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