Anyone here know about fourier series?

@Lord_Farin OK, then yes.

If $A=\overline A;B=\overline B$ we have $\overline{A\cup B}=\overline A\cup\overline B=A\cup B$.

Thus, the union of two closed sets is closed.

@PeterTamaroff The equality $\overline{A \cup B} = \overline A \cup \overline B$ does not hold for general closure operators (consider e.g. transitive closure; the union of transitive relations needn't be transitive).

@Lord_Farin These are my rules, man

@PeterTamaroff Did you even bother to read my rules, linked above? The disparity should be apparent.

@Lord_Farin Yep, that's why I linked to my rules =D

In which case "closure spaces" become a lot less interesting; might as well talk about "topological spaces".

@Lord_Farin That was my point when I said "one to one correspondence"

Although I concede that a different perspective may sometimes yield a more insightful approach (I think of the direct image functor on sheaves, which is only reasonably easy to express using étale spaces).

@Lord_Farin Hehe, no clue on that parenthesis!

@PeterTamaroff Bam, intimidation. :P

@Lord_Farin To Oblivion with you!

@PeterTamaroff Prefer Skyrim nowadays :D.

hahahaha

@Lord_Farin Haha, they still use that phrase.

@PeterTamaroff AH!!! btw, did you make that challenge?

@Charlie What challenge?

@PeterTamaroff to write a paragh with an alliteration on "P"

@Charlie Oh, I did think about it, but I have no idea what to write it about. It has been a long time since I have written anything.

@PeterTamaroff it's a good opportunity to try ;)

having to edit someone else's bad latex is painful :-/

@MarianoSuárez-Alvarez Not to mention those cases where one is almost done when someone else submits an edit... Those are the worst -- rage quit alert!

Is it known the exactly age of the "Ishango bone"?

@Charlie "Exactly" is certainly impossible.

Yes! math.stackexchange.com/badges/157/reviewer

@PeterTamaroff as accurate as possible?

@Charlie Yeah, accurate seems better.

@PeterTamaroff :D do you know?

I don't even know what the Ishango bone is.

it seems to be 25,000 years old

there were lunar markings.women marked their cycles and thus began to mark time

the new is-zero-a-natural-number question is taking its natural course

a crash course

"There: I just counted the blue unicorns in my office!" HAHAHA!

Off to bed. Goodnight everyone.

@Lord_Farin Goodnight, F!

@PeterTamaroff it's possible

@PeterTamaroff have you been using LSD lately, Pedro?

@MarianoSuárez-Alvarez We do not use $0\in\Bbb N$ at the UBA; I heard!

Forgive me, Pedro

@Charlie Night.

Some people do

The wrong people

@Lord_Farin sleep tight!

@MarianoSuárez-Alvarez Ha! But why?

"A natural number is a number that arises natural ly."

Hmmmm.....

"A circle is a a line in a circular shape."

so, it seems that who created the bone was a female

@Lord_Farin It means that it's possible that his office has unicorns of another colors.

@MarianoSuárez-Alvarez Do you happen to know when all the multivariable calc was developed?

Can someone help me figure out what this question is asking for? (Not how to solve it). An equivalence relation on $\mathbb{R}$ is determined by the subset R of the set $\mathbb{R}\times\mathbb{R}$. Determine which of the axioms (reflexive, symmetric, transitive) are satisfied for the set $\{(s,s) | s \in \mathbb{R} \}$ and the empty set.

Say, who stated the Inverse Function Theorem, the Implicit FT, and so on?

@AlanH those two sets represent two relations. Do you understand that?

I'm reading the Cambridge mathematics syllabus, it says that I should learn induction, including a proof of the binomial theorem with non-negative integral coefficients.

What is that?

I understand what a binomial theorem with non-negative integral coefficients is, but I don't know what should be proved.

Is it the pattern in the Pascal triangle that should be proved?

@GustavoBandeira How to prove the statement of the theorem.

@robjohn Don't equivalence relations usually say something like a ~ b iff a and b are conjugate elements, for example. I don't see how that is in the problem. Or perhaps you could clarify, please.

why does each term have the coefficient that it does?

@robjohn And do it via induction?

@AlanH $aRb$ is the same as (a,b) being in the relation

@GustavoBandeira that would be one way

@AlanH Yes, an equivalence relation is usually determined by $a\sim b\iff a$ and $b$ share some property.

@robjohn $aRb$? =\

@PeterTamaroff but he has to show whether this relation is symmetric, reflexive, transitive

@AlanH R is the relation... like $\sim$ you just used

@robjohn Aha. I was just remarking something.

@AlanH Read "a stands in relation to b" or "a is related to b".

@robjohn so you're saying the problem is asking to show that the relation $\{(s,s) | s \in \mathbb{R}\}$ is an equivalence relation? O.o

@AlanH I don't want to use $\sim$ since that implies it has certain properties.

@AlanH no, it seems to be asking you to show that the relation given by the set of pairs $\{(s,s):s\in S\}$ has or doesn't have each of those properties.

an equivalence relation has all three.

@robjohn ohhhh like to find some property to relate the two? like working it backwards?

or not...

@AlanH no. For example, a relation $R$ is reflexive if $aRa$ for all $a$

@Robjohn okay okay, I think I get what it's asking

does the relation given by $\{(s,s) | s \in \mathbb{R}\}$ have that prooperty?

Thank you

@robjohn It seems it would only satisfy reflexive

@AlanH well, does $aRb$ imply $bRa$?

@robjohn oh right, and symmetric. but not transitive

@AlanH well, if $aRb$ and $bRc$ does $aRc$?

@robjohn (s,s), (r,r), you can't get (s,r) unless s = r, no?

@AlanH that is not what transitive says.

transitive says if $(s,r)$ and $(r,t)$, then $(s,t)$

@robjohn but the set we're looking at is (s,s)

as in you can't have (r,t) unless r =t, right?

@AlanH yes, so that means the if $(s,r)$ is there then what?

@robjohn then $s$ has to equal $t$?

If $(s,r)$ is there then $s=r$, no?

yes

in fact, the relation given there is $=$

sorry, but where is "there"?

@AlanH there is in the relation $\{(s,s) | s \in \mathbb{R}\}$

oh

@robjohn okay, so then you do have transitive in the sense that rRs and sRt then rRt, then r=t.

@AlanH if $(r,s)\in R$ and $(s,t)\in R$, then $(r,t)\in R$ it happens that $r=t$ yes

@robjohn okay thanks a lot

@robjohn Could you clarify some notation?

@robjohn Are you there?

@PeterTamaroff yes?

@robjohn OK. This is as follows:

I have to establish sufficient conditions on $f,g$ that ensure the system $$\begin{cases}x=f(u,v)\\y=g(u,v)\end{cases}$$ can be solved for $u,v$. Now, if we define $\Phi(u,v)=(f(u,v);g(u,v)$, the inverse function theorem ensures that $\Phi$ is $\mathcal C^1$ one some open $\Omega\in\Bbb R$ and if ${\bf J}_\Phi(u_0,v_0)\neq 0$ on some $(u_0,v_0)\in\Omega$, there is a local inverse $\Phi^{-1}$

Apostol says "If the solutions are $u=F(x,y)$ and $v=G(x,y)$ show that:

I got something similar, though, and I don't understand his notation.

I got

$${\left. {\frac{{\partial F}}{{\partial x}}} \right|_{\Phi \left( {x,y} \right)}} = \frac{1}{J}{\left. {\frac{{\partial g}}{{\partial x}}} \right|_{\left( {x,y} \right)}}$$

and so on...

@robjohn

(Using the Chain rule on the matrices of the differential of $$\Phi\circ \Phi^{-1}(x,y)=(x,y)$$

where the $${\left. {} \right|_{\left( {x,y} \right)}}$$ means "evaluated at".

@robjohn Are you there?

@PeterTamaroff what does $\Phi(u,v)$ mean?

@robjohn I defined it as the function $\Bbb R^2\to\Bbb R2$, $(u,v)\mapsto(f(u,v),g(u,v))$

@PeterTamaroff ah, there was a missing paren and a semicolon that confused me

@robjohn Oh, OK.

Note that I am writing $\Phi^{-1}(x,y)=(F(x,y);G(x,y))$.

Anyways, all I am asking is the notation, mixing $u,v,x$ and $y$.

@PeterTamaroff That notation is quite confusing, I agree

@robjohn Good! Makes me feel better =)

@PeterTamaroff I usually follow a different way of looking at the same thing

@robjohn Ah?

It is the same thing essentially, but it gets rid of a lot of the intermediate functions, which are the root of the confusion, imo

@robjohn Hmm... OK.

Think of $\Phi(x,y)=(\Phi_1(x,y),\Phi_2(x,y))$

Now he wants me to do the above with $f(u,v)=u^2-v^2$ and $g(u,v)=2uv$

Hey guys. Is this place similar to #math on IRC?

@robjohn Aha.

@JoshuaCiappara What is IRC?

Internet Relay Chat.

@JoshuaCiappara No idea.

Then define $$\frac{\partial(\Phi_1,\Phi_2)}{\partial(x,y)}= \begin{bmatrix}\frac{\partial\Phi_1}{\partial x} &\frac{\partial\Phi_2}{\partial x}\\ \frac{\partial\Phi_1}{\partial y}&\frac{\partial\Phi_2}{\partial y}\end{bmatrix}$$

Well, I was hoping for a hint on a problem. Suppose X = A union B, where X is complete and B is a union of closed nowhere dense sets. I want to prove that A cannot be embedded in a similar such union.

are there any nice general ways to determine of two semidirect products are isomorphic? (I realized while answering math.stackexchange.com/questions/405040/… that I needed a very ad-hoc method for it there, and it was pure luck that it turned out to work)

It feels like an application of the Baire category theorem, but I'm not sure how.

@robjohn Ah, I see, he uses his $\partial(\text{ BIG STUFF })$ thing.

Like $$\frac{\partial(f,g)}{\partial(x,y)}$$

Similar such union within X, I mean.*

@PeterTamaroff then the derivative of the inverse is the inverse matrix

@JoshuaCiappara You can edit your messages within a time range here.

@PeterTamaroff yeah, that is why I said it is pretty much the same thing

@PeterTamaroff: Sorry, what do you mean?

Oh!

I see.

@robjohn I see.

@PeterTamaroff You just prove things using matrix algebra. and the chain rule

@robjohn Aha, that's what I did.

I can prove that A is dense, but I'm not sure how that helps.

All those equations are $D^{-1}=\dfrac 1{\det D}{\rm adj}D$

@PeterTamaroff indeed

@JoshuaCiappara I don't think I can help, but surely Brian Scott can. He's boss in Topology.

@PeterTamaroff he'd have a hard time doing that in $\mathbb{R}^3$ without matrices

@JoshuaCiappara If your ask on main he'll probably answer.

@robjohn Hehe, yes.

@robjohn Anyways, I need to find the inverses when $f(u,v)=u^2-v^2$ and $g(u,v)=2uv$. These happen to fulfill the CR equations.

The Jacobian is $4(u^2+v^2)$.

So this is locally invertible everywhere outside the origin.

Okay, thanks anyway guys.

@PeterTamaroff pretty easy, I think. ping me if it is not :-)

@JoshuaCiappara Could you write your problem in one message?

@robjohn This is $f(z)=z^2$, is it not?

@PeterTamaroff yep