@TheGreatDuck This sounds interesting... please continue (despite my saracasm)

Pascal's triangle lets you calculate $(1)+(1+2)+(1+2+3)+\dotsb+(1+2+\dotsb+n)$ pretty easily

yay triangular numbers

triangles 101

Combinatorics says that $\sum\sum\sum\dots\sum1$ is pretty easy to calculate as well

Apologies for potato quality

@AkivaWeinberger Love that pic!

lol

Blue is triangular numbers, red is sum of blue

if you pick the points at the other ends of the lines, then you have picked the farthest most point. However, rays can be infinitely extended. Therefore, you can extend the point further out. Therefore, it is not the end of the triangle. Therefore, the ends of pascals triangle are not points. Therefore, the bottom of pascals triangle can never be constructed. Therefore, pascals triangle is not a polygon.

Now can we stop calling it pascals triangle?

No.

^

Any finite truncation of it is a triangle. That's good enough for me.

@AkivaWeinberger Where'd you get the pic? Is there a bigger version of it i can download?

@Jeff I made it a while ago. I found a picture of Pascal's triangle on Google Images, and colored in blue and red bits in Paint.

Interestingly, Wikipedia gives the earliest description of the arrangement of binomial coefficients into a triangular structure as 975 AD.

@Semiclassical Please continue.

Found the original

@AkivaWeinberger It just bothers me how the blue $1$ isn't "collinear" with the other blue hexagons.

(math.stackexchange.com/a/980991/166353)

The history on Wikipedia is here, if you want: en.wikipedia.org/wiki/Pascal%27s_triangle#History

@SteamyRoot That was on purpose.

I'd quote further from that but eh.

link is simpler

It makes it clearer that the red is the sum of the blue.

It also includes a reference to Pingala in the 2nd century BC. But there's not enough of the relevant texts surviving to know whether the 'triangular' aspect was recognized back then as well.

I say that we rename it to be Pascal's Stack.

No.

Triangle is fine.

and it is turtles all the way down

it eeez not a triongal!

Pascal's quadrant? :P

I guess I'd call it Pascal's triangles, though. Plural

(said with a really weird accent)

Since there's really an infinite number of row truncations, each of which is a triangle.

@Semiclassical prove that you can construct an infinite number of equilateral triangles in the euclidean plane?

That's kinda trivial

you can tile the Euclidean plane with equilateral triangles

I mean, if you want countably infinite it's enough to take sides of integer length.

That too.

Hm, does anything interesting come out of making a Pascal's tetrahedron?

But I mostly just think this is a garbage discussion. People having been calling it triangular for centuries, across various cultures.

The name is settled.

@SteamyRoot really? I didn't know you sat through eternity and did it!

@BalarkaSen ffs.

I love this chatroom.

@Dodsy than marry the chat room. Not the people in it though. That'd be polygamy. Marry the HTML page.

@AkivaWeinberger How would one do that? My guess is yes.

to have a tetrahedron, you'd presumably need to have a recurrence relation over three indices

It's useful for when one writes equations within 3-space

it's no longer equalities and inequalities

it's molecular algebra

nvmd

i confused myself

Hm. How many words of length $n$, composed of the letters R, U, and O, have $j$ Rs and $k$ Us?

by that I mean: for pascal's triangle, one is taking advantage of the recurrence relation $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$.

@AkivaWeinberger write a brute force program that scans websters.

so the nth row is written in terms of elements of the previous row.

It'd be $\dfrac{n!}{j!k!(n-j-k)!}$, right?

@AkivaWeinberger It's a multinomial coefficient, yeah.

@Semiclassical That's got to satisfy a recurrence relation similar to the one you just posted.

Point.

With three terms on the right.

@AkivaWeinberger if you want to check which words that contain R's contain j's then you need to scan a dictionary....

I imagine it can be done.

words in this context don't have to be dictionary words.

they just have to be combinations of letters.

that's what a word is in the combinatorial sense.

but it's words that contain R's U's and O's and we ask how many contain J's

So, yeah, pretty sure the Pascal's tetrahedron would be to trinomial coefficients what the Pascal's triangle is to binomial coefficients.

No, it's not. Reread.

@TheGreatDuck No, $j$ Rs

ugh

latex is annoying

In less light news. This DC shooter incident is just...ugh.

You're annoyoing

JK

ooook.

OK, so:

@Semiclassical Did something happen while filming Wonder Woman?

$$\binom{a+b+c}{a,b,c}=\\\binom{a+b+c-1}{a-1,b,c}+\binom{a+b+c-1}{a,b-1, c}+\binom{a+b+c-1}{a,b,c-1}$$

i never heard of any shooters

Not what I mean.

DC = District of Columbia

O.O

there's been an actual shooting?!?

aka this: nytimes.com/2017/06/14/us/…

What happened?

I thought you were being sarcastic about film shooters. :/

it happened a few hours ago.

@TheGreatDuck Yes, on Congresspeople!

Some context first.

Among other things that happen in D.C., there's apparently a pair of congressional baseball teams

It might have been planned for a charity event, I forget.

Anyways. The republican team was practicing this morning, so there were a number of reps/senators and their staffers.

oh ok

Apparently a guy walked onto the field, asked if they were republicans or democrats, then pulled out a semiautomatic pistol and started shooting.

i was thinking that the regular baseball team for DC got shot at

i.e. Some MLB team associated with Washington D.C.

wow

Luckily, one of the congressmen was a high ranking member and therefore had their protection detail with them. So they took down the shooter pretty quick, and he's since died.

But even with that, several people were injured (none fatally, it sounded like) and about 50 bullets were fired.

It sounds like that, had the protection detail not been there, it could well have been a massacre.

holy

"50 bullets"

I posted that in the wrong place

sorry

I'm in two chats at once.

it wouldn't surprise me if the gunman used 50 bullets just for the sake of irony.

I think that was just an estimate.

Actually, I'm remembering wrong. I think it was a rifle.

But eh.

@MaryStar period ??

It was probably also fortunate that one of the representatives on staff was a medical doctor.

I hate news like this.

Don't like hearing about it

@Dodsy Yeah, it's pretty f*d up

@Hippalectryon In parts of Europe, they use commas and periods in the opposite way that we do

so that a thousand and a tenth would be 1.000,1 rather than 1,000.1

@AkivaWeinberger Oh sure but why not write "," and "." ?

@Hippalectryon 1,5 with period 5 is 1,555555.....

I dunno, ask her

Do you have an idea @Hippalectryon ?

@MaryStar What's the general definition of $x$ Period $y$ ? At the end is it (1.916 Period 6):5 or 1.916 Period (6:5) ?

$1.5\times1.5\bar5 +(7.65\bar5 -4.3)-1.916\bar 6:5$

@MaryStar What was the colon : for?

So yeah, this is all pretty horrible.

Yeah

With the one bright light being that it could have all been a hell of a lot worse.

@AkivaWeinberger It's division

@Hippalectryon the first one

$1.5\times1.5\bar5 +(7.65\bar5 -4.3)-1.916\bar 6/5$

@MaryStar Alright, what have you tried ?

Do we have to write them as a fraction ? @Hippalectryon

@Semiclassical Back to math, the above identity shows that it is the multinomial coefficients

Right.

@MaryStar That's one way to do it

$\binom{a+b+c}{a,b,c}$ is the value of the entry on the $a+b+c$-eth row, with distances of $a$, $b$, and $c$ to each of the edges

(all zero-indexed, so that the top entry is $\binom0{0,0,0}=1$)

Ok, to do that do we divide the number by 9? I mean $1.5\bar5 = 1+\frac{5}{9}$ ? @Hippalectryon

@MaryStar Yep

How can you tell if a limit does not exist?

@Hippalectryon Ok, thank you!! :-)

@Dodsy Got a concrete example ? That's a vague question

hm

$f(x)=\frac{x^2+3x-1}{x-2}$ as $x->2$

So the top function does not factor

What happens if you do a straight substitution?

but the bottom is equal to 0

The denom is equal to zero.

What is the numerator at $x=2$?

and the top is 9

so 9/0

So, I think conventions vary

but I think if the limit isn't any finite value, it's usually considered not to exist.

Any limit of the form $\frac90$ goes to infinity, so here it does not exist.

But if the top function were $x^2 -4x +4$

then the limit would be zero.

oh wait

right?

Yeah

okay. So check if the denom is = 0

if the top function isn't factorable or able to be rationalized then limit does not exist

Well, check if the denominator is 0, and if so, check if the numerator is zero

If the numerator isn't zero (while the denominator is), then the limit doesn't exist

oh.

that makes snse.

Thanks Akiva

Here's a question that I'm having trouble understanding

how do you write limits?

$\lim_{h\to 0}\frac{f(4+h)-f(4)}{h}=8$ What does this indicate about the graph of the function $f(x)=x^2$

@Dodsy That the tangent of $f$ at $x=4$ has a slope of 8

Well I know the answer

But why

$f'(x)=2x$ $f'(4)=2(4)$

And the limit notation finds the slope of the tangent

It's the limit of the slope of the secant

Not as beautiful as my drawing :P

right

A joke:

What's E.T. short for?

Little legs, he's got