Mathematics - 2018-09-14 (page 1 of 2)

Secret

00:01

$A-B=\varnothing$

Mancala

Countable union of disjoint closed sets is closed?

Leaky Nun

@Mancala no

Mancala

@LeakyNun Even that every point has a neighborhood that intersects only finitely many of those closed sets?

Leaky Nun

yes, even that

consider {1,1/2,1/3,...}

every point has a neighbourhood that intersects no other points

Jake S

Ok, I've returned with a bit of a silly question. I've determined that the bases for the Images of two linear maps are (0,1,0) and {(1,1,0),(0,1,1)} respectively. But how do I prove that their direct sum is the vector space of degree 2 polynomials?

Do I necessarily have an error in the second basis somewhere?

Mancala

00:16

@LeakyNun Do you know what you need for the result to be worth it?

Jake S

I definitely made a mistake somewhere I'm not realizing. Hm

Leaky Nun

@Mancala no idea

Mancala

@LeakyNun Thanks!

Leaky Nun

I mean, it's closed if it's closed, what do you want me to answer

Rithaniel

00:40

Question: Is it possible to have a union of uncountably many open intervals of the real numbers such that are all disjoint?

MatheinBoulomenos

@Rithaniel no

Rithaniel

Okay, excellent, that plays into my hand perfectly. Is there a proof of this anywhere?

MatheinBoulomenos

I can sketch you a proof: the reals are second countable, as is every subspace of the reals. If there was a disjoint union of uncountably many open intervals, that subspace would have a open cover which has no countable subcover which is impossible as second countable implies Lindelöf ("every open cover has a countable subcover")

user193319

@Mancala Not sure if this is what you're interested in, but the union of a locally finite collection of closed sets is closed.

Locally finite collection

In the mathematical field of topology, local finiteness is a property of collections of subsets of a topological space. It is fundamental in the study of paracompactness and topological dimension. A collection of subsets of a topological space X is said to be locally finite, if each point in the space has a neighbourhood that intersects only finitely many of the sets in the collection. Note that the term locally finite has different meanings in other mathematical fields. == Examples and properties == A finite collection of subsets of a topological space is locally finite. Infinite collections...

Secret

I wonder in the surreals if it is possible to find uncountable disjoint intervals that contains only infinitesimals and irrationals but no rationals

Rithaniel

00:56

Thank you for the help, @MatheinBoulomenos . I'm new to the topic of topology, so I didn't know about Lindelof.

Secret

If you zoom that far down, then surely because of the fact that between every irrationals that are countably many rationals and between every rationals there are uncountably many irrationals that it should be possible to have intervals that contains infinitesimals and irrationals but having all rationals excluded?

Rithaniel

surreals?

Secret

Surreal number

In mathematics, the surreal number system is a totally ordered proper class containing the real numbers as well as infinite and infinitesimal numbers, respectively larger or smaller in absolute value than any positive real number. The surreals share many properties with the reals, including the usual arithmetic operations (addition, subtraction, multiplication, and division); as such, they form an ordered field. If formulated in Von Neumann–Bernays–Gödel set theory, the surreal numbers are a universal ordered field in the sense that all other ordered fields, such as the rationals, the reals, the...

It has nothing to do with your question

Rithaniel

Ah, gotcha. But now I've got something to research over the weekend.

Mancala

02:07

@user193319 that's what I need!

rschwieb

02:26

@AdarshKumar Why would it be lost forever? You will obviously still know about it, and you could publish whatever you want with it. Why is it so important to keep a poorly received post alive? Also, closure does not mean a post will be deleted for certain...

EnjoysMath

03:58

P = NP

0

Q: Smallest grammar algorithm treating irreducible grammars like a Rubik's Cube.

I assume you know what the smallest grammar problem is. It's basically, how can we, given an input string $s$ construct a grammar $g$ that expands to the single string $s$? Can we compute the smallest such grammar efficiently is a major open problem that I do not wish to solve. However, maybe ...

maybe will work

Any questions, I'll keep this chat open for a bit ..

I smoked some early harvest and came up with that

It's my best work so far on the problem imo, and quite a simple approach, and reminiscent of existing research: Rubik's cube's group theory

user131753

04:33

@MatheinBoulomenos You mean all $X,Y$ and $Z$ are locally Hausdorff?

Faust

blarbg

user131753

Why would we assume that?

Leaky Nun

05:19

hi @KasmirKhaan

@MatheinBoulomenos what does Spec(k[X1,X2,...]) look like?

Kasmir Khaan

Hello leaky :=

Matein is sleeping atm :D

for sure

and am off to Uni in few min :)

@LeakyNun see you later :'D

Akiva Weinberger

06:04

News is messed up right now

Specifically, Fox News

Tobias Kildetoft

@AkivaWeinberger What do you mean "right now"?

Akiva Weinberger

Fair

(I don't mean current events, though, I mean Fox can go f-ck itself)

Mary Star

06:33

Hello!!

Does someone of you have an idea about my question:

1

Q: Existence and uniqueness of solution

We have the initital value problem $$\begin{cases}y'(t)=1/f(t, y(t)) \\ y(t_0)=y_0\end{cases} \ \ \ \ \ (1)$$ where the function $f:\mathbb{R}^2\rightarrow (0,\infty)$ is continuous in $\mathbb{R}^2$ and continuously differentiable as for $y$ in a domain that contains the point $(t_0, y_0)$. Sh...

differential-equations

Tobias Kildetoft

07:14

I am watching the laudatio for Scholze from ICM. The topic is really interesting, but the speaker is about as monotone as it is possible to be.

ÍgjøgnumMeg

08:08

@Tobias Nice, I haven't seen it

Tobias Kildetoft

@ÍgjøgnumMeg It is on the ICM Youtube channel now.

They have finally started uploading some more talks, though the plenary lectures are still missing.

ÍgjøgnumMeg

am currently sat at my desk at work so I can't watch lol, but I'll take a look on my lunch break

I saw his introductory lecture when he received his professorship

Tobias Kildetoft

The laudatio is not by him (naturally).

ÍgjøgnumMeg

Indeed

Makes sense

Dunno why I mentioned his lecture haha

Tobias Kildetoft

his own lecture is also there now (though it has been on the uni-Bonn channel for a long time)

ÍgjøgnumMeg

08:12

Fair, I enjoyed watching that; I don't know anything at all about algebraic topology so it was nice to see it linked to Galois theory in that lecture

Tobias Kildetoft

I mean the lecture he gave at the ICM here.

ÍgjøgnumMeg

Lawl oh

it's too early

Leaky Nun

08:38

@ÍgjøgnumMeg you should learn algebraic topology

Akiva Weinberger

@ÍgjøgnumMeg How?

What video is this?

ÍgjøgnumMeg

@Leaky Certainly, but I haven't gotten around to it yet!

@Akiva Peter Scholze introductory lecture

or some variant of that title

@Akiva Peter Scholze Antrittsvorlesung is the title, it's in German though I think

(Title is actually Zahlen und Geometrie)

mercio

@Semiclassical since SU(2) is compact, can't you just pick a positive definite symmetric quadratic form on V then average it over SU(2) to get an invariant one ?

Akiva Weinberger

@ÍgjøgnumMeg Aw

Zahlen = $\Bbb Z$ = numbers?

ÍgjøgnumMeg

@Akiva exactly

Akiva Weinberger

08:46

What's it about?

(For the German-impaired)

ÍgjøgnumMeg

some kinda of dictionary between fundamental groups and Galois groups

kind*

or at least analogy between them

and how there's a kind of Galois correspondence for covering spaces

Akiva Weinberger

Ah yeah

ÍgjøgnumMeg

Inclusion reversing correspondence between fundamental groups and covering spaces

Akiva Weinberger

Right

That's kinda just an analogy, though, isn't it?

ÍgjøgnumMeg

(someone who knows anything about Algebraic Topology will be able to confirm if I'm talking out of my anus)

Akiva Weinberger

08:52

Nah you're good

ÍgjøgnumMeg

lol good

Akiva Weinberger

Hatcher brings it up and mentions how it's kinda similar to how Galois works

ÍgjøgnumMeg

Cool :)

I found it interesting (given that I only know a thing about Galois theory(

)

Akiva Weinberger

> This is strikingly reminiscent of Galois theory, with its correspondence between field extensions and subgroups of the Galois group. - Hatcher

ÍgjøgnumMeg

Ah nice

@Akiva you might be able to watch it with subtitles

Akiva Weinberger

08:56

I assume he goes into more detail than Hatcher does

(Hatcher just kinda mentions that it's similar and leaves it at that)

I didn't find anything with English subs

ÍgjøgnumMeg

ah that's unfortunate, there's definitely an option to watch it with auto-generated subtitles but these are unlikely to be very accurate

Zahlen und Geometrie. Antrittsvorlesung Prof. Peter Scholze

►

Akiva Weinberger

Formulas don't need to translate though so that's good

mercio

11

Q: Finite extensions of rational functions

I know that finite extensions of $\mathbb{C}(x)$ correspond to finite branched covers of $\mathbb{P}^1$, and this leads to an abstract characterization of the absolute Galois group of $\mathbb{C}(x)$ as the profinite completion of a free group of rank $\# \mathbb{C}$ via Riemann existence (or som...

algebraic-geometry galois-theory

Akiva Weinberger

@ÍgjøgnumMeg Autogenerated German subs though

mercio

terrytao.wordpress.com/2018/08/28/…

ÍgjøgnumMeg

08:59

@Akiva you can change it to autogenerated English subs

Tobias Kildetoft

@ÍgjøgnumMeg That sounds like a recipe for disaster. First it has to "guess" what is being said, then it has to "guess" that those words should be in English

mercio

6

Q: For which polynomials $p\in\mathbb{C}[w]$ are the branches of the inverse $p^{-1}$ expressible using algebraic operations?

The collection of all degree-$n$ polynomials in the variable $w$ (call this set $\mathbb{C}[w]_n$) can be identifies with $\mathbb{C}^{n+1}$ by the bijection $F:\mathbb{C}^{n+1}\to\mathbb{C}[w]_n$ defined by $$F:(a_0,a_1,\ldots,a_n)\mapsto w=p(z)=a_0+a_1z+\cdots+a_nz^n.$$ Let $\mathcal{A}_n\subs...

complex-analysis algebraic-geometry polynomials galois-theory

so yeah, galois groups and fundamental groups can be very related sometimes

Alessandro Codenotti

@TobiasKildetoft I think it just adds a google translate layer on top of the autogenerated German subtitles so it's probably going to be somewhere between completely unintelligible and very bad

Akiva Weinberger

@ÍgjøgnumMeg Can you? Not on iPhone at least you can't

Tobias Kildetoft

@AlessandroCodenotti Right. Especially seeing as many of the terms will be math specific, so a direct translation will be meaningless (you will probably end up learning a lot about bodies).

Alessandro Codenotti

09:09

Algebraically closed bodies, another unrealistic beauty standard

Akiva Weinberger

Rooms also, probably

mercio

perfect bodies

Akiva Weinberger

What is a "body" in German math

Tobias Kildetoft

So will we have rooms over bodies or rooms about bodies? (I assume that the word used in German is über)

@AkivaWeinberger Körper = field

mercio

what is the english term for room ?

Tobias Kildetoft

09:12

space

Alessandro Codenotti

To be fair space is among the meanings of Raum in German

mercio

there also are the bodies of the ideal classes

Alessandro Codenotti

There is this anecdote that in Göttingen they have a lecture hall dedicated to Hilbert, so it is an actual Hilbertscher Raum

Fargle

Platonists rejoice, Hilbert spaces truly do physically exist!

...or lament.

mercio

lol

Rudi_Birnbaum

09:27

Körper can be field or body

Platonische Körper

is it even "Platonic bodies" ?

Tobias Kildetoft

@Rudi_Birnbaum Platonic solids

Rudi_Birnbaum

yeah ...

Akiva Weinberger

@ÍgjøgnumMeg What does this mean?

Tobias Kildetoft

@AkivaWeinberger Which part? Overall, it says that there is a well-defined Frobenius for each prime $p$ in the Galois group of the polynomial.

(the statement is somewhat incomplete, as it does not say anything about this Frobenius in relation to anything else)

user193319

The problem I am working on is about the space $X = \Bbb{R}\ \{\mbox{y-axis}\}$, the quotient space of $\Bbb{R}^2$ obtained by identifying the $y$-axis with a point. What exactly does this space look like? What do elements in the set $X$ look like? When they say the $y$-axis is identified with a point, are they saying that for every $y \in \Bbb{R}$, $(0,y)$ is sent to some point $(a,b)$?

Akiva Weinberger

09:35

Oh, but it doesn't define what ${\rm Frob}_p$ is?

Tobias Kildetoft

@AkivaWeinberger Right

mercio

I don't think it is well defined iirc it gives a conjugacy class

Leaky Nun

@AkivaWeinberger galois groups and prime numbers
Theorem (Frobenius, 1896)
Let $P(x) = x^n + a_1 x^{n-1} + \cdots + a_n$ be a polynomial with rational coefficients. According to Galois there is an associated Galois group $G$, which permutes the solutions of $P(x) = 0$. For (almost) every prime number $p$ there is an (almost) well-defined element Frob_p in G, the Frobenius automorphism.

Akiva Weinberger

Ah, thanks

Tobias Kildetoft

@LeakyNun Ahh, right, I had missed the important word "almost" (fast).

Leaky Nun

09:37

is this related to ramification?

mercio

for the first almost yes

ÍgjøgnumMeg

probably, Frob exists for unramified primes I believe?

Akiva Weinberger

@user193319 Imagine squeezing the $y$-axis to a point so you're left with what kinda looks like a sideways hourglass

user193319

@AkivaWeinberger So does that mean $(0,y) \mapsto (0,0)$ for every $y$?

Alessandro Codenotti

@TobiasKildetoft doesn't work for slow primes

mercio

09:41

lol

ÍgjøgnumMeg

@Alessandro ouch

Akiva Weinberger

@user193319 It's not mapping points to points in the plane

You know how a plane is homeomorphic to an open disc?

user193319

Sure.

Akiva Weinberger

Imagine squeezing the disk so that the vertical line in it becomes a point

and your disk becomes an infinity symbol shape

user193319

I see. But aren't we still in the plane?

Akiva Weinberger

09:44

$\Bbb R^2/\{y~\rm axis\}$ isn't really a subset of $\Bbb R^2$

Leaky Nun

@AkivaWeinberger don't you believe in the axiom of choice?

well, alright the axiom of choice fails in the category of topological spaces

Akiva Weinberger

(unless you meant $\Bbb R^2\setminus\{y~\rm axis\}$, in which case you're just deleting the $y$-axis)

user193319

So what do the elements of $\Bbb{R}^2/ \{y-axis\}$ look like?

No, it is a quotient space.

Mithlesh Upadhyay

Please vote to undelete this post:

Leaky Nun

it looks like $\Bbb R^2$ but suddenly $(0,2) = (0,\pi) = (0,3) = (0,e)$

2

user193319

09:46

E.g., one part of the question asks whether it is Hausdorff.

Mithlesh Upadhyay

math.stackexchange.com/questions/1656686/…

Akiva Weinberger

@user193319 Each point in that space is an equivalence class of points in the plane

Leaky Nun

Hausdorff / closed = Hausdorff, I believe

Mithlesh Upadhyay

This post has 10,920 times views and 2 years, 7 months old.

Akiva Weinberger

Most of the equivalence classes contain just one point, but one of the equivalence classes contains the entire y-axis.

@MithleshUpadhyay Why do you want this so much

Mithlesh Upadhyay

09:47

Please vote to undelete this post:
https://math.stackexchange.com/questions/1656686/how-many-rectangles-can-be-observed-in-the-grid
This post has 10,934 times views and 2 years, 7 months old.

Rudi_Birnbaum

@mercio how to continue? I just started the mag. resp part.

Leaky Nun

nobody likes my meme :(

user193319

So the equivalence relation is $(x,y) R (w,z)$ if and only if $x=w$?

Akiva Weinberger

@LeakyNun ?

Leaky Nun

2 mins ago, by Leaky Nun

it looks like $\Bbb R^2$ but suddenly $(0,2) = (0,\pi) = (0,3) = (0,e)$

ÍgjøgnumMeg

09:49

@Leaky en.wikipedia.org/wiki/…

Mithlesh Upadhyay

@AkivaWeinberger , I'm trying from my end, I think this is a good post.

@AkivaWeinberger , whould you like to vote to undelete?

mercio

@Rudi_Birnbaum dunno I have been thinking about that hole from yesterday

Rudi_Birnbaum

@mercio good!

Leaky Nun

@user193319 the equivalence relation is: (x,y)R(w,z) iff [(x=w=0) or (x=w and y=z)]

Akiva Weinberger

@user193319 $(x,y)R(w,z)$ if and only if (a) either $(x,y)$ and $(w,z)$ are the same point or (b) $x=0$ and $w=0$ (that is, both points are on the $y$-axis)

Leaky Nun

09:50

rip

Akiva Weinberger

In general, $A/B$ (where $A$ and $B$ are spaces) means you quotient by the equivalence relation where all the things in $B$ are equivalent to each other, and everything else is only equivalent to itself

Mithlesh Upadhyay

@mercio ,

@TobiasKildetoft ,

Akiva Weinberger

For example, if $A$ is a closed disk, and $B$ is its boundary, then $A/B$ is topologically a sphere:

Leaky Nun

@AkivaWeinberger well I say $S^3/S^1 = \Bbb CP^1$

user193319

Ah, I see...But where does the idea of a topology on the quotient come into play? I don't see what the topology on $A/B$ is.

Leaky Nun

09:52

@user193319 it's the coinduced topology

user193319

What does that mean?

Akiva Weinberger

@user193319 Do you remember how a topology is defined on quotient spaces, in general?

Leaky Nun

given a function $f:X \to Y$, if I have a topology on $X$, then I can make a topology on $Y$ by stipulating that $T \subseteq Y$ is open iff $f^{-1}(T) \subseteq X$ is open

Induced topology

In topology and related areas of mathematics, an induced topology on a topological space is a topology which makes the inducing function continuous from/to this topological space. == Definition == Let X 0 , X 1 {\displaystyle X_{0},X_{1}} be sets, f : X 0 → X 1 ...

wow this really feels like the direct image functor

user193319

Ah, I see.

Akiva Weinberger

Quotient space (topology)

In topology and related areas of mathematics, a quotient space (also called an identification space) is, intuitively speaking, the result of identifying or "gluing together" certain points of a given topological space. The points to be identified are specified by an equivalence relation. This is commonly done in order to construct new spaces from given ones. The quotient topology consists of all sets with an open preimage under the canonical projection map that maps each element to its equivalence class. == Definition == Let (X, τX) be a topological space, and let ~ be an equivalence relation on...

user193319

09:54

So in my case, $X = \Bbb{R}^2$ and $Y = \Bbb{R}^2 / \{y-axis\}$?

Leaky Nun

right

user193319

And $f$ is the map that sends $(x,y)$ to its equivalence class?

mercio

yes

user193319

Ah, I see. Thanks everyone!

MatheinBoulomenos

10:37

@user170039 I meant what I said: locally compact and Hausdorff. And one would assume that so that $C(X \times Y,Z) \cong C(X,C(Y,Z))$ holds

Leaky Nun

@MatheinBoulomenos locally compact Hausdroff space has exponential object, is what you're saying?

wait that isn't the exponential object

MatheinBoulomenos

locally compact Hausdorff are commonly used sufficient conditions such that $C(X \times Y,Z) \cong C(X,C(Y,Z))$ hold

Leaky Nun

I see

MatheinBoulomenos

and yes, this makes the category of locally compact Hausdorff spaces cartesian closed i.e. exponential objects exist

Akiva Weinberger

@ÍgjøgnumMeg Knoten means knots, right?

ÍgjøgnumMeg

11:05

@Akiva yis

Secret

11:37

2 hours ago, by Mithlesh Upadhyay

@mercio ,

diedoddiediediredird

Removed painting

Unremoved painting

exremoved painting

hyroglycphic cardinal

cardinology

Deploying demover

mercio

what

user 2646

i think he's rambling, again

Secret

chat.stackexchange.com/search?q=Mithlesh+Upadhyay&room=36

Next time, he will be swept to the dungeons like gateprep

and The Demover will ensure he will never came back

user 2646

11:53

live and let live pal

you know where the ignore button is

2

Jake S

12:54

excuse me, does anyone know of a very basic resource on projections? somewhere that I can read to understand why the trace of a projection matrix gives me the rank (preferably without even touching eigenvalues as I should be able to construct this proof before we get to diagonalization)

Evinda

14:09

Hello

Do you have an idea about my question?

3

Q: Equilibrium solutions and stability

I want to find the equilibrium solutions and determine their stability. $(1)\left\{\begin{matrix} \dot{x}=-y-x(1-\sqrt{x^2+y^2})^2\\ \dot{y}=x-y(1-\sqrt{x^2+y^2})^2 \end{matrix}\right.$ I also want to check the behavior of the solutions of $(1)$ when $t \to \infty$. There are five possible ans...

differential-equations

user131753

14:36

@MatheinBoulomenos But that's not what I had in mind. I wanted $X,Y$ and $Z$ to be arbitrary topological space.

MatheinBoulomenos

@user170039 see: ncatlab.org/nlab/show/… for weakening the assumptions

MatheinBoulomenos

14:56

@LeakyNun the Spec functor is right adjoint to the global section function, so it sends colimits of rings to limits of Schemes $\mathrm{Spec}(k[X_1,X_2,\dots])=\mathrm{Spec}(\varinjlim\limits_{n \in \Bbb N}k[ X_1, \dots X_n])=\varprojlim\limits_{n \in \Bbb N} \mathrm{Spec}(k[X_1, \dots, X_n])=\varprojlim\limits_{n \in \Bbb N}\Bbb A^n(k)$

Evinda

Hello @LeakyNun
I have a question.. We want to have $\int_0^{\frac{\pi}{4}} (\tan{x}-1) f(x) dx=0$.

We know that $\tan{x}-1<0$ for each $x$ in the interval $(0, \frac{\pi}{4})$.

So in order the integral to be 0 , it has to hold that $f=0$ in the interval $(0, \frac{\pi}{4})$, right?

Leaky Nun

@Evinda is $f$ continuous?

@MatheinBoulomenos I see

Evinda

@LeakyNun Yes, $f: [0, \frac{\pi}{4}] \to \mathbb{R}$ continuous

Leaky Nun

I don't think $f$ has to be zero

Evinda

Why? @LeakyNun

Leaky Nun

15:03

$f$ can be partly positive and partly negative

Evinda

So isn't there a condition that f has to satisfy so that we can have

$\int_0^{\frac{\pi}{4}} (\tan{x}-1) f(x) dx=0$ ? @LeakyNun

Leaky Nun

well if $f=0$ then certainly that integral is zero

Evinda

Yes but it doesn't hold that if $f \neq 0$, $\forall x \in (0, \frac{\pi}{4})$ then the integral is non-zero... And it also doesn't suffice for f to be contnuous to determine if the integral can be zero or not... Or am I wrong? @LeakyNun

Leaky Nun

right

Vrouvrou

@LeakyNun Bonsoir, s'il vous plait est ce que l'image d'un ensemble dénombrable par une fonction continue est dénombrable?

Leaky Nun

15:16

what does countability have to do with topology?

Vrouvrou

why it must have a relation with topology ? I just ask for example is $f(Q)$ countable where f is continuous ?

Leaky Nun

mais tu as ajoute "continue" aux proprietes de la fonction

Vrouvrou

donc sans continuité ca marche ?

@LeakyNun ?

Semiclassical

@Evinda If the eigenvalues are $\lambda = -1\pm i$, then a generic solution (for the x part) would be of the form $x(t)=A e^{-t}e^{-it}+Be^{-t}e^{it}$

in which case $x(t)\to 0$ as $t\to\infty$. So wouldn't that be asymptotically stable ? (not unstable)

Leaky Nun

@Vrouvrou oui

Evinda

15:32

What happens with $e^{it}$ and $e^{-it}$ ? @Semiclassical

Semiclassical

well, you want x(t) to be real

Vrouvrou

@LeakyNun on a pas besoin ni de bijection ni de continuité

Semiclassical

so you'd better choose a linear combination of A,B appropriately

Leaky Nun

right

Semiclassical

but e^it = cos(t)+i sin(t) and e^-it = cos(t)-i sin(t)

so in the simplest case you need A=B to get z(t) = 2A e^-t cos(t)

(it's actually a bit more involved than this, since A,B can be complex. once you take that into account, you'll find that it gets modified to $z(t) = C e^{-t} cos(t)+D e^{-t}sin(t)$ where $C,D$ are appropriate constants)

So you'll have an overall exponential decay, along with some oscillatory function coming from $e^{\pm i t}$

Vrouvrou

15:42

@LeakyNun comment le démontrer ? si E est dénombrable donc il existe une bijection entre lui et N, comment montrer qu'il ya une bijection entre $f(E)$ et N?

Jasper Loy

Four consecutive edits, very rare.

Leaky Nun

@Vrouvrou oh f(E) peut etre fini

il y a personnes qui traitent ensembles finis comment ensembles denombrable

Jasper Loy

@LeakyNun My brown square matches your brown walls...

Vrouvrou

si dans notre cours si c'est fini alors c'est denombrable

Leaky Nun

si c'est fini alors il n'y a aucune bijection entre N

@JasperLoy interesting

mercio

15:45

in french, "dénombrable" means "finite or countable"

Evinda

Ok @Semiclassical If we had the eigenvalues $1 \pm i$, how would we deduce the stability of the equilibrum solution?
In general, having found the Jacobi matrix and then the eigenvalues, are both x and y expressed as linear combinations of $e^{\lambda t}$ and $e^{-\lambda t}$ ?

Vrouvrou

oui le truc de la bijection on le laisse pour E infinie

@mercio voila merci

Semiclassical

if you had $1\pm i$, then instead of $e^{-t}$ you'd have $e^{t}$

which would blow up as $t$ increases

hence, unstable

@Evinda They are. But they're not independent combinations, since $x,y$ are related through the (linearized) differential equation

For instance, if you write $x=Ae^{-t}e^{it}+Be^{-t}e^{-it}$, then $x'= -y-x\implies y=-x-x'$

in which case you can see by inspection that $y$ will be a linear combination of the two solutions, but with coefficients in terms of A,B

Vrouvrou

si on suppose que f(E) est enfin comment trouver la bijection avec N @LeakyNun

Evinda

Ah yes, right... But in order to check the stability of the equilibrum solution, if one of x or y tends to $\infty$ we immediatedy deduce that the system is unstable, and if one tends to 0 , we also have to check the other, right? @Semiclassical

Semiclassical

15:53

well, once you know about that e^{-t} factor in the solution of x(t), you know that it'll be in y(t) as well

so the fact that x(t)->0 goes hand-in-hand with y(t)->0

Evinda

Ah yes, right!!! @Semiclassical

Semiclassical

the basic point is that negative real parts for eigenvalues signify stability, not unstability

Leaky Nun

@Vrouvrou sait on que chaque sous-ensemble de N est denombrable?

Vrouvrou

oui

Evinda

I see... Asymptotic stability, since $t \to \infty$, right? @Semiclassical

Semiclassical

15:56

right

Evinda

And if $(x,y)$ is any other solution of the problem, what can we say about the limits $\lim x(t)$ and $\lim y(t)$ ? @Semiclassical

Semiclassical

the only thing that'll change are the coefficients

the e^(-t) factor will persist regardless

Evinda

Why? We have calculated the Jacobi matrix at the point (0,0)... :/ @Semiclassical

Vrouvrou

@LeakyNun ?

Leaky Nun

on va construire une fonction g : f(E) -> E

d'accord

Vrouvrou

15:59

ok

Semiclassical

yeah, but your (linearized) ode is of the form v'(t) = A v(t)

Leaky Nun

si y in f(E), alors il existe x in E tel que f(x) = y, d'accord

Semiclassical

I guess the more significant point is that, by linearizing around x=y=0, you're only going to get information about the Jacobi matrix at (0,0)

Vrouvrou

oui donc g est surjective@LeakyNun

Semiclassical

you could linearize about any other point, but it wouldn't be too helpful since (0,0) is the only stationary solution

Leaky Nun

16:02

@Vrouvrou non, g n'est pas surjective

I haven't even defined g

Evinda

@Semiclassical Is the ode linear althought it contains $x^2$ and $y^2$ ?

Vrouvrou

@LeakyNun oui pardon

Semiclassical

the original ode, no

Leaky Nun

si y in f(E), alors il existe x in E tel que f(x) = y, d'accord

Vrouvrou

oui

Semiclassical

16:04

but then $x=Ae^{-\lambda_1 t}+Be^{-i\lambda_2 t}$ isn't a solution of the original ode

What it solves is the linearized ode that you get by tossing out the nonlinear terms

Leaky Nun

alors on choisit un x et definit que g(y) = x, d'accord

Vrouvrou

ok

Leaky Nun

alors f(g(y)) = y pour tous y in f(E)

Vrouvrou

oui

je ne voix pas ou on va arriver

@LeakyNun

Leaky Nun

si g(y1) = g(y2), alors f(g(y1)) = f(g(y2)), alors y1 = y2

donc g est injective

g : f(E) -> E est injective

E est denombrable,

donc f(E) est denombrable

Vrouvrou

16:18

pourquoi injective suffit

?

@LeakyNun

Leaky Nun

then the image of g is in bijection with the domain of g

g : f(E) -> g(f(E)) est une bijection

Vrouvrou

mais g(f(E)) = E???

@LeakyNun

Leaky Nun

non

g(f(E)) est un sous-ensemble de E

Vrouvrou

je n'ai pas compris

@LeakyNun

Evinda

16:37

I am a little confused now... The x and y that we found correspond to the equilibrum solution... Can we express the general solution of the ode in term of the above? @Semiclassical

Semiclassical

@Evinda the equilibrium solution is just x(t)=y(t)=0

the x(t), y(t) solution found earlier is not a solution to the original ode. What it solves is the linearized ode, which you get from the original ode by tossing out nonlinear terms

In this case, x’(t)=-y-x and y’(t)=x-y constitute the linearized system

If you plug the solution to that linearized system into the original system, you’ll find that it’s not a solution to the latter

Secret

16:52

ugh, what does those cross terms do

rschwieb

17:17

@Secret Whoa, what's the structure in the context?

mercio

i think it's some kind of surreal numbers

or hyperreal numbers

rschwieb

Hmmm

@Secret @mercio What is the text taken from?

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$Mathematics$ Mathematics