling about that" hahaha

Haha, he's so great.

Everything is scrumptuous or not

@Chris'ssis You must show me how I am desperate to know. I tried everything I know (which isn't relatively much but you get the point).

@AlexYoucis I'd like to keep chatting but I don't want to flood the room - do you mind emailing?

@Alizter all you need is a bit of imagination. Try to see what happens beyond those figures, how they can be seen. When you look at such a limit you don't have to see a static image, but a dynamic image, you need to see things turning in alternative forms that you can manipulate the way you want to.

@Mike Yeah man, or if you have Gchat?

I do, lemme dig up my google address

@Alizter I give you a hint - how could you approximate gamma when $x$ tends to $0$?

Lemme know when I can delete that

i do not understand the extended riemann hypothese I think ...

can someone help .

?

why do you guys get silent when im around ? :/

sigh. goodnight.

@Chris'ssis Well I want to say the answer is 0. And I said this before

@Alizter Use d'Alembert criterion.

my question does not get much activity.

@Chris'ssis I do not understand. Is that just not a convergence test?

@Alizter Let $(w_n)_n$ be a sequence of positive numbers , and $\displaystyle \lim_{n\to\infty} \frac{w_{n+1}}{w_{n}}=l$, finite or infinite, then $\lim_{n\to\infty} \sqrt[n]{w_n}=l$.

I did not know that :O

@Alizter Well, you always have alternatives to that (type of) limit.

@Chris'ssis I wouldn't call that D'Alambert's criterion.

@PedroTamaroff I think its full name is Cauchy-d'Alembert criterion. (actually, I'm sure 100%)

Geez, I created a lot of stuff today ... but now I'm far too tired to keep one eye open. I'm out.

@Mike @anon This is clear yes?

Looks fine to me

@TedShifirin will like it. He likes symmetry.