The h Bar

0celo7

23:00

Google is now asking if I always want to translate portugese

Bernardo Meurer

I might drop out of college tbh

I don't want to do this anymore

0celo7

Holy shit is this one exam?

Bernardo Meurer

It's a prep problem set

I didn't find an actual exam

0celo7

honestly this isn't bad

Bernardo Meurer

It has very little to do with my classes

Secret

23:02

The disappearance of the Clovis people may be due to impacts

Bernardo Meurer

@0celo7 Found an exam

math.tecnico.ulisboa.pt/~jpnunes/CII/exames/T1-12-11-16.pdf

These are two in one, two versions of the same test

(Like we had A and B in Analysis I)

0celo7

@BernardoMeurer Just do the homework

Bernardo Meurer

::grunts::

0celo7

@BernardoMeurer I have no sympathy if you don't do the homework, I'm sorry

Bernardo Meurer

I do it

I just hate it

0celo7

23:05

This is standard calc 3

Secret

Using liquid repellant coatings in tanks to better control liquid tanks in microgravity for space fuel applications

Bernardo Meurer

Well, I didn't last semester cause it was some crazy shit so I solved exams since 1997

0celo7

Well Chub n Tuck is crazy

I would diss his area of research but I think it's cool :(

Bernardo Meurer

What's his research?

Chub n Tuck was totally nuts

0celo7

Operator theory and integral operators

I'm currently learning about holomorphic operator theory

Secret

23:16

Now that's interesting, if we can amplify this energy uncertainty in a local region, it might be enough to create a region where time dilates slightly less than elsewhere. Now pair them together, we might have time travel. Otherwise it is interesting to see time itself became more blurry

0celo7

@BernardoMeurer Do you know what a smooth function is yet?

heather

23:30

@0celo7 yes =)

DanielSank

@heather hi!

Physics Meta

0

Q: Does questions pertaining the depiction of dimensionally abnormal space belong here?

I want to ask a question along the lines of "What would we see if there are four accessible spatial dimensions, and we are 'lifted up' then rotated in the fourth axis?" It will be hard to explain/assess this question without details, so please be patient: In Flatland, there exist fictional flat ...

heather

@DanielSank, hello! How are you?

DanielSank

@0celo7 One with as many derivatives as I want.

Continuous derivatives.

@heather I am good. How are you?

0celo7

@DanielSank Ok, you'll be my victim. Do you know what a closed set is?

heather

23:33

@DanielSank Pretty good. Finished my service hours for a school project today, so I'm happy about that.

DanielSank

@0celo7 One whose complement is open ;-)

@heather Nice.

0celo7

@DanielSank *complement. Yes, that's correct.

heather

oh, funny story. i finished a program last night, and tested it and it was working. then my mom comes over, so I'm like, "mom, give me a number!" intending to show her the program. She gave me the one number that resulted in an error.

0celo7

Would you say that closed sets are nice or nah?

DanielSank

@0celo7 Well, no closed set ever spilled beer on my rug, so I have no beef against them. That said, I'd generally rather chill with an open set.

@heather What was the number?

0celo7

23:34

@DanielSank The point is that there are some really nasty closed sets

Like the Cantor set

DanielSank

@0celo7 Yes.

That's why I said I'd rather hang out with open sets.

heather

@DanielSank 32

DanielSank

@heather nice

ACuriousMind

@0celo7 And the complement of the Cantor set is nice?!

0celo7

@DanielSank Right, but check this. Given $C\subset\Bbb R^n$ closed, there's a smooth function $f:\Bbb R^n\to\Bbb R_+$ such that $C=f^{-1}(0)$.

@ACuriousMind It's a countable union of intervals, so yeah.

DanielSank

23:36

@0celo7 ::Reaches for pencil to draw picture::

Bernardo Meurer

@0celo7 Nope

DanielSank

@0celo7 That's kinda funny.

Lemme think about this for a minute.

@0celo7 Ok I'm not impressed.

I think I've even seen an explicit form of this function in the context of partitions of unity.

0celo7

You have a good memory, but it's not really explicit.

It's a horrible infinite sum.

DanielSank

@0celo7 I remember it being not so bad.

Bernardo Meurer

Don't talk about infinite sums near me

DanielSank

23:40

I'll have to crack open Munkres's Analysis on Manifolds if I remember.

Bernardo Meurer

I get triggered

ACuriousMind

If you are willing settle for continuous it's very easy ;)

DanielSank

I do recall thinking that partitions of unity is a very beautiful idea.

Bernardo Meurer

I have PTSD

DanielSank

@ACuriousMind Yes indeed.

Dat kink tho

0celo7

23:41

$$f(x)=\sum_{i\ge 1}\frac{(r_i)^i}{2^i C_i}h\left(\frac{x-x_i}{r_i}\right)$$

where $h$ is the standard bump function

and the rs and Cs are suitably chosen

@DanielSank What would you pick?

DanielSank

@0celo7 I have no idea what you're asking me.

0celo7

What function would you pick to do it?

DanielSank

To do what? To be a partition of unity?

0celo7

No, $f^{-1}(0)=C$

$f$ continuous

ACuriousMind

...he's asking which continuous function you would pick

DanielSank

23:44

Oh, ehh, like I said, I'd go look at Munkres and refresh on partitions of unity.

0celo7

For continuous there's an easy one. The distance from $x$ to $C$.

DanielSank

I think he uses some kind of exponential function. I really don't remember any more.

I remember being surprised that there are smooth functions that can be constant over an interval.

ACuriousMind

@0celo7 Yep, that's what I was thinking of.

DanielSank

Or was it just continuous...

Bernardo Meurer

OK guys, here's my piece of advice for the day:

Do not dance and shave

0celo7

23:45

@DanielSank Hmm? Bump functions

Bernardo Meurer

I almost sliced half my face of because of Death Grips

ACuriousMind

@DanielSank That's because we physicists often intuitively confuse smooth and analytic ;)

0celo7

@ACuriousMind today my QM prof took a limit by taylor expanding first

ACuriousMind

You want to think in terms of Taylor series, but there's no Taylor series that gives such a function

That actually came up earlier today already

0celo7

When I asked him if he could just use Hospital, he sincerely thought that is what he was doing

DanielSank

23:47

@ACuriousMind What's the diff?

0celo7

@DanielSank One admits a Taylor series expansion about each point in the domain, one is just smooth.

ACuriousMind

@DanielSank Analytic functions are those who have a Taylor series that converges to them, smooth functions just are infinitely differentiable

DanielSank

To me, "smooth" means either 1) Has as many continuous derivatives as needed for ensuing statements to be true, or 2) Has infinite continuous derivatives.

@0celo7 I did not appreciate that those were different.

0celo7

1) is "smooth enough" in the rigorous world ;)

ACuriousMind

@DanielSank They are the same over the complex numbers, but they differ over the reals

0celo7

23:49

For complex numbers the crazy thing is that ONE derivative implies smooth AND analytic.

DanielSank

I don't understand complex numbers.

0celo7

And that you only have to check two simple equations! (Cauchy-Riemann)

ACuriousMind

It's actually physically very important that these are different, because it means that smooth-but-not-analytic contributions can be invisible through perturbation theory!

DanielSank

If I have a function $f:\mathbb{C} \rightarrow \mathbb{C}$, I don't really understand what $df/dz$ means.

0celo7

@DanielSank clearly it's a derivation of the germ of functions

ACuriousMind

23:50

@DanielSank It's just $\partial_x + \mathrm{i}\partial_y$ :P

DanielSank

@ACuriousMind Heh, I once spent half a day failing a statistical mechanics exam because I was trying to Taylor expand a function about an essential singularity.

Or something like that.

@ACuriousMind Hmmmm, perhaps.

ACuriousMind

There's a nice exponential $\mathrm{e}^{-1/x}$ whose Taylor series about 0 is just 0 - you can't see it by expanding at all, and it turns out almost all of the "non-perturbative" effects in QFT are of this form and hence invisible to naive perturbation theory.

@DanielSank Ah, I actually botched the sign, but it wasn't a joke, this is called the Wirtinger derivative

0celo7

Yeah but no one defines complex differentiation like that.

DanielSank

@ACuriousMind Yeah that was it.

0celo7

I don't have a good feeling for why complex differentiation is so much better than real differentiation. I guess it's because you have $f(z)$ instead of $f(x,y)$, so one less argument?

Maybe?

Who knows

DanielSank

23:59

A complex differentiable function is like a vector field with zero curl and zero divergence.

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