@Secret but you can take a graph and make an equation for it

@AlexKChen what do you mean. There's ways to know exactly what a graph is doing just by looking at the polynomial equation

No I mean:

This is number heist. Give me all of your countable numbers!

@SohamChowdhury Have you watched that movie

it's pretty good

@Dodsy If I just give you this graph and not the equation, it is in general very difficult to see the coefficients and what powers of x are involved

this is because all the terms of powers of x are all superimposing together, and some kind of deconvolution (if it exists) is needed to separate it back into its components

@SohamChowdhury I probably never talked to you about your tastes, but might have paid attention if you were talking to someone else

Well you can factor it

But yeah quintics are tough

there is so much good cinema that I've "put on hold", but I have no idea when I'm postponing it for

and factorisation is in general hard for anything above cubics

Well you do long division of polynomials

postpone them all and watch Tarkovsky

when we die

@TheGreatDuck $\alpha < \omega_1$

@Secret can you factor a 4 digit number in base 10?

@Dodsy You're surely a high school student ?

@TheGreatDuck prime factors? Not without brute forcing it

@TheGreatDuck Why this question ?

can you do it easily if the factors are small enough?

Yup

@AlexKChen that doesn't really matter, I'm right.

actually, that's a good idea. I need something to do tomorrow

@AlexKChen plug 10 into a polynomial and factor the number. That will generally give you the factors if there wasn't too much carry over.

@Balarka Stalker? I've actually had that downloaded for a while

Good choice

(I've had that, and Synecdoche: New York)

Stalker, Solyaris, Nostalghia - three of my favorites from him in no particular order

@TheGreatDuck Yeah, I know that and a few polynomial tricks. But how that's related ?

i haven't watched that other but it's on Mike's top 5 list

@AlexKChen someone mentioned that they couldn't factor above 4 degree polynomials

Well, good luck factorising a general polynomial of degree $\geq 5$.

lol

that's impossible

Not exactly that. You can't find a general algorithm to express the roots of polynomials with big degrees, or that's what I am supposed to memorise without proof.

cyclic in the streets, nonsolvable in the sheets

err

given $\alpha : A \to B$ rings, for every $B$-module $N$ there is an $A$-module $\alpha_\ast N$.

i am not going to try and produce a mental imagery for that tagline

$\alpha_\ast (M \otimes_B N) \cong \alpha_\ast M \otimes_A \alpha_\ast N$ seems reasonable

@Danu: For higher codimension, the second fundamental form is a normal-bundle valued thing, so saying that two manifolds have "the same" second fundamental form requires care. But there is such a theorem for submanifolds of $\Bbb R^n$ (or, I imagine, of any space form — but for general ambient manifolds, there aren't any motions!).

@AlexKChen if a polynomial has a leading coefficient of 1 and all integer coefficients and a root at a rational number, then must that root also be at an integer number?

greetings, a Balarka

@Soham should be true

hi @Soham

Hi, a @Ted

hello, @Ted

good to ... encounter you

I'm sure it's a brief encounter.

isn't it early-ish where you are?

No.

@Balarka, you know that movie?

I have heard of it

but not being a movie buff I haven't seen it

You're way more of a movie buff than I am.

i just love the mind melting things

yeah, call me weird

@Demonark @Astyx: "The rivalry between the weather forecasters was ..." (7/2) NETPAGIUH

@TedShifrin that was a nice movie, romance right?

well, we've known you're weird for quite a while

@TheGreatDuck the question sounds good, and I think yes. But why are you asking ?

back from the 40's, Zee ...

@AlexKChen because you're asking about something that is very broad. How many roots/solutions does an Nth degree polynomial have?

@TedShifrin I seen it on that channel that shows old movies, I remember the ending when she realized her husband is a great man

I haven't seen it in ages, Zee

my favorite romantic drama is The Fly

LOL

@TheGreatDuck I have memorised the answer to be n, but I have no idea why.

...

x^2 = -1

^^how many solutions. Identify them.

I'll be back later ...

bye

@TedShifrin :o

bye

@TheGreatDuck Alex has a good point, and is not worthy of ellipsis. That an n degree polynomial has n roots (upto multiplicity) over C is a nontrivial theorem of mathematics.

It is """obvious""" but there is no simple proof.

Obvious? O_O You aren't Ron Maimon

@BalarkaSen yeah, but they seem to indicate that they don't believe it is true.

Hello, given is plane which contains points Q(0,0,-3) and R(1,0,-1). That plane is tangent to surface given by z = xy - 1/x in point P. How one can find P's coordinates?

[Random] Consider instead of taking middle 1/3 of [0,1] repeatly to make the cantor set, take middle 1/n each time. Investigate what happens when $n\to \infty$. Do we get y=x or a nonwhere continous strictly increasing function

@Dartek12 learn derivatives

come back then

I am not saying I don't believe it's true, but that to prove Olympiad question in polynomials, I have to use FTAlgrbra without having any idea how it works.

Well learn the proof

Which is pretty demotivateing.

the [random] is hwo I stumble myself upon egyptian fractions

@TheGreatDuck I know derivatives, but i've got some probles with this one, shall I be looking for dot product of grad z and QR vector that is equal to zero?

In various points of mathematics education you have what we call "black boxes" - things you have to learn to use that require too much background to actually prove / understand the proofs. For you, right now, it's FTA. For me a few years ago it was a topic in algebraic topology called "spectral sequences". The point is that eventually, you can learn them, but sometimes you have to wait. :)

Dtldarek's fantastic answer here: math.stackexchange.com/questions/1285186/… (Most others are crap) suggesting me to not learn the proof now. So I think I should wait.

@Dartek12 i don't know. You mentioned a tangent plane.

@AlexKChen mb.math.cas.cz/full/135/1/mb135_1_5.pdf

@MikeMiller I find the idea completely crap that you should use a proof without understanding it, in mathematics. It messes up with my brain.

hi chat

You either learn a theorem or you don't learn it. There's nothing intermediate between them and there's no point complaining.

Sorry about that.

Of course that's true.

fine

go learn set theory

and ZCF

and start by proving the properties of real numbers

Hello. Is there a non-empty variety without a free algebra?

is it? I couldn't make it past the first section

but that might be because I don't have much patience for logic and set theory

@AlexKChen I was hoping you might take it from a grad student that this happens through your whole life. You do learn, and sometimes usage helps you learn.

What your last line actually tell ?

I was very similar once upon a time. As a new convert, it's pretty hard to forgo the zealous "math is proof and I can't compromise on the rigor by skipping the proof" thing though.

He means you can sometimes not learn the proof of a theorem and use it. So you end up learning how to use a theorem instead of how to prove it.

@AlexKChen what is the definition of the limit?

(So you do learn something instead of learning nothing)

Perhaps it's part of math "growing up".

some theorems have proofs that never come in handy again

yeah i still don't know the proof of Sard's theorem

oh I guess Daminark told me that a few days ago

Sard's theorem was what i had in mind lol

@SohamChowdhury I agree

but of course i have been on the extreme end of nonlinear learning

everyone just prove everything!

including the riemann hypothesis

Sorry, but I can't understand the point of half of @TheGreatDuck's questions.

@BalarkaSen Ctrl-F "tendrils"

That quote's a classic !

@AlexKChen the limit is needed to prove the fundamental theorem of algebra

Oh, oh, I have heard MM use the tendril analogy

I have seen the exact word "Ctrl-F'ing classic" somewhere, pretty sure.

and go off on a tangent on hyperbolic geometry

@BalarkaSen MM ?

@SohamChowdhury Convert ?(!). Also, not exactly the rigor - but also the underlying intuition behind it.

@Alex do you know Liouville's theorem?

@AlessandroCodenotti Nope. (High school student)

@AlexKChen perhaps it was me

Nevermind then, using that is the easiest proof of the FTA that I know :(

@Soham our cm's saying something on tv. i have serious trouble digesting the english

there's an "almost algebraic" proof of fta due to artin that's quite nice

Redo Mandarin, @BalarkaSen ?

You still need an analytic lemma in the Galois theoretic proof, right?

yes

Aluffi wisely says, "You can't escape using the completeness of $\Bbb R$"

Just saying, don't visit my Mathematics profile. Your computer will hang. (Really)

NoScript, friendo

bahahaha

There's a great proof of the FTA using winding numbers

Essentially, if you have a loop in the complex plane that doesn't go through the origin, the "winding number" tells you how many times that loop goes around the origin.

(Can be negative if it goes around it clockwise)

(Also, the loop is allowed to pass through itself)

That's my usual proof.

The important thing is that, if you deform the path, the winding number doesn't change (as long as the path doesn't go through the origin at any point during the "deformation")

Hull0

So, speaking to my original question, the answer I'm getting is this: "Memorize the proof of fundamental theorem of algebra (or read it somewhere), and then while manipulating polynoimals, push symbols around and keep doing it and keep doing it until some centuries years later you build some good intuitions. " Right ?

no

learn number theory

:p

The problem is that you probably don't have the background to read the proof of the fundamental theorem of algebra. But otherwise yes.

i don't really know to be honest. Just continue in your classes and you'll get there.

I don't have the background to read the proofs.

O.O

Yeah, I know some (very very) elementary number theory. It's useful and provides very interesting motivations in some contexts, but in most cases, I don't see the necessity.

@AlexKChen encryption

The proof of the FTA that I started outlining is very visual and interesting in its own right, to be honest

I'm back, I'm sure everyone is thrilled about that.

To my understanding, quite a lot of algebra is motivated by number theory. Also, it's just fun

I like number theory.

@Akiva which proof? Winding numbers?

Yeah.

Yup, some number theory questions are really fun, because you can see it in lot of ways.

@EricSilva Hey man

Ah, that's a good proof, I think it's pretty illuminating

@AlexKChen Do you know what the path $e^{ix},~0\le x\le2\pi$ looks like in the complex plane?

If you're willing to black box the claim that a polynomial is an open mapping, you can also do it that way

Is that approach via the argument principle? @akiva

Uni cir in com pla @Aki ?

@AlexKChen that's not why number theory is fun. It's cause you can prove a lot of stuff in it with little higher level math. :p

@AlexKChen english please

@AlexKChen Cor.

@Daminark What's open mapping ?

(Correct)

The image of an open set is open

@Daminark Why do we need that?

ohi @Dodsy

So you're trying to find the infimum of $|p(z)|$

@AlexKChen Now, what about $e^{inx},~0\le x\le2\pi$, for $n$ an integer?

Think of this as a path in the complex plane

The point $e^{inx}$ moves about as $x$ changes

Outside of some large enough disk, you can guarantee that $|p(z)|$ is huge, so the infimum now becomes a minimum attained inside the disk. Assume that this polynomial has no zeroes, then $0 < \mu = \min |p(z)|$

What's nice about the argument principle version of this is that it provides a practical algorithm for root-finding

@Daminark That's a completely different proof, isn't it?

@AkivaWeinberger Speedy rotation along the unit circle ?

@AlexKChen Yes; specifically, going around the unit circle $n$ times.

Now, the thing is, if you go to the image plane, you get that the image of the polynomial is somehow "tangent" to the disk of radius $\mu$, but that geometry is impossible because polynomials are open mappings

(Or, if $n$ is negative, going around the unit circle $|n|$ times in the opposite direction)

Oh yeah it's completely different

Daminark is a troll?

:( no

;)

Write down the relevant contour integral, compute it numerically, and read off the number of roots

@AlexKChen So, now, consider the function $f(x)=x^3$. I want to put in a path as an input, and get a new path as output

Ooops sorry gotta go @AkivaWeinberger.

Specifically, I'll input $Re^{ix},~0\le x\le2\pi$. And I'll get $f(Re^{ix})=R^3e^{3ix},~0\le x\le2\pi$ as output. @AlexKChen

If you know that there's only one root in a region, you can then modify this approach to get the exact location

@AlexKChen Oh, OK, maybe I can finish this conversation later

And Balarka

How are you, son.

The point will be, "if I apply an $n$-degree polynomial to a large circle, I'll get a path with winding number $n$. So, if there are no roots, the same should happen for a small circle. This will lead to a contradiction"

@Dodsy you are right Daminark is 100% a troll

But I'll flesh out that argument later when you have time

gasps @EricSilva

:o)

@Dodsy more or less pretty much literally almost fine

Go back to your bridge daminark.

ik him in real life i can confirm

@AkivaWeinberger I think this would probably be hard for most people to see in such a quick discussion

I know :(