Mathematics - 2013-10-16 (page 1 of 3)

Pedro Tamaroff

00:01

@DanielRust Dude, not cool.

KAY. Daniel.

Daniel Rust

@PedroTamaroff It was directed at a concept not a person. I don't think any reasonable person would take offense to that.

Pedro Tamaroff

@TedShifrin Read today what the tensor product of two vector spaces is.

Wonder what others definitions are there.

anon

now do the tensor product of two algebras

Ted Shifrin

Cool ... LOL@anon

Pedro Tamaroff

@anon I called you a while ago.

Ted Shifrin

00:06

Can't he warm up on rings?

Daniel Rust

@PedroTamaroff That was the biggest hurdle for me when I started looking at TQFTs

Pedro Tamaroff

@DanielRust TQTISUFYIUYF whut?

anon

@PedroTamaroff I haven't seen breaking bad.

Pedro Tamaroff

@anon Ha, not that!

Ted Shifrin

@Pedro: Come grade my exams. I'm tired.

Daniel Rust

00:07

topological quantum field theories. It's not as exotic as the name suggests :P

Pedro Tamaroff

I think I have a proof, I think, of the following.

@anon

A matrix $D$ is diagonal if and only if it commutes with every polynomial in the sense that $(P(A))_{ij}=(P(a_{ij}))_{ij}$.

@TedShifrin What are you grading?

@TedShifrin You can buy some Skittles and align them at each exercise and eat one for each exercise you grade. You know, positive reinforcement!

Ted Shifrin

Exam on derivatives, continuity, linear equations, rank, etc.

Daniel Rust

@TedShifrin sounds like a mixed bag of concepts

Ted Shifrin

LOL @Pedro ... You want me fatter for our tennis match?

Pedro Tamaroff

@TedShifrin Rainbows don't fatten! They bring joy.

Ted Shifrin

00:10

Well, it's 1/3 our course, @Daniel

Daniel Rust

@TedShifrin I guess we just modularise our courses a bit more (maybe not a good thing)

Pedro Tamaroff

@anon Halmos defined the tensor product $V\otimes W$ as the dual of the set of all bilinear forms in $V\oplus W$.

anon

that's good

Pedro Tamaroff

@anon Oh! Wasn't expecting that. =)

@anon He said there is also a more "formal" viewpoint.

That is, a more symbolic definition.

Ted Shifrin

Go universal mapping property!

anon

00:13

there are UPs of course, but the symbol-crazy definition is as a quotient of the free vector space generatd by the underlying set of $V\times W$ by the appropriate bilinearity relations

Daniel Rust

Universal properties tend to be the easiest definitions (sometimes not as illuminating)

Ted Shifrin

@Daniel: Mine is a somewhat rare course that integrates linear alg, multivar cakc, and analysis.

@Pedro can get tenser and tensor ...

Pedro Tamaroff

@TedShifrin HAHAH.

Daniel Rust

@PedroTamaroff Not long now til you're working with Monoidal Categories!

anon

there is also a quantum-mechanical view of the tensor product

Pedro Tamaroff

00:15

@anon I have to eat now, but quick question.

anon

if X and Y are sets of states of two disjoint physical systems, then the free vector spaces generated by X and Y are the possible superpositions of states, and the tensor product represents the superpositions of states of the composite system

Pedro Tamaroff

The notation for $[x,y]$ as the "bracket" between the linear form $y$ and the vector $x$, i.e. $[x,y]=y(x)$; has been replaced by $\langle x\mid y\rangle$; right?

Not working.

Kevin Driscoll

@DanielRust Speaking of QFTs.... are you at all interested or have heard about this somewhat recently announced result for N=4 super yang-mills?

anon

@PedroTamaroff <y|x>; things on the left act on things on the right

Kevin Driscoll

Although this is a CFT, dunno if that has anything to do with TQFTs

Daniel Rust

00:18

@KevinDriscoll there's also some nice discussion about the 'amplituhedron' on MO... I can follow nearly none of it :P

Kevin Driscoll

@DanielRust One of my professors was one of the pioneers of periodic orbit theory for QFTs (he does nonlinear dynamics now) and he seems incredibly skeptical

Pedro Tamaroff

@anon Halmost writes it the other way around.

Kevin Driscoll

I don't know anything, so I default to skepticism as well

Pedro Tamaroff

Weird, because one can naturally read "y evaluated at x".

With <y|x>.

anon

well, some authors are idiosyncratic

Daniel Rust

00:20

@KevinDriscoll From what I've read, it seems to have been a bit of a premature announcement, and also a bit over exaggerated

Pedro Tamaroff

@anon Me leaves, for a while.

anon

you can also write <y|A|x> for operators A

Ted Shifrin

Bon appétit, @Pedro.

Kevin Driscoll

@anon @pedro $\langle x | y \rangle$ to me means represent the object $y$ in the '$x$ basis'

Pedro Tamaroff

@anon Oh, man. I'm in for a treat. =)

Kevin Driscoll

00:21

@anon @pedro welcome to physicist who knows nothing about rings or hilbert space tries to do quantum mechanics!

anon

heh heh

Kevin Driscoll

$A |y\rangle$ $\langle y |A$ mumble mumble.....self-adjoint operators.....same thing....mumble mumble

Ted Shifrin

I leave y'all to your mumbo jumbo.

Kevin Driscoll

@Ted does not approve of our algebra

Ted Shifrin

I don't mind kets.

Daniel Rust

00:24

Sounds like a similar gripe i have with the word 'functional'..... it's just a function!

anon

it's pretty fun using string diagrams

Kevin Driscoll

@DanielRust From the physicists perspective, it was certainly over-hyped. No physicist actually cares about N=4 super yang-mills

Anthony Carapetis

I thought mumbo-jumbo was synthetic non-commutative geometry over reductive n-topoi

anon

@TedShifrin so you're not a bra-burner?

Ted Shifrin

smacks @Anthony

Anthony Carapetis

00:25

ok I deserved that

Ted Shifrin

Not exactly, @anon. <grin> you?

Kevin Driscoll

@anon does it make you angry if I write $\langle x | x^{\prime} \rangle = \delta(x-x^{\prime})$?

anon

not at all

Kevin Driscoll

okay, just checking

Not sure what I'm allowed to say with delta 'functions' around here

Daniel Rust

it's a distribution really isn't it?

(not done much physics)

Kevin Driscoll

00:28

It is

Don't ask a physicist what a distribution is. They don't know

Daniel Rust

haha

Anthony Carapetis

it's what they call things they want to call functions to stop the mathematicians from yelling at them

Ted Shifrin

In math it's confusing because there are different meanings if distribution. Not as bad as "normal."

anon

just read two consecutive sentences in a text that both started with "But ..."

Daniel Rust

some of our naming conventions are bad.... but then you get some awesome ones like BanAnaMan :)

Kevin Driscoll

00:30

@Anothony but then there are 'tempered distributions' that any physicist would always call a function

Daniel Rust

Well strictly speaking I think distributions are functions.... just not between the spaces physicists think they are

Kevin Driscoll

@DanielRust Certianly not on the Hilbert space

Daniel Rust

@KevinDriscoll I think wikipedia does a good job of defining a distribution as a linear map from the space of 'test functions' on your space to the reals.

It does require quite a bit of topological groundwork though

Pedro Tamaroff

@anon Eye twitch.

Mariano Suárez-Alvarez

Hi all

Pedro Tamaroff

00:38

@MarianoSuárez-Alvarez Hey.

Daniel Rust

@MarianoSuárez-Alvarez evening

Pedro Tamaroff

@MarianoSuárez-Alvarez I was telling the guys I picked Halmos' book on "Finite Dimensional Vector Spaces". Heh, the title is peculiar.

Mariano Suárez-Alvarez

why peculiar?

Kevin Driscoll

@DanielRust I've never understood what that really means. Or rather, I know what it means but I don't see how it fits into what I actually do. If $f: \mathbb{R} \rightarrow \mathbb{R}$ then when $f$ appears in an equation it is clear that the function takes some (presumably physical) inputs and gives some answer

Pedro Tamaroff

@MarianoSuárez-Alvarez Most bookes are just titled "Linear Algebra."

Mariano Suárez-Alvarez

00:40

boooooring

Pedro Tamaroff

@MarianoSuárez-Alvarez HAHA exactly.

I like the title.

Kevin Driscoll

but for $\delta(x)$ the case is very different. It maps test functions to the reals. But these inputs have nothing to do with the physical aspects of the problem. They can be arbitrary, and usually not such

input are actually ever input into the relationship

Bitrex

Well, I accomplished nothing today. At least the Red Sox won.

Kevin Driscoll

@DanielRust So I realize that distributions don't actually have domain $\mathbb{R}$, but I don't understand what their actual domain means for physics

Pedro Tamaroff

@MarianoSuárez-Alvarez Halmos proves that a vector space is reflexive iff it is finite dimensional, but then says that "...[this] would shock most of the experts on the subject. The reason is that the customary and fruitful generalization of the concept of reflexivity to inf dim spaces is not the simple minded one given in (h)" (h) is the assertion that every element of $V^{\ast\ast}$ is an evaluation.

Mariano Suárez-Alvarez

00:44

Indeed

Daniel Rust

@KevinDriscoll I think it's just notational in order to make everything rigorous. You can take the usual idea of a function and map everything in to the framework of distrubtions. So you only gain computational ability, and lose, I guess, some intuition about what a function is.

Pedro Tamaroff

@MarianoSuárez-Alvarez What is the "the customary and fruitful generalization of the concept of reflexivity"?

Mariano Suárez-Alvarez

One rarely studies infinite dimensional vector spaces

They are usually endowed with a topology

anon

replace dual with "topological dual" in the definition

Mariano Suárez-Alvarez

and one restricts attention to the topological dual $V^'$, the set of continuous linear functionals

Pedro Tamaroff

00:45

@MarianoSuárez-Alvarez Aha...?

Ah.

@MarianoSuárez-Alvarez \prime

Mariano Suárez-Alvarez

$V'$ itself has a natural topology

so one can consider $V''$

Kevin Driscoll

@Daniel Yeah and I think this is what most physicsts/physics students do. I still don't understand it from either perspective though

Mariano Suárez-Alvarez

and we say that the topological vector space $V$ is reflexive if this $V''$ is isomorphic in the usual way to $V$

Pedro Tamaroff

@MarianoSuárez-Alvarez OK.

Mariano Suárez-Alvarez

the idea being that there are usually MUCH less continuous linear functionals that linear functionals

Daniel Rust

00:47

@KevinDriscoll I think maybe I'm too out of my depth to offer any insights :).

Mariano Suárez-Alvarez

sothat $V'$ has some hope of beeing of the same size as $V$

(in the usual, purely algebraic case, $V^*$ is always larger that $V$ if $V$ is inf dim; you can find a few proofs of this on MO)

Pedro Tamaroff

@MarianoSuárez-Alvarez Right. I have been told that in general if $B$ is a basis for $V$, then $V\simeq K^{(B)}$ whilst $V^\ast\simeq K^{B}$.

Mariano Suárez-Alvarez

no one alive says whilst

Pedro Tamaroff

@MarianoSuárez-Alvarez =D

Kevin Driscoll

@MarianoSuárez-Alvarez @PedroTamaroff Is undead, so no problem there

Daniel Rust

00:50

whilst we're on the subject of words we no longer use.... I'm bringing back hitherto

Pedro Tamaroff

@MarianoSuárez-Alvarez Rumour has it that the new Algebra I programme has scratched proofs of statements. Is that true?

Mariano Suárez-Alvarez

thitherto is cooler

Daniel Rust

hitherto is a great word :D

Mariano Suárez-Alvarez

@PedroTamaroff If they did, they did not tell me

Pedro Tamaroff

@MarianoSuárez-Alvarez Play the fool. =D

Bitrex

00:51

don't the Brits use whilst

Mariano Suárez-Alvarez

whitherto durst we go?

the Brits are not alive

don't tell them. though

Bitrex

:(

Daniel Rust

It's true, I am dead

Pedro Tamaroff

@Bitrex My english eduaction was mostly brit english.

Mariano Suárez-Alvarez

better dead than undead

Bitrex

00:53

@PedroTamaroff Why don't you speak 'MURICAN

Pedro Tamaroff

@Bitrex I don't have an accent, if that is what you're wondering.

@MarianoSuárez-Alvarez Right. All this Walking Dead fanaticism has been a problem.

Mariano Suárez-Alvarez

@PedroTamaroff but it is not obvious that $K^{(B)}$ and $K^B$ are not isomorphics: therein lies the rub.

See mathoverflow.net/questions/13322/… for a couple of arguments

Pedro Tamaroff

@MarianoSuárez-Alvarez Hehe, I see.

Mariano Suárez-Alvarez

and thereupon I flee ye all to conjure dinner

Pedro Tamaroff

@MarianoSuárez-Alvarez The naïve argument would be that the former has way less elements than the latter, yes?

@MarianoSuárez-Alvarez HAHAHAHA

Mariano Suárez-Alvarez

00:56

Yup.

Pedro Tamaroff

Stahp.

Bitrex

@PedroTamaroff This is how to speak 'murican: youtube.com/watch?v=jXhQhd_vq5U

Pedro Tamaroff

@Bitrex This.

@MarianoSuárez-Alvarez Bye byes.

Bitrex

@PedroTamaroff Sounds like a Southern Kermit the Frog.

Pedro Tamaroff

01:25

Does anybody know the origin of the symbol $\S$?

Ethan

someone got bored and started making squiggles

Anthony Carapetis

@pedro en.wikipedia.org/wiki/Section_sign

Mariano Suárez-Alvarez

Section sign

The section sign ( , HTML §, TeX \S) is a typographical character used mainly to refer to a particular section of a document, such as a legal code. It is also called "double S", "hurricane", "sectional symbol", "the legal doughnut". The likely origin of the section sign is the digraph formed by the combination of two S glyphs (from the Latin ). When duplicated, as §§, it is read as the plural "sections" (e.g. "§§ 13–21"), much as "pp." (pages) is the plural of "p.". It is frequently used along with the pilcrow (¶), or paragraph sign. Like the dagger (†) and double dagger (‡)...

Pedro Tamaroff

@MarianoSuárez-Alvarez I have to add it to my list of "I cannot write it down", along with $\zeta$ and $\xi$.

At least in "I cannot write it as I'd like." Heh.

Mariano Suárez-Alvarez

Usually, peoplee write it like a integral-over-a-curve sign

Pedro Tamaroff

01:31

@MarianoSuárez-Alvarez But, it is a double S. Fava can write $\xi$. =)

@MarianoSuárez-Alvarez What is the translation for "brackets"?

As in $\langle \; \varphi\mid x\;\rangle$, $\varphi\in V^\ast,x\in V$.

Mariano Suárez-Alvarez

corchetes

Pedro Tamaroff

@MarianoSuárez-Alvarez No se habla de "Cors" y de "Chetes", no? =P

Mariano Suárez-Alvarez

god no

Pedro Tamaroff

@MarianoSuárez-Alvarez HAHAHA.

chubbycantorset

Is anyone here familiar with graph theory?

Bitrex

01:44

@PedroTamaroff Did you ever watch Destinos? No of course not, you grew up speaking Spanish, I guess. It was a telenovela series that we watched in high school in their vain attempt to get us to learn the language.

You might get a laugh out of it: youtube.com/watch?v=LjSoOPkuBDM

Pedro Tamaroff

@Bitrex I'll take a peek in a while. Listening to Lenny.

chubbycantorset

No? :(

Pedro Tamaroff

@chubbycantorset Not me.

@FernandoMartin Bop.

Fernando Martin

How's it going @PedroTamaroff?

Pedro Tamaroff

@FernandoMartin Not bad.

Enjoying Halmos' exposition.

You?

Fernando Martin

01:54

Just came back from class

Pedro Tamaroff

@FernandoMartin Number Theory?

Fernando Martin

and Projective Geometry

Potato

Why do people frequently use the typewriter font with grey background for quotes? It looks horrible. Why not just use quote marks?

Pedro Tamaroff

@Potato ?

Potato

@PedroTamaroff Yes?

Pedro Tamaroff

01:55

@Potato Where?

Potato

Most recently on the comment by Izkata on the question here: math.stackexchange.com/questions/527248/…

But I've seen it in many other places.

Pedro Tamaroff

@Potato I do it because it looks nicer, to me, than italics.

Doesn't work here. =P

Potato

It looks very weird. Why not use quote marks? That's what they're there for.

Pedro Tamaroff

@Potato Well, there's a word for that.

$\Large{\text{Subjective.}}$

Potato

Oh, sure. It's just something I've never seen elsewhere, and it seems very out of place to me. Just registering my displeasure.

Pedro Tamaroff

02:00

@Potato Displeasure registered =)

On the other hand.

Grey background plus italics looks nice.

Potato

I agree it looks fine in questions. Without the Courier font it's much less offensive.

Bitrex

02:14

Anyone here know any German? What does "Einzelhandelskaufmann" mean?

Is it something obscene :(

Google Translate is coming up empty

Kevin Driscoll

damn agglutinating languages....

Pedro Tamaroff

@Bitrex Google gives "Retail merchant."

Bitrex

Ah

for some reason it wasn't coming up for me

Pedro Tamaroff

I think you should split as "Einzel-handelskauff-mann".

Bitrex

Thanks

Pedro Tamaroff

02:17

Kauff means "buy".

Mann means "man". =)

Einzel has to do with "one".

Bitrex

I think I may have come up with a way to do that river problem with Euler-Lagrange that isn't well, wrong

Ted Shifrin

@Bitrex: I've spent a lot of time on the problem today (including coding it in Mathematica to see what the solutions look like). I think the only interesting goal is to have a geometric interpretation of the solution. There won't be a closed form. I think one of the answerers gave a Snell's Law interpretation that seems correct and as informative as possible.

@Pedro: You should practice writing your various "squiggles" (that's what I call $\xi$ and $\zeta$ in class)

Bitrex

02:40

@TedShifrin The idea I had is that there's a way to modify the Euler-Lagrange equations so that one of the endpoints of the integral you want to minimize is not prescribed, but is subject to a constraint

Ted Shifrin

I don't think of E-L as a natural setting for piecewise-linear functions.

Bitrex

Yeah, probably not.

Though you can derive Snell's law from E-L.

Ted Shifrin

Yes, true.

Kevin Driscoll

THe only E-L interpretation I can really think of is equivalent to the Snell's law answer

because its basically minimize the weighted arclength (which is a functional) where i say weighted to emphasize that the velocity in the different media is different

in particular the blue stuff should have "0" velocity

Pedro Tamaroff

03:24

@TedShifrin How was yer meal?

Fernando Martin

Is anybody here in the mood for some number theory?

Pedro Tamaroff

I can always try, but I don't know much about what you're studying.

Fernando Martin

I'm working on problem 2

It's almost done, it's what I was talking about yesterday

Basically I solved the problem assuming that $m|(p-1)$

Ethan

@FernandoMartin character sums?

Fernando Martin

Yes Ethan

Now, in the general case, I'd like to replace $m$ with $(m, p-1)$

this is easy to do with the RHS

but not with the LHS

Ethan

03:31

I can't read spanish

Fernando Martin

I can translate what I wrote

Ethan

Have it in english?

Fernando Martin

No, but I'll write it down

Just a sec

The exercise says: "Let $p$ be a prime. Prove that $\sum_t \chi(1-t^m) = \sum_{\lambda^m=\varepsilon} J(\chi,\lambda)$"

Where $\chi$ is a non-trivial character

Ethan

And whats $J(\chi,\lambda)$

Fernando Martin

The Jacobi sum associated to $\chi$ and $\lambda$

The exercise is almost solved, I only need a small reduction

from the general case to a particular case. I'm fairly sure it's easy

Pedro Tamaroff

03:39

@FernandoMartin Esa es la tipografía que usa Mariano.

Ethan

@FernandoMartin eh I don't think I can help you sorry, I haven't really studied that deep into this stuff

Fernando Martin

@Ethan: no problem, thanks anyway

Sí?

Es Palatino

La que me gusta mucho (y no se cómo se llama) es la que usan en las guías de Topología

acá, por ejemplo

Vaughan Hilts

Hey folks, got a question for you guys :)

Ethan

shoot

Vaughan Hilts

I've got the rational root test which works on integer co-efficents: purplemath.com/modules/rtnlroot2.htm

How can I adapt this to work with rational co-efficents?

Can I?

Ethan

03:42

Yes

Vaughan Hilts

I assume I need to try all combonations of some combonation of the factors of the ratio's

Fernando Martin

Multiply by the greatest common denominator

Ethan

@VaughanHilts Just multiply it through by a common denominator shared by all the coeiffients, the resulting polynomial will still have the same zeros

Vaughan Hilts

What if they don't have one?

Fernando Martin

They always have one...

Vaughan Hilts

03:43

Well, I guess they have to share at lest one.

Ethan

They will always have one

Vaughan Hilts

That is, they all multipled by each other.

Stupid question, forget I asked that... :)

Fernando Martin

just multiply all the denominators together

Vaughan Hilts

Yep, just realized that. How foolish of me :)

So I'd multiply them all together, then find the GCD between them, and multiply that all by the tops, right?

Fernando Martin

Let's say the polynomial is $p_n/q_n x^n + ... + p_1/q_1 x + p_0/q_0$

multiply that by $q_n\dots q_0$

Vaughan Hilts

03:47

Yup, I understand fully now. Excellent. :)

Fernando Martin

you are now left with a polynomial with coefficients in $\mathbb{Z}$ which shares the same roots as your old polynomial

Vaughan Hilts

And then I can apply my rational root test

Fernando Martin

Exactly

Vaughan Hilts

You guys are awesome :)

Ethan

You don't really have to multiply through by any coefficient though, just divide the constant term by the leading rational coefficient and proceed just as you would were they integers

Vaughan Hilts

03:51

Would that really work, though?

Ethan

yes heres an example,

Vaughan Hilts

I'm listening. :)

Ethan

@VaughanHilts nvm you would have to change it slightly

Just multiply through

Vaughan Hilts

Once they all have common denominaotrs, they're not qute integers, hough

Ethan

Once they have common denominators

you can get rid of them

Vaughan Hilts

03:56

Even with the x's attached, huh?

Well, I guess that makes sense.

Since it's all one factor attached to the numerator.

Ethan

Your just multiplying the polynomial by a constant term, it doesn't change its zeros

Vaughan Hilts

Yup, that makes sense. :)

Ethan

why are you using the rational root theorem?

Vaughan Hilts

I've been tasking with finding all real roots of a polynomial with rational co-efficents

This seemed to be the easiest route

Ethan

just use a calculator

Vaughan Hilts

04:00

It's a computer program.

I'm writing that calculator. ;)

Ethan

just because the polynomial has rational coefficients doesn't mean it will have rational roots

Vaughan Hilts

I can exclude any imaignary ones.

As long as I get the real ones.

Ethan

the test only works in finding the rational roots

Vaughan Hilts

Well, it has be rational, right?

Ethan

not necessarily the real ones

Vaughan Hilts

04:02

The only other option is irrational, I guess

Is it possible to have an irrational root?

Ethan

yes

Fernando Martin

@VaughanHilts: What roots does $x^2-2$ have?

Ethan

consider the polynomial $x^2-2$

Vaughan Hilts

Hah, well that's shitty. :)

Ethan

It has roots at $\pm \sqrt{2}$

Vaughan Hilts

04:03

Actually

I misread the question

I only NEED the rational roots :)

Ethan

@VaughanHilts if its for some practical application I think you would probably be better off finding them numerically

Vaughan Hilts

It's not, or I'd be doing it numeraically.

Fernando Martin

Then you're all set

Ethan

@VaughanHilts the rational root test is pretty brute force

Vaughan Hilts

It'd for an assignment, unfortunately.

So I need to try it for all factors in the numerators

And then try all the negative factors, as well

Ethan

04:06

school?

Fernando Martin

nah

you just need to try a few factors

Vaughan Hilts

@Ethan It's a school assignment, yup. :)

Fernando Martin

If $p/q$ is a root, then $p$ divides $a_0$ and $q$ divides $a_n$

just factor those two numbers and try all the combinations

Ethan

@VaughanHilts was thinking of applying to several schools in the UK, though im sol now sense the deadline for ucas fall apps was today lol

Vaughan Hilts

Call and see if there's still room?

In canada at least, if you miss the deadline you can still apply if there's slot sleft

Ethan

04:15

I am from California anyway, so its fine just trying to keep my options open lol

Gustavo Bandeira

@robjohn Can you take a look at this?

Vaughan Hilts

This might sound stupid.. but if I multiply all their denominators together... what do I do about the numerator? :)

I just multiply them each by every denominator before that wasn't their own, right?

Ethan

here vaughan

multiply all the denominators together to get some natural number $a$, then multiply the entire equation by a

Vaughan Hilts

The whole co-efficent?

Ethan

@VaughanHilts Say I have the equation $$\frac{x^3}{6}+\frac{2x^2}{3}+\frac{1}{2}$$

Multiplying 6,3 and 2 together to get 36

then multiply the entire equation by 36

Vaughan Hilts

04:22

Sure.

Ethan

and apply the rational root theorem

Fernando Martin

Just to be nitpicky, that's not an equation.

Ethan

what is it

a polynomial in Q

Vaughan Hilts

So I'd multiply each each term by 36/1 then

Fernando Martin

yup, a polynomial with rational coefficients

Gustavo Bandeira

04:23

@FernandoMartin Who told you..?

Ethan

@FernandoMartin trying to ease down on the terminology for vaughan

Fernando Martin

ok then

Gustavo Bandeira

@Ethan NO! You were trying to troll him! Tell the truth!

Vaughan Hilts

xD

Ethan

i should probably be doing work for one of my classes now, im just too lazy tho

Vaughan Hilts

04:27

So I'm doing -2/3 x^3 + x^2

So that means multiply everything by 3/1

Ethan

@VaughanHilts 3*1, yes

Vaughan Hilts

Sheesh, you're a monster. :)

Gustavo Bandeira

@Ethan Holy shit! You solved 400+!

Ethan

I waste too much of my time lol

Kevin Driscoll

Just earned my tumbleweed badge....... don't mind me, jus tpayin my dues

Ethan

04:34

I should put that on my common application lool

list my badges on math se

Gustavo Bandeira

@KevinDriscoll TumbleCrystalFuckingWeed.

Vaughan Hilts

Hmm, @Ethan what happens if my trailing co-efficent is zero

The factors of 0 don't exist :)

Ethan

if its a polynomial in $x$ divide the whole thing by $x$, and add $x=0$ as one of the zeros

Vaughan Hilts

Ah yes, I guess if the trailing (base constant) is 0 it'll always go through 0

Hm, still stuck -______-;

-2/3 x^3 + x^2 has soltuions 3/2 and 0

Fernando Martin

04:53

factor out 0 first

Vaughan Hilts

Multiplying across, i get a constant of 0 and a leading co-efficent of -2

Ethan

Factor out the largest power of $x$

Fernando Martin

if the constant term is $0$ then $0$ is a root

Vaughan Hilts

That much I understand.

My only other problem is applying rational root test to brute the 3/2 one

Fernando Martin

Then factor out the lowest power of $x$

the polynomial you're left with has now a non-zero constant term

and you can use the rational root test there

unless the original polynomial was $x^n$; in that case the only root is $0$

Vaughan Hilts

04:56

Ugh, so complicated in code :)

Fernando Martin

if you tokenize the polynomial and save it as an array of rational numbers it shouldn't be much of a hassle

Vaughan Hilts

Well, I have it as an array of rational numbers as is

I just need to divide out by the highest co-efficent, right?

Fernando Martin

lowest power of x

if you have an array $[q_n,...,q_0]$, with the last $k$ terms equal to $0$, just move the array $k$ positions to the right and fill the remaining positions with zeroes

Vaughan Hilts

heh, you make it sound so easy.

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$Mathematics$ Mathematics