@Chris'ssis you derived them ? :D

It's a non-standard way of visualizing Pascal's triangle, so that adds confusion.

@r9m :D

@TedShifrin I made it from the picture i keep referring to, that's why

@Hippalectryon, right now we are looking at a 'discrete' function. I.e. $f:\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$. It's possible to turn this into a more 'continuous' function since binomial coefficients generalize to take on real-valued arguments. See here: en.wikipedia.org/wiki/…

So the intensity is a measure of $\binom{x-y}{x}/\max$.

Then you logarithmatized the $x$-axis?

We don't need that, @Kaj. Let's stick to integers and use Stirling's formula.

Point being, it may or may not make it easier to explore this curve with this perspective.

Let's figure out the function and then approximate it by Stirling ..

I just want to understand the discrete function :P

Same

My formula can't be right. Agh.

@Chris'ssis Okay .. great !! :D .. now I'm gonna watch some gangster movie and go to sleep .. gdn8 :-)

@r9m I also go to sleep. I wanna see the new 300 (movie) ... Is it nice?

@Chris'ssis Nice ?! .. its bloody action packed imo :D .. and I like eva green's character in that movie :)

@r9m :-))) I need to watch it.

Ah, it should be $\binom{y-x}x$.

No, still wrong. GRR. I need dinner.

I give up for now.

Want me to star that ? xD

Re: Stirling's formula....there's another case where $\pi$ comes up without obvious explanation. What on earth does the ratio of a circle's circumference to its diameter have to do with $n!$?

@KajHansen This is getting interesting :)

It certainly is @Hippalectryon. I just wish I knew more math :P

Me too :)

anyone know an efficient algorithm to determine which of n sectors a vector is? the most obvious case being 4 sectors, so simply the quadrants of a circle, but preferably for any number of sectors

I have the feeling there is a very elegant trigonometry trick I'm missing here ;-)

Interesting question @hugovdberg. Let me think for a second.

but all solutions I can find so far are for the quadrant case, and simply based on checking for the signs of the coordinates

Here's something that's definitely faster than that:

Say you divide the Cartesian plane into $n$ sectors.

You can think about then dividing the graph of $\cos(x)$ into $n$ equal pieces on the interval $[0, 2\pi)$.

Nope, that's no good. Blah.

@KajHansen prntscr.com/3v7227 I see no link at all :/

@Hippalectryon, careful to restrict your bounds to something useful. $\binom{x}{y}$ makes sense in this context only if $x>y$.

@KajHansen Yeah, just pay attention to the continous half of the graph

Not super important, but we're looking at some junk.

@KajHansen But idk how to remove the other one with mathematica

Still thinking @hugovdberg.

@KajHansen that's my problem as well :P I'm also playing around with the atan2 function in matlab, that's at least in a domain between -pi:pi, but for odd cases of n there is a problem around the -pi/pi

@DanielFischer

Ok, this might work @hugovdberg:

@PedroTamaroff ?

Divide the cartesian plane into $n$ sectors. Then think about dividing the graphs of $\sin(x)$ and $\cos(x)$ into those same sectors.

Now, given an arbitrary vector in the plane, we can compute the cosine of the angle it makes with the $x$ axis by taking the dot product of the vector given with the vector $<1, 0>$ and dividing off by the magnitude of the given vector, if it isn't unit.

Hmmm...that restricts things to two possible sectors.

Since cos(x) will take on the same value twice for any given value on the interval.

Which will narrow the possibilities down to two sectors.

@DanielFischer I was trying to prove the closed graph theorem.

From there, you'll have to check what quadrant the sector is in.

Yuck. There must be an easier way.

One part is easy, the other I reckon follows from OMT.

@PedroTamaroff That's the easy way. One can also prove it (at least for Banach and Fréchet spaces) without using the OMT, but that's not so nice.

@DanielFischer Should I check some known book for a proof?

@PedroTamaroff You should be able to figure out the proof of the CGT from the OMT yourself, you're smart. Give yourself some time and think about it.

@DanielFischer OK. Thanks.

@KajHansen

Hey there Pedro

@KajHansen How is it going?

Not too bad. Chat's been discussing this question for a while: math.stackexchange.com/questions/843031/…

Oh wait, I see you've been here for a while. I guess I'm kind of out of it.

@KajHansen Heh, at any rate I didn't pay much attention to it.

LOL, I think they spend more time trying to think of what to call people in word problems then they do about the actual math: math.stackexchange.com/questions/843032/…

Oh interesting. It turns out that Ehrenfest was a theoretical physicist.

@KajHansen I was watching the Ramsey Theory videos.

haha, hopefully I didn't bore you to death. The first couple are kind of bad until I got the hang of being in front of the camera and whatnot.

@KajHansen Heh, I see.

@KajHansen I hoped so as well, but for now I can't think of anything better either

Did you think about the matrices problem?

I thought about it, but no luck.

Not since that one day. Like I said, I have definitely proven the existence of such an $n$, but I didn't use Ramsey numbers nor have I been able to establish any sort of upper bound given a value for $m$.

@Kaj: The $\sqrt \pi$ comes from $\int e^{-x^2}dx$ :)

@TedShifrin, I was thinking about that too when I wrote that comment. Weird stuff huh?

I'm out now :) see you

Unrelated to $\pi$, but more black magic: $\displaystyle e = lim_{n \rightarrow \infty} \sqrt[n]{lcm(1, 2, 3, ..., n)}$

I realized whilst cooking dinner that to make @Hippa's stuff work we need $N$, with $0\le x,y\le N$ and $\binom{N-y}x$. :)

What is $N$ in this case?

@DanielFischer I saw the solution to the $\displaystyle \lim\limits_{n \to \infty} \sqrt[n]{lcm(1, 2, 3, ..., n)}$ .. :D AWESOME !! (once again I must insist ... plz consider about writing a book :D ) .. and also .. please lemme know if there is any progress on $\binom{3m}{3m}\pmod{3^m}$ :-)

The number of rows, @Kaj.

@r9m Typo there.

@PedroTamaroff where ?

ugh ... I see it :}

@r9m Thanks. I have to admit, I haven't yet really looked. But now my conscience is bad enough that I'll take a couple of sheets of paper to scribble on tomorrow [well, today, it's 2 a.m. here, after I've slept I mean].