Mathematics - 2017-07-18 (page 5 of 5)

Ted Shifrin

22:00

I see that @Alessandro's amoeba is here, too

Maks

Hi @TedShifrin

:D

Ted Shifrin

heya @Maks

isaac9A

I would like to know what the group operation for the dihedral group is

Ted Shifrin

What do you mean, @isaac9A?

Composition of functions, if I think of it as geometric symmetries.

isaac9A

@TedShifrin What is the binary operation or function that takes two elements of D2n and returns an element also in D2n?

Ted Shifrin

22:04

I just answered that.

You tell me how you're defining the group. I told you how I am.

Semiclassical

Hi chat

Ted Shifrin

hi Semiclassic

Typhon

5

Q: I would like help identifying the rigorous classification of this 'surface' geometry based on my interpretation of 3D models.

I want to try and identify a geometric structure I thought up while doing some weird stuff with making things walk on the surface of a 3D model and trying to incorporate backface culling into the surface geometry itself. See, in computer graphics each side of a polygon or triangle are considered ...

looking for advice on who to award the bounty

isaac9A

@TedShifrin so if r is the element gotten by rotating clockwise 2pi/n degrees in D2n and s is the element gotten by reflecting then rs is just "do r" then "do s"?

i.e. function composition

Ted Shifrin

Precisely ... if you're thinking of actual geometric symmetries of a regular $n$-gon. Yes.

isaac9A

22:08

the set of elements in D2n are geometric symmetries of a regular nn-gon

Akiva Weinberger

It could be the reverse order theoretically

Ted Shifrin

blah

Akiva Weinberger

Some books will have rs as "do s" and then "do r"

It really doesn't matter all that much, though

Ted Shifrin

oh, that would be what I meant.

I didn't read carefully what isaac said.

isaac9A

@AkivaWeinberger awesome thanks! also sr = rs^-1 correct?

Ted Shifrin

22:10

@isaac: your way, if you try to do it with matrices, for example, you'll have to operate on vectors on the right. I should have read more carefully. You ordinarily want $rs$ to mean $r\circ s$, so you start with $s$ and then do $r$.

Akiva Weinberger

@isaac9A That sounds about right

Did you learn what an isomorphism is yet?

isaac9A

Yeah

Akiva Weinberger

It doesn't matter which direction you interpret "rs" to mean, because the resulting groups are isomorphic

isaac9A

i.e. every group of order four is isomorphic to either the cyclic group of order 4 or the klein 4 group under some function

Typhon

@AkivaWeinberger I proved that the rational coefficient polynomials have a division theorem

*monic

Akiva Weinberger

22:12

Yay

Typhon

I believe I can prove my statement now

Akiva Weinberger

Division theorem meaning they're a Euclidean domain, yeah?

Typhon

it is as simple as follows

@AkivaWeinberger yeah

Ted Shifrin

DogAteMy: True, but it drove me nuts when Herstein (for example) wrote his functions acting on the right: $(x)f$, etc.

Akiva Weinberger

??!!

Typhon

22:15

1. all quadratic rationals are solutions to some rational second order polynomial 2. the multiplication of non-monic integer coefficient polynomials is a non-monic integer coefficient polynomial. 3. Need to relate the "1" to "2" in such a way as to cause a contradiction if I suppose not.

Am I on the right track @AkivaWeinberger

Akiva Weinberger

(That's actually a chess notation; it means "??!!")

4

Typhon

@AkivaWeinberger eeerrr.... that's the same thing.

Akiva Weinberger

@Typhon Yeah, I think so. Though note that there actually is an exception to 2; if you multiply two things with leading coefficient -1, you get something monic.

PVAL-inactive

Pretty certain I make a move like that every time I try and play chess.

Akiva Weinberger

Like $(-x+2)(-x^2+x)$ or something

Typhon

22:17

@AkivaWeinberger true, but those are monic if we consider algebra to be a thing.

Akiva Weinberger

I'm 80% sure I know what you mean.

Typhon

-x + 2 = 0

therefore

x - 2 = 0

therefore x - 2 = -x + 2

in the sense of polynomial equivalency

Akiva Weinberger

I don't quite think it works like that

Typhon

how so?

polynomials are uniquely defined by their solution sets

Akiva Weinberger

Up to multiplication by constants.

Typhon

22:19

@AkivaWeinberger those are all equivalent

Akiva Weinberger

Oh, OK, sure I guess

I didn't know what "equivalent" meant

Typhon

1/2 x^2 + 1/2 x + 1/2 is still a monic polynomials

it is equivalent to x^2 + x + 1

Akiva Weinberger

I don't think so

Pretty sure "monic" is not a property preserved by that equivalence

Typhon

they both have the same solutions

Akiva Weinberger

Doesn't matter

Typhon

22:20

I would argue that it is a sort of monicity

or perhaps monicity simply means that the leading coefficient divides all other coefficients?

Akiva Weinberger

@Typhon Perhaps. But then the notion of monic-ness becomes a useless concept in Q[x]

Typhon

@AkivaWeinberger but indeed it is.

it is only a minor shortcut in some cases

Akiva Weinberger

Fair

In any case, you do seem to be on the right track

Typhon

I would argue that monicity is primarily useful for Z[x]

okay thanks

and I'd post a Q & A on here regarding this stuff...

but I'm currently showing off the stuff to some other people and transcribing proofs from that place is.... tedious.

Akiva Weinberger

I'm pretty sure it is called "quadratic irrational" and not "quadratic rational," by the way

Typhon

22:23

@AkivaWeinberger quadratic rational is my name for things of the form a + bx where x is a solution to a quadratic monic integer coefficient equation and a and b are rationals.

it is a made up term

I literally thought it up

Faust7

i finnally figure out wth a binary function was

Typhon

and irrational makes little sense. They are fields and they are done by appending to the rationals

@Faust7 it is another name for multivariate, right?

@AkivaWeinberger the proofs are being written on these

minecraft.gamepedia.com/Sign

Daminark

@TedShifrin Lol, I remember when Herstein first introduced permutations and was like

Akiva Weinberger

@Typhon That's equivalent to saying "it satisfies a quadratic with rational coefficients"

Faust7

well mostly but the context of the guys question i finnaly understand what he was saying =P

Daminark

22:26

Yeah we're not even gonna be consistent about mappings in general, but permutations are on the right, "as algebraists do"

Akiva Weinberger

Proof: Let $\bar\alpha$ be its conjugate. Then $(x-\alpha)(x-\bar\alpha)$ is a quadratic with rational coefficients @Typhon

that $\alpha$ is a root of

For the converse,

Typhon

@AkivaWeinberger I know, but I prefer to phrase it in the same form as the quadratic integers and quadratic rings. XD

and let the solutions to polynomials thing be a proof

I am well aware of that, in fact.

Akiva Weinberger

(Cont'd) we'd need to show that if something is the root of a (possibly non-monic, possibly non-integer) quadratic, then it can be written as $ax+b$ where $x$ is the root of a yes monic yes integer quadratic

which is not too hard I think

Typhon

@AkivaWeinberger oh yeah, the reason I need to show the quadratic rational solutions to integer coefficient monic polynomials is because I seek to prove that all rings Z[\sqrt{b}] are integrally closed when b = 1 (mod 3)

I was able to prove for quadratic equations

but.... higher order is beyond me so I assumed that maybe all solutions are solutions to quadratic polynhomials

if that makes sense

Akiva Weinberger

"Quadratic irrational" seems to be a generally accepted term, though it looks like it excludes rationals

Typhon

22:31

that is completely different than my thing

that's like the idea of algebraic integers

and algebraic rationals

Akiva Weinberger

Yeah, the naming does seem a bit inconsistent

but it's the things that are roots of quadratics with rational coefficients, and as we just discussed, it's equivalent

Typhon

if x satisfies the polynomial p, then a + bx satisfies the polynomial p((x-a)/b) which has rational coefficients

^^all you need to state to prove your claim

Akiva Weinberger

Oh, whoops

Typhon

fixed

Akiva Weinberger

Well the converse is still a thing I guess

Typhon

22:33

i mean, there's a little algebra missing there

but it is good enough

Akiva Weinberger

Yeah OK

Typhon

i mean, you can expand p I suppose.

regardless

for each x, there is a separate quadratic field and ring

the article you linked lumps them all as the same set

along with your statement of "they are just solutions to rationally coefficient polynomials"

those are insufficient to capture the idea of them being separate fields

it lumps them all together

:p

Akiva Weinberger

By "the quadratic ring," you mean Z[x]?

Or Q[x]

Typhon

Z[x]

Akiva Weinberger

Oh, OK, good

Typhon

22:36

cause Z[x] is a ring and not a field

Akiva Weinberger

Yeah

Typhon

oops

whereas Q[x] is both

anyways

@AkivaWeinberger I seek to prove that Q[\sqrt{b}] only has Z[\sqrt{b}] solutions in monic integer coefficient polynomials when b = 1 (mod 4).

any suggestions?

aside from using the polynomial stuff and proving for the quadratic equations?

@AkivaWeinberger hello?

Akiva Weinberger

@Typhon I'm confused

Typhon

@AkivaWeinberger ok

let me be more rigorous

Akiva Weinberger

$x^2-x-1$ has a root of $\dfrac{1+\sqrt5}2$, but it's not in $\Bbb Z[\sqrt5]$

and it's in $\Bbb Q[\sqrt5]$

Typhon

22:46

im sorry

Akiva Weinberger

and $5\equiv1\pmod4$

Typhon

b = 3 (mod 4)

Akiva Weinberger

Ohh

So you just want to show that stuff like the above can't happen when $b\equiv3\pmod4$

Hm. The fact that the root of $ax^2+bx+c$ is in $\Bbb Z[\sqrt{b^2-4ac}]/2$, and that $b^2-4ac\not\equiv3\pmod4$, seems relevant

Typhon

a is 1

Akiva Weinberger

Sure

Typhon

22:49

and there are no b's in z/4 such that b^2 equivalent to 3 mod 4

Therefore, you wont find any quadratic counterexamples.

Akiva Weinberger

That's what I said

Typhon

I desire to go further.

I know but i was already saying it and I got distracted.

Z/2 is modular arithmetic 2.

did you mean Z[1/2]?

Akiva Weinberger

I meant all the stuff in that set divided by two

Typhon

oh

well Set/n is modular arithmetic notation

as in set element = _ (mod n)

Akiva Weinberger

I've usually seen $\Bbb Z/n\Bbb Z$ or $\Bbb Z_n$

Typhon

22:51

oooh

maybe it was subscript

darnit

yer right

Akiva Weinberger

The stuff in $\Bbb Q[\sqrt b]$ are all quadratic stuff (roots of possibly non-monic quadratics with integer coefficients)

Typhon

anyways, my problem is that i can prove it (and have) for quadratic polynomials. What now?

Akiva Weinberger

so the same logic applies, no?

Just with $a$ not necessarily equal to $1$.

We still have that $b^2-4ac$ is never $3$ mod $4$.

Typhon

uuuh

wat

I wish to prove it for non-quadratic polynomials that are monic

Akiva Weinberger

Oh

Typhon

22:54

not non-monic quadratic polynomials

Akiva Weinberger

Whoops

Got it

Typhon

slaps @AkivaWeinberger in the back of the head

brains dude

Akiva Weinberger

Ow!

Typhon

:-)

Akiva Weinberger

You have no right to do that!

You're not Ted!

Typhon

22:54

I watched NCIS

that gives me the right

[NCIS] - Headslap Compilation

►

Akiva Weinberger

—So that's why you want to show that "root of quadratic (in $\Bbb Q[x]$)" and "root of monic (in $\Bbb Z[x]$)" imply "root of monic quadratic (in $\Bbb Z[x]$)"

Typhon

um what

John Locke

Hey friends! I'm back. With another proof problem if you guys are available to help or want to take a break from your current math problem :) math.stackexchange.com/questions/2363237/…

Typhon

no

John Locke

Okay.

Typhon

22:56

that's not what I seek to prove

Akiva Weinberger

@JohnLocke He was replying to me

Typhon

also no @JohnLocke

John Locke

ahh sorry for interjecting!

Typhon

@AkivaWeinberger is both an option?

John Locke

Oh.

Akiva Weinberger

22:57

Maybe later

Typhon

yeah maybe later

John Locke

Okay, thanks!

Akiva Weinberger

@Typhon But wouldn't that imply what you're trying to prove now?

And isn't that what you were trying to prove earlier today?

Typhon

@AkivaWeinberger no I wish to show that root of quadratic with rational coefficients and root of monic with integer coefficients (irregardless of being in Z[x]) implies being root in monic with integer coefficients as that completes the proof.

you were slightly off.

I think

@AkivaWeinberger It is. I'm still stuck. You said to start with a division theorem.

for polynomials.

I got that covered. Now what?

Akiva Weinberger

@Typhon What? "Integer coefficients" and "in $\Bbb Z[x]$" are the same

Here $x$ is just the variable of the polynomial

Typhon

23:00

@AkivaWeinberger In Z[x] means it is a quadratic integer which is what we are trying to prove.

XD

unless you meant a different Z[x]?

Akiva Weinberger

Z[x] is the set of polynomials with *integer coefficients

Typhon

lmao

that makes more sense now

Akiva Weinberger

In any case

We need names

Typhon

yeah,proving that would serve as using Mjolnir to pound in plastic toy nails.

Akiva Weinberger

Call our quadratic polynomial $q(x)$ and our Monica polynomial $m(x)$

Typhon

23:02

"Monica"

Akiva Weinberger

(Leaving it in)

Fargle

mimic polynomial*

Typhon

Monic

not mimic

Akiva Weinberger

$m$ clearly has degree $\ge2$

Typhon

@AkivaWeinberger of course.

Fargle

23:03

>thinks I'm serious

Akiva Weinberger

@Typhon Clearly you missed the conversation about this from earlier

Simply Beautiful Art

Let us define $f(x)$ to be the Dirichlet function:
$$f(x)=\begin{cases}1,&x\in\mathbb Q\\0,&x\notin\mathbb Q\end{cases}$$
How do we prove that:
$$\forall x\in\mathbb Q\implies \lim_{h\to0}\frac{f(x+h)-f(x-h)}h\text{ exists}$$
$$\forall x\notin\mathbb Q\implies \lim_{h\to0}\frac{f(x+h)-f(x-h)}h\text{ doesn't exist}$$

Typhon

ooooh

are we about to use the division theorem?

Akiva Weinberger

@SimplyBeautifulArt Two cases. $h$ is rational or it isn't

Typhon

@SimplyBeautifulArt sup maam?

Akiva Weinberger

23:04

For the first one at least

I think I see how to do the second one

Simply Beautiful Art

@AkivaWeinberger I don't know. Taken from Wikipedia

Akiva Weinberger

The idea is that irrational+irrational might be rational and might be irrational

@SimplyBeautifulArt Don't know what?

I was saying how to do the first one

Simply Beautiful Art

@AkivaWeinberger Oh, nvm

Akiva Weinberger

@Typhon Yeah divide $m$ over $p$, remember that they have a common root

Simply Beautiful Art

Ah yes, the first one is easy, I do see it now

Akiva Weinberger

23:05

Let's call it $\alpha$, actually, why not

The irrational we're thinking about

Typhon

@AkivaWeinberger I'm not sure how dividing is helpful?

Akiva Weinberger

What does it get you?

Remember the formula for the division algorithm

It's like $m(x)=A(x)p(x)+R(x)$ or something, right? @Typhon

Typhon

let n be the difference of the orders of m and q. Then q = x^nm + q-x^nm?

@AkivaWeinberger yeah....

Akiva Weinberger

And the degree of $R$ is less than the degree of $p$

Typhon

yeah....

Akiva Weinberger

23:07

which means it's either linear or constant

First I want to say that $R$ has to be zero

Typhon

hold on

let me think for a moment here

Simply Beautiful Art

If $x$ is irrational and $h$ is rational, then
$$\frac{f(x+h)-f(x-h)}h=0$$

Akiva Weinberger

@Typhon You do that. I'll walk home and catch you in 10-15 minutes

Typhon

i'll be gine then

i think I screwed up in my division theorem proof

I proved there is an r with order lesser than q for one of my cases

f********

Simply Beautiful Art

If $h$ is irrational and $x+h$ is rational, then $x-h=(x+h)-2h$ is irrational, so
$$\frac{f(x+h)-f(x-h)}h= \frac1h$$

So as $h\to0$ here, the limit fails to exist

Typhon

23:11

@SimplyBeautifulArt examining funky derivative operators I see.

Simply Beautiful Art

:P

Typhon

@SimplyBeautifulArt That's my forte, not yours.

don't steal my thunder.

Simply Beautiful Art

:P
:P

Typhon

overuse of emojis is my thing as well

stop stealing mah thunder

Akiva Weinberger

23:31

..)

Julius

Hi. I'd like to find such $f$ that $f(x)f(y)=g(|x-y|)$ for some nonconstant $g$ and all natural $x$ and $y$, including zero. Any suggestions? Perhaps there isn't one

Simply Beautiful Art

@Julius If it holds for all $x,y$, just set $y=0$.

To find $f(0)$, set $x=y=0$

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$Mathematics$ Mathematics