@TedShifrin true

It's weird because PDEs is a second year mandatory course instead

@Rudi_Birnbaum -oid

The ODE course you're talking about is probably dynamical systems, basically, @Alessandro.

@Alessandro PDEs are just more interesting, I guess

OK, thats also what I thought

No, not really.

like how "void" is derived from "v"

morning

thanks-- lol

Dynamical systems is beautiful stuff that gets into manifolds/topology in the end.

@TedShifrin I'm not sure, I didn't take it :P

morningoid

nope, @Faust. It's afternoon, even here.

oh i guessed right

Here ODEs is a third year course since it requires analysis, but in the past it was kinda easy because a lot of students went in not knowing enough linear algebra, and also I think the lowest level analysis doesn't cover Arzela-Ascoli (which mid and high level do) so they often spend way too much of the class reviewing

2/e-1 is negative and that's impossible

Demonark: We're not talking about that sort of ODEs here.

Now that we have an LA class, this may change

because it can't cross the line y=x-1

mercio: Common sense should allow you to eliminate a few but not all.

1/e seemed too small :s

e way too large

Still guessing is dangerous and expensive unless you're 75% sure or something.

@Daminark having a mandatory LA class sounds like a good thing. it just shows up everywhere

anything above 1 was also impossible because if you go backwards from there you go up

OK, I'm out of here. Y'all misbehave without me.

and then i take as much time taking a guess than i would need for solving it completely lol

Oh I remember talking about integrating factors once here actually, you multiply through by $e^x$ in this case? So you have $y' + y = x$, then $xe^x = e^xy' + ye^x = (e^xy)'$

if not more

@Daminark right

Bye @Ted

See you @Ted!

bye Ted

NOw you have to remember integration by parts, Demonark ... or else just guess and correct.

bubye

and if it were 1/e it would need to cross the line x=y lower than that and that's impossible

so by elimintation i only have 2/e left

and that took me 8 minutes

But yeah so to finish this off, $\int xe^x = xe^x - \int e^x = (x-1)e^x + C$, so $y = (x-1) + Ce^{-x}$, $y(0) = C-1 = 1$ so $C=2$, meaning $y = (x-1) + 2e^{-x}$, so $y(1) = 2/e$

@Daminark I don't understand. I don't mean to offend you, but isn't integration by parts taught in the first year? I'm just curious.

We never covered much ODEs, we just followed Spivak which mentioned ODEs when talking about exponential and trig functions

hmm

Sanity check: no algebraically closed field can be ordered, right?

And it was just like, oh exp is the unique solution of y' = y, and analogous stuff for sin and cos

@AlessandroCodenotti yup, you're sane

In analysis we did a bit more ODEs but it was just proving existence/uniqueness a bunch of times + Lyapunov functions

Right after asking I realized it's obvious if one remembers that orderable fields are formally real, thanks for confirming though!

@Leaky oh oh wait integration by parts? For some reason I read integrating factor

Yeah I know integration by parts

@AlessandroCodenotti all ordered field is char zero, so everything is separable and $i$ exists, then follow the proof for $\Bbb C$

maybe use the fact that ACF_0 is complete

then you can just argue that $\Bbb C$ is not ordered.

all ordered fields admit a finite-dimensional divsion algbra (analogous to $\Bbb H$), which contradicts the fact that $H^2(G_K,K^\times)$ is trivial if $K$ is algebraically closed

Is being orderable a first order statement in the language of fields?

I guess you can bring in a new symbol for the ordering predicate, I'm not sure

I don't think it is

@MatheinBoulomenos what is $G_K$?

ah, that's just notation for the absolute Galois group of $K$

in an ordered field, every square is positive, so if it's algebraically closed, every number is positive, and so it can only have 1 element

I'm so used to that notation that I didn't remember that it isn't standard outside of NT

Using orderable iff formally real (-1 is not a sum of squares) it should be possible to express being orderable with infinitely many sentences

@MatheinBoulomenos but $K$ is algebraically closed?

@LeakyNun right

But not being orderable seems harder if there exist fields of arbitrarily large Stufe

so $G_K = 1$?

yeah

I can order the field $\Bbb Z/p\Bbb Z$ just fine

elements of $H^2(G_K, {K^{sep}}^\times)$ (for any field $K$) can be represented by finite-dimensional central division algebras over $K$

I see

@LeakyNun not in the sense of an ordered field

If $F$ is orderable the biggest orderable extension of $F$ in $\overline{F}$ is just the real closure of $F$, right?

up to isomorphism ?

Maybe I actually want $F$ ordered and the order on the extension to preserve the order on $F$

@AlessandroCodenotti maybe tell us what you actually want

@mercio yeah

@LeakyNun I just did :P

how is the real closure defined again ?

doesn't the definition of relative real closure require an order on $F$ to start with? While orderable is just for abstract fields

makes that seem unlikely

e.g. there are archimedean and non-archimedean orderings on $\Bbb Q(x)$. That's not a disproof, but somehow I don't believe it

Apparently there are many definitions of real closure among which is exactly being the maximal extensions that preserves the order so that works

@AlessandroCodenotti oh I missed that

my answer commented on the original question with just orderable

Yeah I realized orderable was too vague

@LeakyNun I asked you question about torsion is two statements are logically equivalent or not? I got answer no in stack exchange what are your thoughts about that?

don't trust me

Why are you saying like this sir?

mathonline.wikidot.com/… on this proof how does equation * follow?

I agree with equation 1, and I also think that m*(E) \leq \sum(l(I_n)).

@user100000000000000 universal properties of infimum

I wonder how absurd this pages feels to onlookers en.wikipedia.org/wiki/Picture_(mathematics)

All of mathematics feels absurd to some onlookers :P

Just so you know, the French word dessin is usually used for this, I believe.

@TedShifrin !

re Leaky

I concur with your "don't trust me"

nice :P

you agree with me :P

wait... you don't agree with me

ok this is a paradox

Maybe not quite Russell's.

@LeakyNun cool!

hi @LeakyNun - just curious, what can Lean do right now? Is it meant to be an automatic prover or a verifier right now?

it is a verifier that has many tactics (programs that can generate simple proofs)

any main highlights so far? (e.g. did it produce a simpler proof of known theorems?)

I have to physical quantities, the is $\Re(I(x))$ and the other one $\Im(I(x))$ for $I(x)=\int f(x,y)^* \nabla f(x,y)dy$, with some nice $f(x,y)$. I wonder if that rings a bell in someone? The reason I ask is I search for a connection between the two functions, exceeding that.

One of the connections I suspect, is that the zeros of the two are in some way related, however I know that they are not equal.

hi @Piggy

@Pig a main highlight would be the definition of perfectoid space

@Rudi_Birnbaum you have two physical quantities

hi @TedShifrin

@Rudi: This is confusing. So $x,y$ are real, but you're treating the two components of that real integral as parts of a complex number?

Hey everybody!

rehi Demonark

@TedShifrin x,y real but f complex

Thanks @LeakyNun - so sounds like the main highlight right now is we are able to define fairly advanced mathematical objects in computer

right

Oh, if $f$ is complex, what does $\nabla f$ mean? And what is the star?

nabla is the scalar product of the gradient with itself

* is complex conjugation

OK, now I'm lost. Is there a gradient missing in the first term?

So you're just doing $\int \|\nabla f(x,y\|^2\,dy$?

oh, that's horrendous notation.

@LeakyNun I understand the logic in what you wrote, but what plays the role of b in my question?

x could be vectors, but for simplicity do not need to be

OK, but I don't understand the integrand at all now.

@TedShifrin Multiply the complex conjugate of f with the "second derivative" of f

second derivative? Seriously? You said scalar product of the gradient with itself. That's very different.

@TedShifrin $x$ could also be a vector $(x_1,x_2,x_3)$

but for simplicity needs not

I understand that. That's the least of my worries.

(d/dx)^2 = d^2/dx^2, no?

That's not at all taking scalar product of the gradient with itself. That's $(df/dx)^2$ ...

@user100000000000000 $\mu^\ast(E+a)$

For a function of several variables, the second derivative becomes a symmetric matrix. So I am not sure what you're doing.

@TedShifrin Don't think so. $(d/dx)2 f(x) \ne d^/d x^2 f(x)$

@Rudi_Birnbaum wtf

I'm telling you that scalar product of the gradient with itself is $\|\nabla f\|^2$, not second derivatives. So your words were wrong.

SOrry

I am wrong!!

Its just the gradient!!

But now I'm telling you that second derivative becomes a matrix ... so what does it mean to multiply $f^*$ by the matrix?

no square

<--- gives up :P

lol

OK.

nothing second, no square just grad!

Sorry!!

maybe rudi meant a laplacian instead of a nabla ?

oh god

laplacian and hessian

So we're just doing $f^*\nabla f$ and getting a vector output. So $I(x)$ an ordered pair of complex functions?

I confused myself I wrote gradient but then read it as laplacian ...

and in the book it was a wronskian

I'm not sure we understand yet what's going on.

... lol

just take it as d/dx to begin with and $x\in \Bbb R$

So you don't mean the gradient of $f(x,y)$. You mean just the $y$ derivative? Just the $x$ derivative?

Is this coming from a book or paper? i'd like to see the original.

the x derivative

OR I could forget i ever saw the question.

I rewrite it

You're doing $\int f^* \dfrac{\partial f}{\partial x}\,dy$?

yes

What do we know about $f$?

bounded, integrable, differentiable,

I don't think it matters that the second factor is df/dx

@LeakyNun Why does it make sense to talk about

since you are integrating w.r.t y

Compact support? I don't know where this lives.

It matters if I want to differentiate under the integral sign, @mercio.

$f: \Bbb R^2\to\Bbb C$

@LeakyNun for all b when b is m^ast(E+a). I thought that quantity was fixed here

The integrand is basically $\frac12\frac{\partial}{\partial x} \|f(x,y\|^2$.

I'm substituting $b=\mu^\ast(E+a)$ into 2

it's called universal elimination

@TedShifrin yes thats another way writing it

This is reminding me of an energy computation for the heat equation.

It would be nice if $f$ satisfies some nice PDE or something ...

$x$ position, $y$ time ...

To be honest its an $\hat{H}f=E f$

Then when I differentiate $I(x)$ it turns into a second derivative which turns into a time derivative if I have the heat eqn.

oh hey, schrodinger equation?

Oh goody, @Semi is here!!!

lol

and $\int ||f(x,y)||^2 dy$ is the density

(No, that wasn't sarcastic)

the density is symmetric in $x$ and $y$

<--- gets a martini and hopes @Semi will figure this out instead of me.