@GFauxPas oh my, you're so cute

@Lucas my school actually has a lot of Brazilian students so maybe think about applying here

falls in love with @GFauxPas

I'm not cute, I'm smart and scary D:<

@EricSilva I'm a bit concerned about economical matters

that's fair

Is inflation as bad in Brazil as in Argentina

@AkivaWeinberger $p(\sqrt{x})$

@Typhon Not a polynomial, in general

@AkivaWeinberger darnit. wasn't sure if x being a solution made it a square

p(x)^2?

No, plugging in $x^2$ gives you $p(x^2)^2\ne0$

@AkivaWeinberger I have no idea of how economy works but AFAIK inflation is a lot lower than it was months ago, about 7%/year

@AkivaWeinberger is it possible?

Yes @Typhon

hrmm

i give up

@Typhon $p(\sqrt x)p(-\sqrt x)$

raises eyebrow

It's easy to see that it works, but less easy to see that it's always a polynomial

(It works 'cause plugging in $x^2$ gives you $0\cdot p(-x)=0$)

of course of course

but the idea is: Consider the related polynomial $p(x)p(-x)$.

im thinking

If we show that every exponent (of $p(x)p(-x)$) is even, we're done, right?

yeah sure but let me think please

:p

you don't need any of that

none of that is needed

the product of a number and its conjugate is always rational

period

those two polynomials are conjugates of each other

clever

Yeah essentially

i mean x would have to be an integer to make the appeal to conjugate

but meh

clever solution

now here is one for you

@AkivaWeinberger if $p(x) = 0$ and has rational coefficients, then find a $q$ such that $q(x^n) = 0$

n is any integer

bonus points for doing n is any rational

So the generalization

indeed

dont bother with n being any real number

that is evil

Well $\sqrt x$ and $-\sqrt x$ are the two square roots of $x$

mmhm

@Typhon It's probably impossible for any real number

go on

@AkivaWeinberger indeed

I mean, there exists a real number $n$ such that $2^n=\pi$. There's a polynomial $2$ satisfies but not $\pi$.

(Specifically log base 2 of pi)

heh

pi doesnt even exist transcendentally.

But in any case if $\zeta$ is like $e^{2\pi i/n}$,

want a hint?

then the $n$th roots of $x$ are $\sqrt[n]x$, $\zeta\sqrt[n]x$, $\zeta^2\sqrt[n]x$, etc

think about other conjugates

oh

so I would do $p(\sqrt[n]x)p(\zeta\,\sqrt[n]x)\dotsb$

beats me

i was just gonna multiply the n-tuples of conjugates

Yeah I think it's the same

it is

I think those are the other conjugates

but... finite in product

dear god

It is finite

what have thy done

I just didn't write the end :P

@AkivaWeinberger i know.

it was a statement

$\dotsb p(\zeta^{n-1}\sqrt[n]x)$

i was thinking $c_1c_2c_3 \cdots c_n$

In any case, for rationals, note that if the exponent is $\frac1m$, then just $p(x^m)$ works

heh

so if the exponent is $\frac nm$ then take the thing from before and replace $x$ with $x^m$

can you write it out?

the formula?

$p(x^{m/n})\cdot p(x^{m/n}\zeta)\cdot p(x^{m/n}\zeta^2)\dotsb p(x^{m/n}\zeta^{n-1})$

my eyes

my eyes

they are blinded by your math skillz!

It's not even that complicated of a formula

lol

@AkivaWeinberger i know, but you managed to do it.

what is zeta anyways?

It's a primitive $n$th root of unity

oh

duh

My name is N. Idiot

i thought you were pulling out the zeta function

the zeta function

No no no no no

little known fact, the zeta function $\zeta(n)$ agrees with the $n$th roots of unity for $n$ sufficiently large

What do you mean @GFauxPas

(Doesn't mean anything)

OK

where $n>2$ is an even prime

>:(

@Typhon Hm, I guess if $p(x,y)$ is a polynomial in two variables (like $1+x+xy^2$ or something)

then $p(\sqrt x,\sqrt y)p(-\sqrt x,\sqrt y)p(\sqrt x,-\sqrt y)p(-\sqrt x,-\sqrt y)$ is also a polynomial

(in two variables)

a polynomial in $x$ and $y$ or in $\sqrt x$ and $\sqrt y$?

@GFauxPas so 56! is an nth root of unity?

@GFauxPas In $x$ and $y$

Otherwise it's trivial

wolframalpha.com/input/?i=is++56!++an+even+prime%3F not sure @Daminark

I mean we know from Grothendieck that 57 is prime

If it's even, then $\zeta(57) = 56!$ might be a root of unity

Test: ${\newcommand{\wnadwnawb}{\rm sin}\wnadwnawb}\wnadwnawb$

$\wnadwnawb$

These tests are sinful

You're confusing $\zeta$ and $\Gamma$

57 is not even

and it isnt prime

wolframalpha.com/input/?i=is+57+an+even+prime%3F WA doesn't understand my question :(

@Daminark GLaDOS voice At the end of testing, you will be baked $\vphantom{and then there will be}$ cake

THE CAKE IS A FUCKING LIE!

XD

If the laws of physics no longer apply in the future, God help you.

i have a $\pi$ tho

@Semiclassical :D

and I bet that $\pi$ is transcendental.

@Semiclassical it is algebraic

You know, I've been thinking. If love gives you lemons, don't make lemonade.

Make life take the lemons back!

Get mad!

just complete the circle

I don't want your damn lemons, what am I supposed to do with these!

@AkivaWeinberger no. It's turn them into flaming lemons and hurl those fucking things back at life's house

Well, I finally took the plunge and downloaded Python.

@Semiclassical beware the serpent. hissssss

Now I just need to figure out what to use it for...

make a snake game

for the irnoy

Demand to see life's manager! Make life rue the day it thought it could give Cave Johnson lemons—do you know who I am?! I'm the man who's gonna burn your house down! With the lemons!

and put an iron in it to create double irony

I actually have a non-math application in mind.

with snakes?

I'm gonna get my engineers to invent a combustible lemon that burns your house down!

(cough)

I like to do the daily cryptoquip in the local paper every so oftne.

with the hint that R=N. (Each letter stands for another one.)

What I'd like to make in Python is a little tool for that.

"In his house at R'lyeh, dead Cthulhu waits dreaming."

Is this an involution on the alphabet @Semiclassical

Like are they switched in pairs

Nah.

solved it

It's a permutation.

it is funny

Well O is S

the pun is real

Q is O

My key into it was to read "LAEL MUQMHU" as "that people"

So is R converted to N or the other way around

R should be replaced by N.

Not the other way around.

Anyways. The tool I'd like to make is a script which loads the quiz and allow me to supply substitutions

say, replacing a black character with a green one once a substitution is added

oh dont use python for that

it isnt gui based

use a pencil and paper

nah

use javascript

I did it in Notepad, actually :P

lol

I'm stuck on trying to see what _ASA__LAN_A'S could possibly be

hint

Yeah, that took a while.

it is a place

My hint is to do the last part as soon as you can.

n + (n/2 + 1/2) + (3n/4 + 3/4) + (9n/8 + 9/8) + \cdots (3^m/2^m)(n + 1)

= n + (n+1)(1/2 + 3/4 + 9/8 + \cdots (3^m/2^m))

looking at the collatz conjecture to see if the worst case scenario must ever reach a power of 2

Last night dream is something that I am not sure if it is a maths dream or a horror dream: it begins with a proof by induction of some conjecture through the natural numbers, which corresponds to some square rooms. Suddenly, the surrounding environment flickers and fade from one to another as I look around and found myself in a small brown and red room with a bathroom on the right and the window outside is a dusty desolate place. I then explore the brown room and then suddenly the wall closed in

@Secret meh

and crush me flat in it

sounds normal?

@Secret did you die?

did it hurt?

@Secret What was the secret of the dream?

badum ch

The scene fade to black before it repeats in a slightly different way as I tried to understand what's going on. Eventually, the existence of that desolate brown red room is because the conjecture failed for a complex number $\lambda$ such that while a product of $f(1)f(2)\cdots f(n)$ commute, $f(\lambda)f(\lambda)f(\lambda)\cdot\cdot$ does not. Thus the brown red room correspond to a product string that is not rearrangeable I.e. $f(1)f(2)f(3)\cdots f(n)f(\lambda)f(\lambda)f(\lambda)\cdot\cdot$,

and each f() in this particular string don't commute

@Semiclassical OK I think I did it except the first word doesn't make sense

The rest of it does though

It is also mentioned that that weird brown red room is the result of all square rooms merged together and that the final string is some kind of infiniteth term of the sequences of fs

What'd you get for OAEBU?

SHARE

B is wrong. That's why the first word doesn't make sense.

(you already should have Y is R)

Oh

plus, what do you do with the objects at the end?

Oh duh got it

(as in, what's something a person might do with them)

yep.

Got it got it

The standard of humor for these is...not great.

Just kinda went through the alphabet there

The dream does make me think of one thing though: Is it possible for nonlinear maps to be eigenfucntions?

Not sure what you mean by that. I'm guessing you mean something other than $Ly=\lambda y$ with $y$ a nonlinear function.

In any case Semi every invertible matrix is the product of elementary matrices

One way to generalize that, though, is to write it as $(L-\lambda I)y=0$.

and those are commutators

so every invertible matrix is the product of commutators

One can generalize that by replacing $L-\lambda I$ with some polynomial in $\lambda$.

Doesn't necessarily mean they're commutators themselves

e.g. $\mathcal{P}(\lambda )y=(A_0+\lambda A_1+\lambda^2 A_2+\cdots+\lambda^n A_n)y=0$.

Wait I misspoke

I think it might be $Ly=\lambda y$ where L or y or both are nonlinear maps

I need everything to have determinant 1

Well, $y$ fails to be linear all the time.

Same argument works, though

Credit goes to the internetz

Wait why are they commutators?

Those have to have trace 0

I mean, trig functions are solutions to a linear ODE but they're not linear maps themselves.

anyways, back in a bit.

And I don't think elementary matrices always have trace 0

@Daminark Why would they?

Have trace zero

$A^{-1}B^{-1}AB$ not $AB-BA$

Trace is additive, and tr(AB) = tr(BA)

Oh that

It's been a long time since I had last touched linear algebra, but for any matrices, $A$, $B$, and $C$, such that $A \cdot B \cdot C$ is defined, is $A \cdot (B \cdot C)$ also defined?

Oh yeah sorry I meant what Semi said

@SalehenRahman Yeah

@AkivaWeinberger thanks!

Ah

This is what an exclamation mark and a question mark look like?!

Nah that's what a question mark and exclamation mark look like

Concatenation is not abelian

Oh

This is what a colon looks like:

I have now deleted all video games.

Kek

Oh @Dodsy so that's why mine are all gone

lmfao yes.

Wai tho? You could've just deleted your copies

I had been playing diablo 2 religiously the past few days.

Heheh

Diablo and religion...

and then my hardcore character died, I realized that I had spent countless hours doing nothing and had not been doing any math.

So, I am back.

Everyone make way.

pushes someone to the side to make way

How are you dami?

I just thought of a three letter word, and…

I occasionally try to log into the text MUD I used to frequent

What is text MUD?

oh wow

I thought I was a nerd.

;)

Just kidding, of course.

It's when you drop your text into this really muddy...

But most of the people I knew are long gone, and a lot of the orgs I was in are too

So usually I just log in and log back out

I was hoping one of the words in that sentence before "and" would be a three letter word @Akiva so I could deliberately interpret it as being that

Rip @Semi

@Semiclassical I know these feels, very sad.

Lots of sentences have punctuation in them

Like that one, for example ^

The word punctuation?

Yeah

Those sentences would be pretty meta.

Probably using that word wrong.

Nah I think it's right

Great! :3

Also, no matter how you write it, the word incorrectly will be spelled incorrectly

My character on there is like...430 years old or so now @dodsy

(25 days irl = 2 years in-game)

@Semiclassical 14.726027397260273972602739726027 years?!?

what were you playing?

eh. That just tells you when I created the character. My activity dropped way down after high school

@Semiclassical oh. I thought you meant you spent that many years in game

so you are 31?

Turned 30 on the 21st

So close enough

@Semiclassical ah ok

so you created it when you were 15-16?

(I could actually look up how much RL time I've spent logged in but uh

I probably don't want to know)

who cares? XD

@Semiclassical me neither

Sounds right. It was spring of freshman year of HS

it is probably more than the collective hours spent playing roblox

by everyone one the planet

#burn

I always have to work backwards to figure out my age at that point

heh

I always buy games and never play them.

@Dodsy me too

But I think that it was spring 2003? So I'd have been just shy of 16

:-(

I'd play a game but there's never any time.

A few years ago I bought every single world of warcraft game + expansion

and then never played it.

Didn't even download it.

wow

Not sure why.

heh

did you want to play it?

Probably not.

:P

oh

well I get games I want to play

you can play it for free basically anyways using private servers.

and then never have time to play them.

that makes sense.

yeah...

sure

i spend time on here yet i have no time to play games

The game I watch a lot of Youtube videos about lately (well, one guy's channel really) is Elite: Dangerous.

mostly because I like space stuff :P

cool

i dont really like watching people play games

reminds me that i dont get any free time to play them myself

:p

Well, I like it because there's typically so much investment of time required in a game like that to get to any of the interesting stuff.

so it's nice to get a window into the stuff that I otherwise wouldn't see

sorry typo: the above two $\lambda g$ should be $0$

So, I'm trying to prove that if you take the derivative operator on the space of polynomials with degree at most $n$, that the invariant subspaces are precisely the space of polynomials of degere at most $k$

That those subspaces are invariant is clear

But I'm not too sure how to show that those are the only ones

@Secret I mentioned it in passing earlier, but what you call a polynomial eigenvalue problem is also known as a 'matrix pencil'/'operator pencil.' (See for instance en.wikipedia.org/wiki/Matrix_pencil or birs.ca/workshops/2013/13w5059/files/Levitin.pdf)

I see

yay terminology :/

The linked wiki page also has a link to a page nonlinear eigenproblems, though it's not that descriptive.

@Fargle do you have any ideas?

@Daminark I've been puzzling over it. I don't have any as yet.

Okay so if it's one-time invariant, it should stay invariant if you keep doing it

So then, take a derivative enough times that you could only have at most a constant

So you have two options

Either you contain all the constants, or you contain none, so these polynomials are all of the form $a_nt^n + \ldots + a_1t$