Mathematics - 2017-09-03 (page 2 of 7)

yes, but in my case i need to show the definition of $\ominus a \implies \ominus(\ominus a) = a$ without appealing to $a + \ominus a = 0$ and instead just use the identity like i did above

Typhon

@salt what I mean is that you give some abstract operator with abstract qualities. Giving it the symbol $-$ attaches a connotation to it which confuses us into thinking it is negation. Therefore, since this is talking about some arbitrary function maybe using a function name such as $f(x)$ would work better?

Akiva Weinberger

It confuses you since you jumped in without looking for or asking for context

@Salt I didn't appeal to it, I proved it, didn't I?

Salt

maybe, lemme gander again

Typhon

@AkivaWeinberger I'm thinking in the context of a formal proof.

like i said

minor quibble

Akiva Weinberger

5:08 AM

@Salt You agree that $a_s+\sum_{n\ne s}a_n=\sum_na_n$, yes?

Salt

yes

Akiva Weinberger

And, by the identities, that gives $a_s+\ominus a_s=0$, right?

Salt

yep

Akiva Weinberger

Since $\ominus a_s$ is in the set, it must be $a_t$ for some $t$

The same argument says that $a_t+\ominus a_t=0$

Substituting it back in, we get $\ominus a_s+\ominus(\ominus a_s)=0$.

Yeah?

Typhon

@AkivaWeinberger i have a question. We can use your 3d model for reference.

i know it's fake

but that's irrelevant

@AkivaWeinberger suppose I had a single point on one of those polygons selected. Understand?

I've picked it

now

Akiva Weinberger

5:16 AM

Sure

Typhon

you can move in any direction to an adjacent polygon so long as that point is one of the vertices of the other polygon

got it?

Akiva Weinberger

Yeah

Typhon

now, let us pick the set of all polygons we can move between

Akiva Weinberger

So you move along edges?

Typhon

no

Salt

5:17 AM

yes, but it was so much more fun to cancel the terms combinatorially

Typhon

you move between the insides of the triangles

the segments are the walls

do they compose the set of all polygons sharing that point relevant to calculating the surface normal at that vertex by taking a weighted average of the surface normals?

that is to say

does it exclude any polygons that might be inside the sphere?

that also share the point?

Akiva Weinberger

I don't understand. Relevant to calculating the surface normal?

Typhon

each of those are a slice of a curve of a progressively smaller sphere

(or rather a polygon approximating a sphere)

scratchapixel.com/lessons/3d-basic-rendering/…

from that^^

@AkivaWeinberger you've never done 3d modelling have you?

Akiva Weinberger

Nope

I don't know the answer to your question

Typhon

ok

take the red black and blue lines as what I said

if i did my algorithm

would I only end up with blue if I started with a blue polygon on the outer surface?

you know what

Akiva Weinberger

5:27 AM

What algorithm

I don't understand

Typhon

what I described?

the whole thing I just said earlier?

O.o

did you forget?

Akiva Weinberger

Oh OK

Secret

To make $(\mathscr{P}(S),\circ)$ a group for arbitrary set $S$, define $\circ$ as follows:
\begin{align}
\prod_{s \in \mathscr{P}(S)} s & = \emptyset\\
\prod_{s \in \mathscr{P}(S)-\{a\}} s & = a^{-1}\\
A \subset \mathscr{P}(S),\prod_{s \in \mathscr{P}(S)-A} s \circ \prod_{s \in A} s & = \emptyset
\end{align}

Akiva Weinberger

Smaller and smaller spheres

I thought you meant the thing from earlier

With the vertex

Typhon

I'm doing the same thing

basically I want to grab all the points in the blue

not the red or black

without actually excluding them

(with some explicit marking)

Akiva Weinberger

5:29 AM

By "it" you mean "them"?

Typhon

sure

XD

I'll just test it

Akiva Weinberger

OH, OK, I get it

Typhon

it's just a shitty matlab renderer anyways

Akiva Weinberger

I don't know the answer

Typhon

if it breaks, who cares?

:p

Akiva Weinberger

5:30 AM

You were not being clear

Typhon

shrugs

there's a lot of documentation on lighting calculation given the normal vectors

but literally jack squat on how to get said vectors

Akiva Weinberger

No code snippets?

Typhon

it's supposed to be a weighted average of all normal vectors of triangle's containing the point

@AkivaWeinberger nah

just ones using surface normals which I don't want

Akiva Weinberger

What's the difference between normals and surface normals?

Oh, you mean face normals versus vertex normals?

Typhon

each triangle has it's own set of three vertices each with it's own normal vectors

one tactic is to give the points the normals to the face instead of the actual one prescribed via surface geometry

Akiva Weinberger

5:33 AM

Doesn't this thing you linked earlier go through vertex normals versus face normals?

Typhon

so of course that's what google gives 90% of the time

idk

Akiva Weinberger

Re: the last image in that page

Typhon

can't read their code for shit

Akiva Weinberger

Typhon

and i think they assume closed surfaces

Akiva Weinberger

5:34 AM

It's the right programming language, I assume

Typhon

"right programming language"

Akiva Weinberger

The same one you're coding in

Typhon

berserk button is pressed

ooooh

(there's no RIGHT language, btw)

no im coding it in matlab

Akiva Weinberger

@Typhon Obviously

Typhon

you ever used matlab?

Akiva Weinberger

5:35 AM

@Typhon Do you know that language, at least? (I think it's Java but I might be wrong)

@Typhon No

Typhon

that's not matlab

it's C++

Akiva Weinberger

Do you know C++?

Typhon

yes

Salt

hmm i got serially downvoted earlier today, then it was reversed by the system, and now it happened again...is this a common thing?

Typhon

that doesnt mean i can read their code, or rather... read it and get anything of value

Salt

5:36 AM

or did i just anger someone?

Akiva Weinberger

@Salt Not that I'm aware of. Maybe take it up with the Mathematics Meta site?

Salt

i figure it'll get reversed again

Leaky Nun

@Salt how can you get serially downvoted if you only asked one question and answered three?

Akiva Weinberger

@Typhon You should go to Stack Overflow, the site that actually deals with programming

Typhon

the problem is that they don't actually give the computation of the surface normal

it's not a programming question though

Salt

5:37 AM

i've made 3 posts

Typhon

that's the subtle issue. It's actually a geometry issue

Salt

and i've attempted 1 answer on one of them

Typhon

it's a question of how to get the normal at a vector of some polygonal surface

Secret

Then:

@LeakyNun $$\forall A \subset \mathscr{P}(S), \emptyset=\prod_{s \in \mathscr{P}(S)}s = \prod_{s \in \mathscr{P}(S)-A}s\circ \prod_{s \in A} s \implies \left(\prod_{s \in \mathscr{P}(S)-A}s\right)^{-1}=\prod_{s \in A} s$$

Typhon

the problem is that at different points it varies depending on the side of the surface your own

and whether it is closed/unclosed self-intersecting/not-self-intersecting causes the problems.

Akiva Weinberger

5:39 AM

> If the object is a triangle mesh, then each triangle defines a plane and the vector perpendicular to the plane is the normal of any point lying on the surface of that triangle. The vector perpendicular to the triangle plane can easily be obtained with the cross product of two edges of that triangle.

Typhon

if only that were true

Akiva Weinberger

That paragraph goes on to describe how to do it in more detail

Do you know cross products?

Typhon

yeah

im a senior in college

the problem is that that only applies to points inside the triangle

consider the whole issue of the derivative at sharp corners

now consider the same, but for normal vectors

ugh

Salt

it might not be serial downvoting, but as it was across all but one of my posts I assume it is, especially since it happened earlier

Akiva Weinberger

Wait, by "surface normal" do you mean "face normal" or "vertex normal" @Typhon

Typhon

5:41 AM

vertex normal

Akiva Weinberger

Ohh

Typhon

yeah

i know how to compute it for a closed surface

Akiva Weinberger

9 mins ago, by Typhon

just ones using surface normals which I don't want

^That's what made me think you meant vertex normals

Typhon

just take a weighted average of all the normals to the triangles in their insides containing the point

Akiva Weinberger

Why doesn't that work for non-closed surfaces?

Typhon

5:44 AM

more formally if triangles $t_1,t_2,t_3,...$ all contain the point $p$ and $N(t)$ returns the normal vector of the plane containing triangle $t$, and $A(t)$ returns the area of triangle $t$, then the normal vector at $p$ is proportional to $N(t_1) + N(t_2) + N(t_3) + ....$ if and only if the surface is closed locally.

Akiva Weinberger

Why can't you just compute $\frac1n\sum N(t_i)$ anyway? ($n$ being the number of triangles with $p$ as a vertex)

Typhon

@AkivaWeinberger let's stick a plane through the middle of the inside of your ball. That can arbitrarily be extended through the backside of your ball to infinity and heavily disrupt the weighted sum. However, note that the surface at the point hasn't changed. We've just disrupted the algorithm by taking something into cosideration which we shouldn't be.

Secret

So, the difference between the finite case and infinite case of $(S,\ominus)$ is that there are inverse elements that cannot be decomposed in terms of sums of $\ominus a_x$

Akiva Weinberger

Why would it disrupt the weighted sum?

Typhon

more formally if triangles $t_1,t_2,t_3,...$ all contain the point $p$ and $N(t)$ returns the normal vector of the plane containing triangle $t$, and $A(t)$ returns the area of triangle $t$, then the normal vector at $p$ is proportional to $A(t_1)N(t_1) + A(t_2)N(t_2) + A(t_3)N(t_3) + ....$ if and only if the surface is closed locally.

Akiva Weinberger

5:47 AM

So by plane you mean a new triangle

that contains $p$ as a vertex

Typhon

@AkivaWeinberger Cause yer adding an extra normal vector that has no relevance to the sum.

yes

@AkivaWeinberger yup

@AkivaWeinberger do you see where the problem arises?

technically

such a self-intersecting surface has more than one normal vector depending on which side of the surface you look at and which triangle

Akiva Weinberger

Why don't you normalize $N(t_1)$ to be a unit vector?

Typhon

@AkivaWeinberger they are normalized.

we just assume they are

no need to do it again

Akiva Weinberger

I guess I mean, why not ignore the $A(t)$ factors?

Typhon

uuuh

those are the weights

Akiva Weinberger

5:50 AM

Oh wait I think I see why

Typhon

for the weighted average

the more area a triangle has the more the correct normal vector points in that direction

Akiva Weinberger

No I meant like why not take a not-weighted average

But I see why

Typhon

@AkivaWeinberger cause then you get garbage

shrugs

Akiva Weinberger

Seems like the normal would change if you subdivide your mesh though

Typhon

why does the arclength formula use a distance formula type thing instead of direct addition?

@AkivaWeinberger on one side

what about the other side?

heheh

Akiva Weinberger

5:52 AM

??

Typhon

every time you split on one side, you get more normals

Akiva Weinberger

I'm still talking about the one vertex

Typhon

sigh

brb

irrelevant blue line

consider the normals in the creases

Akiva Weinberger

I meant like in general

Typhon

(cant find a decent diagram so just assume it narrows to points or whatever)

Akiva Weinberger

5:53 AM

Like say you have a cube

Typhon

i know

ok?

@AkivaWeinberger if it's just a cube, you use the formula i gave before

Akiva Weinberger

You want to find the normals to the vertices of the cube, and it seems to me that that depends on how exactly you triangulate it

Typhon

nah

oooh

hrm

i see

that explains why I was hearing about "angle weighting"

i need to go to bed in a minute though

it's really late and i have to get up to go to church in the morning

:p

Akiva Weinberger

Yeah, go to bed

I should too

Typhon

angle weighting would fix my issue though

i think. It might only need adjacent triangles to work. That would trivialize my algorithm. More efficient too.

Akiva Weinberger

5:56 AM

@Typhon It looks like vertexNormal is a keyword in Matlab

mathworks.com/help/matlab/ref/…

Typhon

im writing the 3d engine from scratch

Akiva Weinberger

Ah OK

Typhon

yeah im not using that data structure

it needs a triangulation thingy

Salt

i'm writing a 1d engine right now, but the physics doesn't seem right

Typhon

@salt lol

3d renderer

not engine

Salt

5:59 AM

i figure i'll just do the 1d case and then use induction to get all the other engines

Akiva Weinberger

You know, in theory, you could write a 4D renderer, it just wouldn't be able to display it

Typhon

that's great for physics

@AkivaWeinberger actually that's falsew

I could project to a 2d plane

it would just look like gibberish, like if we projected from 3d to a line

anyways

Salt

don't they usually take a projection to 3-d then usual projection to 2-d

Typhon

goodnight. thanks for the help

@Salt uuuh what?

Salt

for the 4d case

Akiva Weinberger

6:01 AM

@Salt There's a video game out there that deals with 3D slices of 4D stuff

Salt

when showing, like, a hypercube

Typhon

no clue. I've never done that. I'd write to a 3d pixel-space.

Akiva Weinberger

@Salt That only works if it's wireframe

Typhon

:p

Akiva Weinberger

Hm, not necessarily true actually

It could work

You'd just miss out on a lot of the detail

Salt

6:01 AM

i don't actually know, but i thought they projected twice

Akiva Weinberger

Yeah

Secret

@Salt 4D->3D->2D projection is often highly confusing in a gaming perspective, which is why people tend to do cross sections for 4D games

Typhon

@Salt that was probably something else

Akiva Weinberger

Usually just the wireframe, though, or possibly with the 3D faces partially see-through

Typhon

4D games are an actual thing?

Akiva Weinberger

6:02 AM

Let me find a Wikipedia example

Secret

miegakure.com

this is the most famous one

Typhon

woah. OK I AM SICK OF THIS. literally EVERY SINGLE POST OF MINE IS GETTING HIT WITH THE FLODDING FILTER. ARRGGH XD

Akiva Weinberger

Typhon

annoying

Akiva Weinberger

Like that

@Secret That's not out, yet, is it?

The creators of that came out with I think it's called "4D toys" though

You can find online videos of people playing it in VR

Secret

6:04 AM

oops yeah. Miegakure seemed to be still in development

But 4D toys is available in iphone

Akiva Weinberger

Cool

Typhon

i a saving that link

Secret

What is rare are games and applets with surface culled 4D->3D->2D projections

Typhon

starred

now it is saved for me. I'll find it later. Goodnight and thanks for all the help @AkivaWeinberger. I would've been real confused later.

XD

Akiva Weinberger

TBH technically those are just "projections" @Secret

Typhon

6:07 AM

(when it all started breaking due to the entire formula being flawed)

Akiva Weinberger

The same way $(x,y,z)\mapsto x$, $\Bbb R^3\to\Bbb R$ is a projection

Secret

urticator.net/blocks is an example of those rare examples

Akiva Weinberger

Projecting to a subspace, not necessarily of dimension one down

Salt

well, i thought it was 4d -> 3d engine of some sort -> 2d screen

Secret

The space between the wireframes are actually opaque in the 4D sense in that 4D blocks example

Akiva Weinberger

6:08 AM

@Salt Same result as doing 4D->2D directly

I think

Kinda like 3D->2D->1D can be thought of as projecting onto a line

Salt

i wouldn't know, i thought the 3d engine sort of gave it a familiar perspective

but maybe you can just skip that middle man

Secret

The way 4D->3D->2D works is it first project the 4D object as a perspective projection in 3D, and then said projection is projected onto the 2D screen to display. 4D->2D thus omit the perspective distortion of accomodating the 3D depth information

And so you are effectively projecting a parallel projection of 4D to 3D, into the 2D screen

Salt

do people prefer the 4d->2d way?

Secret

It's in some sense easier to comprehend since distances are not distorted.

But it works well only with some 4D opacity

Salt

well, that's sort of a given, i think

though i'd prefer not to (think)

Secret

6:17 AM

Perspective is best for translucent objects, but it takes time to interpret them properly. And it can be misleading for shapes of low symmetry

In addition, coding cross section is much easier than perspective projections

It is also robust against showing shapes with low symmetry

Salt

oh, i remember seeing, errr, vihart? making that VR for exploring 4-d objects in a 3-d space

BAYMAX

Any help on Finite Difference Method?

like I was thinking that in the case of explicit forward finite difference scheme for solving a parabolic PDE will always give us a solution which is perodic in time?

Salt

Eating 4D Polytopes (Hypernom)

►

it was a numberphile video... of course

Anonymous

@Semiclassical Would you be able to answer this question?

Leaky Nun

6:40 AM

0

A: A rigorous treatment of the seemingly simple d.e. $x^2 y' - y = 0$

Naive solution for some intuition (with division of zero involved): $$\begin{array}{rcl} x^2 y' - y &=& 0 \\ x^2 y' &=& y \\ \dfrac{y'}{y} &=& \dfrac1{x^2} \\ \displaystyle \int \dfrac{\mathrm dy}{y} &=& \displaystyle \int \dfrac1{x^2} \ \mathrm dx \\ \ln y &=& C-\dfrac1x \\ y &=& Ae^{-1/x} \end...

@BalarkaSen @AlessandroCodenotti @TedShifrin @Semiclassical

Daminark

7:18 AM

Checks out. And holy shit

I'm just gonna go to bed on that note. See you!

Jenna Sloan

I don't suppose anyone has a formula that does the following: f(0)=0, f(1)=1, f(2)=288, f(3)=6670903752021072936960, ...

I tried to formulate it, but yielded naught

Salt

got a calculator handy? whats $\log_2 (f(3))$?

$\approx 72$

i have no idea, but f(3) has a really big prime factor 27704267971

$f(3)-1$ is the product of two large primes, $28308476633 \times 235650396823$

it's like some number a crypto-algorithm would use

Anonymous

7:59 AM

@JennaSloan Huh. It should be pretty easy to find such a polynomial function.

Salt

i mean you can make a polynomial fit any finite sequence, but surely that sequence is some exponential one

Secret

9:04 AM

Mar 27 '15 at 21:37, by Ted Shifrin

@Kevin: No, other than saying the set of all comparable metrics. There are uncountably many. :P

(To be investigated) : Topology of the set of all comparable metrics over a topological space X

Hmm, nope, need something larger: The topology of all metric spaces

That is, each metric space forms a subset in the topology, so that there might exists some continuous map between any two metric spaces

Kinda interesting how political science deals with such higher dimensional construct

usukidoll

whoa Topology

Secret

9:36 AM

chat.stackexchange.com/transcript/message/20784598#20784598 Ok so no prime closed form, back to the drawing board

Secret

I really have to procrastinate less on TV and quickly redo all my chemistry calculations, in order to spare enough time to get back to topology and algebra

right now, I am only halfway redone them. The excel data entry is what slows it down

One issue I had and is mostly responsible for all those [Random] is because there is a lot of things I want to try out in maths, but the guilt of spending too much time in maths jeponise my chemistry PhD prevent me to allocate a sizeable enough chunk of time to actually address those maths problems I am interested in

and this result in I can only do partially complete analysis, time capsuled using the [Random] tag

That does not make them any less random though. They are so named because they are free associations and random maths thought captured to be dealt with when I finally have enough time

shikari-shambu

10:16 AM

I have a quick combinatorics question. Say you have unlimited numbers of red, blue and black balls and you have to choose 5. This is an $r$ combination of $n$ objects with repetition so the answer is ${{n + r - 1}\choose{r}}$ or ${{5 + 3 - 1}\choose{3}}$. So far so good. But what is wrong with this reasoning:

Say you have the following sequence _, _, B, _, B, _, B, _, B, _, B, _, _ (i.e. five Bs and 8_s). To divide the sequence into three parts, insert two $|$. Then, you have three sets of Bs and each set corresponds to one colour. There are ${{8}\choose{3}}$ ways of doing this, hence the…

Sorry, that should be ${5 + 3 - 1}\choose{5}$ and ${8}\choose{2}$

Leaky Nun

11:22 AM

metric topologies are boring

prove/disprove

Secret

Mar 28 '15 at 20:42, by David Wheeler

More immediately relevant to you: every (!) group is a quotient group of a free group.

Leaky Nun

@Secret ?

it's called the presentation of the group

Secret

Claim: Quotient group = choose which reduced words in a free group to be set to the identity

user84215

12:23 PM

Suppose that $\omega_{ij}=\sum da_{ik}a_{jk}$ . I do not know why a minus signappears in the exterior derivative of the above: $d \omega_{ij}=- \sum da_{ik} \land da_{jk}$

user84215

A space is needed between "sign" and "appears".

user84215

12:53 PM

My problem has been solved. Thanks for your answers.