I should make you a meme, @Hippa.

@Hippalectryon ever do surface geometry?

of any kind?

@TedShifrin You don't have my picture yet :-)

You have to do distributions and integration by parts, Duck and Hippa.

we'll use your avatar

Y a une invasion Française il paraît! Omelette du fromage!

@TheGreatDuck Some, but just for fun

@Hippalectryon making things walk on the surface of a continuous 3D model is trivial.

or, in other words, a polygon faceted with triangles.

@TheGreatDuck Depends on how it walks

?

@JeSuis Et sinon de ton côté ?

@user379685: I don't know probability off the top of my head. I taught it a few years ago but I don't do it instantaneously. That must be the law of large numbers or something ...

@TheGreatDuck What do you mean by "making it walk" ? What's "it" ? A point ? A robot ?

@user379685: I presume you need to compute mean, variance, etc.

anything

@Hippalectryon M2 l'année prochaine mais je ne sais pas vraiment quoi choisir.

@TedShifrin just the probability

just programming a thing to walk on the surface of a 3d model

as in a video game

not in real life

@TheGreatDuck Making robots walk on the ground is actually a nontrivial task isn't it ?

Ooh

Right, but you can calculate stuff knowing the probability distribution, @user379685.

I'm not talking about physics, dumby.

Find the mean and the variance, then use the law of large numbers.

@TheGreatDuck Well, it depends on the degree of realism you want. Walking perpendicular to the ground isn't always the best way

I mean in a super mario galaxy spatial sense

jumping ain't trivial

I'd like to see Hippa walk perpendicular to the ground.

but walking on geodesics is

@TheGreatDuck I've done some 3D engines in the past :D

@TedShifrin oh ok thanks

Or maybe it's central limit theorem — one of those.

mutters about finite measure spaces and ceases to speak of probability

@Ted Got Mike to answer it.

@Hippalectryon if you're offering your help I sadly cannot accept aside from random questions in terms of casual conversation. The kid's only allowed one person helping him, or rather, mentoring him in my case. I think his teacher was just expecting a sibling or classmate. I guess they didn't know I was... well-experienced.

:p

One concrete way to do the last one is to Fourier transform everything and do the approximations in that space

Also, hey everyone!

That's cheating, a @Balarka. Also, did you see my ping re that affine subspace thing you and I discussed?

@TedShifrin hey, um... one of the things I wanted to make as surfaces to walk on was a flat corkscrew, like a winding staircase. Know of such a parametric equation?

@TheGreatDuck Oh, I didn't get it was a school project

(I'm being a bit vague since I don't actually remember how the details work.)

not for me

but it's the kids project

try helicoid, Duck

thanks

@TheGreatDuck What 3D engine are you using ?

Find n that satisfies the equation $$\frac{5^1}{25^1-1}+\frac{5^2}{25^2-1}+\frac{5^4}{25^4-1}+\dots +\frac{5^{32}}{25^{32}-1}=\frac{1}{4}-\frac{1}{5^n-1}$$

Stop giving us all your homework, @AbdullahUYU. I'm done.

@Hippalectryon it's not a 3D game. It's a 2D game and we're making literally one level have 3d if we have time cause I'm ambitious. ;p

@TheGreatDuck :D

@Hippalectryon game maker 8.0 lite

these are not homework.

I honestly have no clue how I'm going to get 3d to work without the paid version.

and you don't have to do anything

@Ted Which was the affine subspace problem again?

XD

I reminded you up there ^^^ a few pings ago.

@Balarka: this question.

Ah I see.

@Hippalectryon It's a pretty clunky engine. Basically we're making a 2D turn based rpg on a grid, like some of the old table top games. Except there's weird minecraft like block moving stuff. I probably cannot describe it and do it justice.

XD

@TheGreatDuck Got a screenshot ?

not really no. We pretty much just started a couple weeks ago. I mean, I could grab something of the overworlds, but those are pretty dull atm. Most of the graphics is just colored squares.

and the default game maker player sprite. XD

ok :P

@TedShifrin Gotcha. That's a nice perspective.

I still like you geometric ideas (which I improved)

I suppose I have that from when we were discussing bosses

I knew what I was doing was too hard. I upvoted him, although I find his attitude a bit prickly.

but that's just a mockup

it's also missing the block selection hud

XD

since I had it open I made you onw

the lava was redone last night

no animations but I think 64*64 pixel blocks are so much better

and they have no impact on the throughput

@AbdullahUYU Note that $\dfrac{5^k}{25^k-1} = \dfrac{1}{5^k-1}-\dfrac{1}{5^{2k}-1}$. The surviving terms in the sum are $\dfrac{1}{5-1}-\dfrac{1}{5^{64}-1}$ so $n=64$

@Hippalectryon so yeah I guess I do have screenshots, but I'm not going to let anyone play it anytime soon. If we decide to post it anywhere it will be after we do a lot more work after the summer's over. We're just focusing on the main bulk of the project. We figure the teacher probably isn't expecting it to be too massive. After all, they have to grade it. So we're mostly focusing on quality atm.

The joy of telescoping series :)

and also building stuff to make our lives easier making the game

I've wondered why they're called telescoping

@Ted I posted a comment.

@TheGreatDuck Great :D good luck

@Daminark they are called telescoping because you can collapse them like a telescope. The really old ones actually closed up and could shrink in length.

I see, thanks!

@Hippalectryon thanks. Hopefully the kid responds soon. I guess some of the other people in the chat were cursing and so he's not allowed on google. Kid didn't even do anything. Ugh.

a mosquito is making me nuts

@BalarkaSen Mine is making me pizza. :-)

How does one make nuts?

@Hippa and now you're sniping me too

@Hippalectryon anyway, since two triangles make a square, the surface geometry of a 3D model leads naturally into a means of making maps like that running through space and stuff.

if you want to make puns stop being slow

commits to speeding up

did you solve my S2xS2 problem yet

@TedShifrin Sorry if i was rude. I don't want to be thankless. I just wanted to say that they aren't hw.

I have not, but in fairness I also haven't thought about it all too much

"Find a metric of positive sectional curvature"?

lol no

Finding an embedded sphere with non-zero self-intersection number

oh

Sad!

(In contrast to S1xS1)

Lolol

@MikeM we three all screwed up on this one a few months ago when we said every element of H_2(S2xS2) is of the form a[S2]_x + b[S2]_y

hm? I mean, it's true

we just did a shit job of squaring it

yeah true

I think I also screwed up when I made the same statement with T2xT2... the shame

in a different context, with a different person

If a random variable (X,Y) has the density function f(x,y)=x+y, 0<=x<=1, 0<=y<=1 how can i find the distribution of the product XY?

@user379685 Can you find the marginal laws of $X$ and $Y$ ?

Nevermind got it, thanks c:

:-)

ehhh. how do I prove that AB = I implies BA = I? AB = I means B is injective, and injective linear maps are surjective (image of a basis is a basis, which spans all of the space).... then what?

Oh, sure, that means there's a C such that BC = I. BC = BA, C = A. OK.

However could you help me with this: two dimensional random variable (X,Y) has uniform distribution on a disk with radius 1. Find the distribution of the random variable Z that is the distance from the point (X,Y) from the edge of the disk.

is there a closed form for the coefficients of $\frac{x^{n+2}(1-x)^{2n}+2^n}{1+x^2}$?

and yes it is a polynomial

@user379685 Well if you know that the distribution is uniform, what's the probability to be at a distance between $r$ and $d+dr$ from the origin ?

Dear god. Why's it a polynomial?

i know that the distance for a point (X,Y) is 1-sqrt(X^2+Y^2)

@user379685 If you prefer, what's the probability to fall in a given area $A$ of the disk ?

A/pi?

Exactly. Therefore, what's the probability to fall between the circle of radius $r$ and the circle of radius $r+dr$ ?

@BalarkaSen why is a very hard question to answer, but you can verify it by bashing

(r+dr)^2-r^2

@user379685 Right, can you simplify it ? (dr^2=o(dr) so it's negligible)

2dr

Uh ? Are you sure ?

2rdr

Great. Now, as you said, being at a distance $R$ from the origin is the same as being at a distance $r=1-R$ from the edge. Therefore, what's the elementary probability of being at a distance $R$ from the edge ?

(you forgot $\pi$ btw: 2 pi r dr)

to be on the disk of radius R? and you asked me what was the probabilty of being between the circle of radius r and r+dr, so (pi*(r+dr)^2-pi*r^2)/pi1^2

so pi cancels?

yeah, so p(r,r+dr)=2r dr. But that's the probability to be at a distance r from the origin. can you deduce the probability to be at a distance $R$ from the edge ?

2r - r^2

?

Uh ? How did you get that ?

the probability of being in the ring between the edge and some disk of radius 1-R, where R is the distance from the edge is (pi-pi(1-R)^2)pi=2R-R^2

Oh, but what we're looking for isn't the probability of being between the edge and a disk of radius R, but the probability of being at distance R of the edge

i.e. the probability of being between a distance $R$ and $R+dr$ from the edge

i don't know :c

Well, you know the probability of being at a distance $r$ from the center right ? We computed it earlier

yes

well we know the probabilty of being between circles don't we?

And isn't it equivalent to be at a distance $r$ from the center and to be at a distance $1-R$ from the edge ?

yes

@user379685 The probability of being at a distance $r$ is the probability of been between distances $r$ and $r+dr$ when $dr$ goes to 0

@user379685 So, using what we computed earlier, what's the probability to be at a distance $R$ from the edge ?

2(1-R)?

Yep! 2(1-R)dR

That's your probability distribution. You can check that if you integrate it from 0 to 1, you get 1.

shouldn't i represent it with X and Y?

@Sophie induction etc etc? ok i can believe that

i dunno about your question though

@user379685 The probability distribution of finding a point (X,Y) at a distance R from the edge is 2(1-R)

in the wording of the excercise there was no mention of R

Well in the exercise it's called Z isn't it ?

that's the name of the new random variable

and it's also the distance

hmmm

ok but the density function f(Z) is just 2(1-Z) without the dZ?

exactly

ok thank you

Let $\frac{x^{n+2}(1-x)^{2n}+2^n}{1+x^2}=\sum_{h=0}^{3n}a_hx^h$. Then $a_0=2^n$, $a_1=0$, $a_{3n}=1$, $a_{3n-1}=-2n$ and $a_h+a_{h-2}=0$ if $0\leq h<n+2$ and $a_h+a_{h-2}=(-1)^{n+h}{2n \choose h-n-2}$ if $h\geq n+2$

So $a_{n-k}$ for $k<2n-2$ is a polynomial in $n$ of degree $k$

For polynomial $P(x)=x^2+bx+c$, why $P(2)=6$ doesn't satisfy the conditions $P(-3)>0$ and $P(1/3)<0$?

You want it to pass through (2,6) and be a monic polynomial. So you need $P(x)=(x-2)(x-r)+6.$

So what are $P(-3), P(1/3)$ in that case?

How do you conclude that i want it to be monic polynomial?

because you wrote x^2 not ax^2

hmm, i misunderstand monic. $r>1/3$ so $P(-3)=21+5r>21+5/3>0$ and $P(1/3)=-5/9+5r/3>-4/9$. Both seems OK.

@AbdullahUYU Check your work on the first one. Also, I don't see a reason to assume $r>1/3$.

Actually, check both.

back later

without knowing $r$ how can we proceed?

but wait, $r$ is not the root of the $P$.

i tried to show one of the root is in the interval $(-3,1/3)$. And the other is greater than $1/3$. But $r$ is not the root of $P(x)$

Anyway, $P(-3)=21+5r$. $P(1/3)$ should be $\frac{15r+49}{9}$

@AbdullahUYU So you need $P(-3)=21+5r>0\implies r>-5/21\approx-0.238$ and $P(1/3)=15r+49<0\implies r<-49/15\approx -3.26.$ See a problem?