:-/

@BernardoMeurer what's going on

@0celo7 I'm having a beer and eating canned tuna

sounds disgusting

I miss home

Howdy

@BernardoMeurer You'll make it

@SirCumference Of course I will, I have a 6 pack (of beer, not of muscle)

(Thank you though)

I think you might have an addiction...

Just a thought

racist

1 in 3 of your posts involves alcohol

that sounds like he's doing something right

@SirCumference Can you back that up with stat-burp-istics?

Interestingly enough I did burp while writing that

Jeez...At this point your name might as well be Beernardo. (kidding)

groan.

I'll look at the paperwork, I'm willing to change

Did you start drinking young?

@0celo7 the best puns are the worst puns :D

I am young

@BernardoMeurer what new analysis have you learned?

Nothing, I missed class today

WHAT

you're gonna be fucked

Hung over?

What? It's a Wednesday?

I'm so confused

I don't drink my brains off on the middle of the week

I just didn't feel like leaving home

@BernardoMeurer :/

Also today I just had useless classes

Like Programming and Computer Architecture

I just go there to listen to BS

did you not have analysis?

@BernardoMeurer do you know the material already?

And in Analysis II he just talked about balls and stuff

For Programming I know the whole 90% of the course, for Computer Architecture I know about 60%, and right now it's just ultra basic stuff

@BernardoMeurer do you know that already?

No, so I just read it from my intro to topology book

@BernardoMeurer How would you prove that an open ball is open?

How does Griffiths' wording here make any sense?

@0celo7 Check if $\Bbb R^n \setminus A$ is closed?

Each intersection is a stationary state. One block = 1 state

Really, you just said each intersection is 1 state.

Therefore, I see 8 states with every block.

or check for boundary points (correct translation?)

I see now why people complain about Griffiths' wording issues

@BernardoMeurer You want to see if the boundary of the ball is disjoint from the ball?

Could someone help me understand what the heck that means?

There's a way to show it directly using the definition of open set

which are points such that $\forall \epsilon > 0; B_\epsilon(p)\cap A \neq \emptyset \vee B_\epsilon(p)\cap(\Bbb R^n\setminus A)\neq\emptyset$

If there are such points then the set is not open

@BernardoMeurer This is why you should have gone to class.

Why? I got this def. from my friend's notes

A set $S$ is open if and only if $(\forall x\in S)(\exists\epsilon>0): B_\epsilon(x)\subset S$.

Oh, then the class is terrible.

Yeah, your def is more concise

Mine makes sense in my head those

$$\nexists p \text{ s.t. }\forall \epsilon > 0; B_\epsilon(p)\cap A \neq \emptyset \vee B_\epsilon(p)\cap(\Bbb R^n\setminus A)\neq\emptyset$$

You're never going to be able to prove anything with that...

oh, those are boundary points

so what is your definition of "open"??

I say a point $p$ is inside a set $A$ iff $\exists \epsilon > 0 : B_\epsilon(p)\subset A$

That's what I wrote above, just less correct

@BernardoMeurer I think that's the European way of defining it

and $A$ is open iff $\forall p \in A : \exists \epsilon > 0 : B_\epsilon(p)\subset A$

Or at least, the way of defining it without the closure

@BernardoMeurer is that not what I wrote above?

So now prove that $B_\epsilon(x)$ itself is open

Yes, but now I got it from my head

we will prove a nice result using this

@BernardoMeurer that's good

well, $B_\epsilon(a) = \{x\in\Bbb R^n \colon ||x-a|| < \epsilon \}$

@ACuriousMind Why isn't "invariance of boundary" trivial? (It says that the interior and boundary of a topological manifold are disjoint.)

If anyone has time and understands Fermi energy and k-space, feel free to help me out :3

So it's trivial to see all it's points will be interior points

@BernardoMeurer Why?

Because the boundary does not belong to the set

You need to find a smaller ball that fits inside $B_\epsilon(a)$ around $x\in B_\epsilon(a)$ given

Which again are points such that $\forall \epsilon > 0; B_\epsilon(p)\cap A \neq \emptyset \vee B_\epsilon(p)\cap(\Bbb R^n\setminus A)\neq\emptyset$

@BernardoMeurer Why?

Because per the definition of a ball the diameter is an open set in $\Bbb R$

No! That's the point!

Hm?

It's not open by definition.

How come?

It seems pretty open

You just told me the definition of open.

Where does it say there that balls are open?

It says that any set with that property is open.

Nowhere, I'm just saying I think they are

Ok, you should try to prove it

I can tell you how to do it

I'll take a shower first b/c I stink

brb

ok

I KNOW HWO

I was in the toilet

Suppose they were closed

@BernardoMeurer "not open" is not "closed"

That would mean we could find two points in the ball

Shit

I'll be back

Also there are simultaneously open and closed sets :)

clopen yes

such as $\Bbb R^n$, right?

Yes, but no one actually says clopen

OK

@0celo7

Supposed $B_\epsilon(a)$ was not open

That would mean that there would be two points on this balls whose distance is $||x_1-a||+||x_2-a|| = 2\epsilon$

(there is a direct proof, but ok, let's see)

@BernardoMeurer whoa what

?

Why does it mean that?

Because if a ball has radius $\epsilon$ for it to not be open it must contain all points such that $||x-a|| = \epsilon$

AHA!

That's not always true :P

What the fuck :P

There are subsets of $\Bbb R^2$ that fail to have that property

(with the induced metric)

Fuck me

How do I prove it then?

@BernardoMeurer Consider $B_\epsilon(a)$. Let $x\in B_\epsilon(a)$. Let $\eta=\epsilon-||x-a||$. Claim: $B_\eta(x)\subset B_\epsilon(a)$. To show this, we must show that if $y\in B_\eta(x),$ then $||y-a||<\epsilon$. But $$||y-a||\le ||y-x||+||x-a||< \epsilon -||x-a||+||x-a||=\epsilon.$$

Which one do you think would roll down faster?

A bottle with frozen honey

A bottle with honey at room temperature

honey is so viscous I doubt it matters

@YashasSamaga My self esteem

The unfrozen one actually moved faster

@0celo7 Hm, that makes sense

@BernardoMeurer So now we have: A set $X$ is open if and only if $$X=\bigcup_{\alpha\in J}B_{r_\alpha}(x_\alpha),$$ i.e. if $X$ is a union of open balls.

Hmm

Okay, makes sense

Of course, this decomposition is not unique.

@BernardoMeurer One direction is trivial. For the other, take $x\in X$. Then $x$ is contained in some ball, but that ball is by definition in $X$!

@BernardoMeurer Now consider the special case of $\Bbb R$. What is a ball in $\Bbb R$?

A neighbourhood like I saw in Analysis I

Yes, what does one call such a thing?

Eh? A neighbourhood was what I called it

$V_\epsilon(x)$

A semi-line or w/e the name in english?

interval

Ah, yeah, sure

@BernardoMeurer So what is the union of two balls in $\Bbb R$?

A union of intervals?

Eh

I don't know what you want me to say here

There's two cases: it's either a single interval or not

Ah, okay, yeah

So the union of two balls in $\Bbb R$ is either a ball or just two disjoint balls.

Sure, okay

That means that an open set of $\Bbb R$ can be uniquely written as a disjoint union of intervals.

(note I said intervals, not balls, here)

(you might have something like (a,infty))

Why does this matter?

Like, I see it's true but what's the point?

@BernardoMeurer Because I'm not done

So we just have an arbitrary union. But in fact, the union can be made countable.

So you have that a set in $\Bbb R$ is open if and only if it is a countable union of intervals.

That's a very important topological result used in measure theory

Yeah, alright

you might need it one day

Meh-asure theory I'd call it

@BernardoMeurer I'd try to wiggle out of this one by giving a real-world application but I probably can't :P

You can prove something very neat though

A monotone increasing function is differentiable except on a set of measure zero

None of this matters you know, ZFC is wrong

::Smiles::

C is definitely wrong.

@0celo7 Isn't that someone's theorem?

Not sure about ZF

@BernardoMeurer Lebesgue's, why?

Because I believe I saw that already

Well then

The proof requires what I just told you about

It's pretty neat, I hadn't thought about balls in $\Bbb R$ yet

Oops, said balls, hope someone doesn't ban me from the chat!

flagged

Good, I have to go to sleep anyway

@BernardoMeurer if someone does, it won't be because you said "balls" (and implying that it might be is really not helpful)

@BernardoMeurer So what I was saying earlier is interesting too

Thanks for the revision/help

Wait!

AH

Do you know what the closure of a set is?

@0celo7 that's not helpful either, I thought you were warned about that

@DavidZ Slipped my mind, apologies

@0celo7 Lol, you actually flagged that

no

@0celo7 I gotta go, we talk about closure tomorrow

ok

Goodnight man!

night

Cya

@DavidZ can we please have a chat-appropriate word for "balls" that could not be suspendable in any formulation?

Spheres.

@0celo7 "balls" is a chat-appropriate word for balls

Globes.

@DavidZ So why was I suspended?

No idea, offhand, though I remind you we don't discuss suspensions in public

It was an automatic suspension.

Why won't you discuss in public?

Let's ask @Shog9

For one thing, it's not productive. The people with the authority to override suspensions are the SE team. If one believes one has been unfairly suspended, one should take it up with them.

Do moderators not have the ability to overturn 30 minute automatic suspensions?

@DavidZ

@0celo7 authority, not technical ability

What does that mean?

You can, but are explicitly not allowed to?

30 minutes is nothing.

It means that moderators generally recognize that each others' decisions are made for valid reasons and will not override them out of respect for those reasons. If a decision is to be reversed, it should be done because the reasons for that decision are not actually valid, and nobody is better placed to make that determination than the mod who did it in the first place.

It's a sort of professional respect.

@DavidZ The automatic suspension is usually due to random 10k users.

^

i remember this situation. everyone agreed it was a bit...unnecessary.

I'm sure at least one moderator reviewed the decision, in that case. Though @skullpetrol is right, 30 minutes is not much.

@DavidZ And yet such suspensions were used against me in my secret trial & excecution hearing.

So it does matter.

What matters the most is you're back pal.

@0celo7 Counterquestion: Could you please use formulations that cannot be mistaken for crude sexual implications? Honestly, if you don't see that "I'll shove these balls down your throat" will likely get you suspended, there's no point in discussing this issue.

If it was actually secret, how do you know? ;-)

@ACuriousMind The chat is supposed to be accessible to 13 year-olds, and as such one must assume that each chat user is that innocent.

That's not how it works

@DavidZ Yes it is, why am I being punished for other people's dirty minds?

@0celo7 because you are chatting with other people. (To clarify: to whatever extent you are being punished for other people's dirty minds, it is because you're chatting with other people. I don't mean to comment on whether or not it's accurate to say that this qualifies as a punishment.)

@0celo7 I have no idea what that is supposed to mean. Are you seriously claiming "I'm allowed to make innuendo as long as 13-year-olds don't get it"? That's not how respectful conversation works.

@ACuriousMind I wasn't making an innuendo!

(removed)

@skullpetrol No, you were seeing comments that were migrated from a post.

He is not back, unfortunately.

@DavidZ Sigh. Why did you delete that?

@0celo7 Then what was your message about "one must assume each chat user is that innocent" supposed to mean?

I mean, really. Why did you take the effort to delete that!?

@ACuriousMind Because I was innocent!

When you have a hammer ...

that is so big

the introduction and the conclusion are bloat

Feel free to edit it out

I don't feel like reading it lol

whoa

Fair enough

Good comment, then

@0celo7 Ignorantia legis non excusat.

Whatever happened to that mathematician user with the Yugoslavian name? @0celo7

@ACuriousMind Is this a court of law or not?

You can't have secret trials and execution squads then quote law.

That wasn't law, he was casting a spell

@skullpetrol I don't know.

:-P

@DavidZ Machines cannot control magic.

@0celo7 There are no execution squads and this is not a court of law. Nevertheless, I think that that particular legal principle encodes a more general prinicple of human interaction that intent does not matter as much as the perceived meaning.

@ACuriousMind My friend @skullpetrol would beg to differ on the execution squad front. And I refuse to accept that legal principle when we do not have free discussion of suspensions, i.e. as long as there are secret courts.

You can't accept one and reject the other.

@skullpetrol you have the dubious distinction of being in a single-digit position among the top most-suspended chat users.

A low single-digit position

@Shog9 Is that for automatic suspensions or overall?

My point being, you'd have to be a lot more specific.

I'd imagine I have the most automatic, for no particular reason.

So, like just bad luck?

That's your theory?

If I'm finding $\langle \psi | H | \psi \rangle$, where $H$ depends only on $r$ (the radius), what bounds do I use for the integrals? For example, I'm doing this for the hydrogen atom, but I don't know how to decide the bounds for the radius. Is it $r=[0,R]$ or $r=[0,\infty]$?

@Shog9 There are some people who dislike me. Beats me why, but I think they flag everything I say.

Thousands of people chatting on this server, and somehow you manage to step on the equivalent of a rake hidden in the grass more often than just about anyone else?

@loltospoon for the hydrogen atom, it's infinity

@DavidZ why?

@loltospoon If you're integrating over $\Bbb R^3$ you'd do the usual thing, with measure $r^2drd\Omega$ and $0\le r< \infty$.

@loltospoon You integrate over all of space, always, the only reason to do otherwise is if you can leave out parts where the wavefunction is known to be zero. (Or if you have a cyclic dimension or some such nonsense :-P)

@Shog9 Yeah. I find it mysterious too.

@DavidZ ohhhh ok I think I've heard that reasoning before. Ok thanks!

So am I in the low single digits too?

@0celo7 be that as it may, you should probably consider that most moderators would just suspend you at the drop of a hat at this point, purely for efficiency.

At what point, for what?

@loltospoon no problem

@Shog9 I don't understand. We were having a pleasant conversation then you said most people would execute me at the drop of a hat? What kind of thing is that to say to a person?

@0celo7 Think of it this way... You're some random moderator in chat here, and you see a flag pop up. "Hmm, there's a flag, I wonder if this person just made an innocent mistake and... Oh, it's you again."

@Shog9 should I try and contact the Stackexchange team about bringing the sleigh rides back?

@skullpetrol I'd wait for, maybe November

If I want to use the variational principle to estimate the ground state energy of the hydrogen atom, what hamiltonian do I use? Is it $H = -\frac{\hbar ^2}{2m}\frac{d^2u}{dr^2}+(\frac{-e^2}{4\pi \epsilon _0}\frac{1}{r}+\frac{\hbar ^2}{2m}\frac{l(l+1)}{r^2})u$

Ok.

Next holiday is April Fools, and that's not usually a sleigh-ride sorta place

Do you watch march madness?

@Shog9 I sincerely doubt anyone remembers me.

And all the mistakes I make are innocent, anyways. There's a malicious flagger.

Anyway, I have potatoes boiling over so I gotta run, but... Yeah, you're both pretty well ranked on the top-100 troublemaker list for chat, so I wouldn't be too cavalier about it all. Everyone makes mistakes, but most folks don't make the same ones repeatedly.

Cheers!

@loltospoon Yeah, whatever the actual Hamiltonian is. I suppose you could also do it with an approximate Hamiltonian, but that's going beyond the variational principle as it's usually taught.

I'm a bit iffy about this Hamiltonian that I've written, but it's the only reasonable one that I've found in my book.

Cya

@skullpetrol I wish someone would tell me what mistakes I keep making instead of sniping me from the shadows then berating me.

@ACuriousMind Did you figure out your Witten quote?

No, I've been living for most of the evening :P

@ACuriousMind What does that mean?

(for example)

if I understood correctly

stupid computer touching the wrong window >.<

Why is it that in my book they define the variational principle as $E_{gs}=\langle \psi | H | \psi \rangle$, but on this website it defines it as $E=\frac{\langle \psi | H | \psi \rangle}{\langle \psi | \psi \rangle}$

@loltospoon $\psi$ needs to be normalised for the first to work.

@0celo7 so then is the second one simply explicitly showing the normalization while the first one assumes the wave function is already normalized?

@loltospoon probably

I'd have to see your book to be sure, but it does seem reasonable that they might assume the wavefunction is normalized.

@loltospoon It's generally bad to assume the wave function is normalized in this context.

So the second one is better.

Ok guys, I'm really not convinced with what I wrote above. In this other example, the person uses different values of $T$ and $V$ in using the variational principle to estimate the GSE of the hydrogen atom.

He/she doesn't use my obnoxiously large and complex Hamiltonian

@loltospoon let me first see what you wrote

er, what did you write?

@ArtOfCode Greetings.

Hang on

@0celo7 chat.stackexchange.com/transcript/message/35619221#35619221

@loltospoon Uh, that looks like the correct Hamiltonian I guess.

You can use a simpler one if you care less about the accuracy.

@0celo7 ok