The h Bar

Slereah

15:00

I mean I think you can just add it inside the switch after the cases

I think that will make it execute after the break

Bernardo Meurer

Or I can call it from the move function

0celouvsky

@ACuriousMind I think it's more common in the humanities

Bernardo Meurer

@0celouvsky Proprietary malware?

0celouvsky

@BernardoMeurer no, someone took shit from my room

Bernardo Meurer

@Slereah If a function receives an array (and it decays to a pointer) How can I from that function call another function to receive the same array?

@0celouvsky Reb

0celouvsky

15:01

I accused her and she got mad

So I don't think that will work

Bernardo Meurer

lol

Slereah

pass the adress of that array to the function?

Bernardo Meurer

Ah, wait

I did that and it imploded

but it was something else

0celouvsky

@ACuriousMind libanswers.snhu.edu/faq/129372

@Slereah Did you see my essay on Lebesgue integration?

Slereah

I did not

0celouvsky

15:04

what the hell lol

3 hours ago, by 0celou7

@Slereah Let me put it this way: Lebesgue integration is not really useful for the new types of functions you can integrate in practice...no one cares if you can integrate the indicator function of $\Bbb Q$. It is useful for telling you certain operations on sequences are nicely behaved. For example, $f(x)=\lim_n f_n(x)$ is aways Lebesgue integrable if the $f_n$'s are (the integral might diverge of course), but in the Riemannian case this is not true. Furthermore, the Lebesgue integral [...]

start here

Bernardo Meurer

@Slereah What's your github name?

Slereah

Got none

Bernardo Meurer

I don't know why if I call it from inside the function it won't work

@Slereah :(

Balarka Sen

@0celouvsky ...

0celouvsky

@BalarkaSen Yes?

Bernardo Meurer

15:06

@Slereah How do you not have a GH account?

0celouvsky

@BernardoMeurer I don't have one either

Slereah

why would I

Bernardo Meurer

@0celouvsky That's because you want to spite your sister

@Slereah To help me :3

Slereah

not a very attractive prospect

0celouvsky

lol rekt

@BernardoMeurer I don't want to spite her

Bernardo Meurer

15:08

@Slereah You'll regret this when I become a gorgeous woman in a couple years

0celouvsky

How could you become a woman if you're a man now?

Bernardo Meurer

Am I?

We haven't met

0celouvsky

Michelle says you're a man

well, a boy

Bernardo Meurer

And you trust her?

ACuriousMind

lol, this paper by Milnor is great: "One first shows that $\Omega_n^\text{spin}$ is isomorphic to the stable homotopy group $\pi_{n+k}M(\mathrm{Spin}(k))$ of a suitable Thom complex; and then determines these homotopy groups by a formidable computation. No details will be given"

Slereah

15:11

forgot to close your dollars

ACuriousMind

I am now angry that the authors of that other paper would reference this one if it just states the fact they reference it for without proof or any details

Balarka Sen

@ACuriousMind Sounds like Thom-Pontryagin

ACuriousMind

They might as well have just asserted it :P

Slereah

probably trivial

ACuriousMind

@BalarkaSen It does - what irks me why they would reference Milnor just vaguely describing the construction instead of something with more...substance. Has no one ever actually published a more precise computation of the spin cobordism groups?

Balarka Sen

15:14

No idea, actually. Haven't thought about spin cobordisms; what's M(Spin(k))?

0celouvsky

stahp with the topology pls

ACuriousMind

@BalarkaSen I presume it's the "suitable Thom complex"

0celouvsky

I walk into physics and there's TEST 2 up on the projector

my heart exploded lol

turns out we're going over test 2, which we took 2 weeks ago

Yashas Samaga

wrong room, sorry

Balarka Sen

looks relevant: map.mpim-bonn.mpg.de/Spin_bordism

ACuriousMind

15:17

The assertion that all spin manifolds in 7 dimensions are null-cobordant is kinda important for another paper I'm reading, which in turn is kinda important for another...I'm several levels into these dependencies and at the bottom I find Milnor vaguely waffling about, I'm not amused :P

@BalarkaSen lol, "The spin bordism groups up to dimension 8 are given in [Milnor1963a] without proof. Milnor states that this is the result of a formibable calculation"

0celouvsky

Maybe I'm looking at the wrong kind of math but where does this stuff even show up?

Balarka Sen

lol whoops

@0celouvsky Classifying bordisms (upto structures) is a pretty classic problem

0celouvsky

All this algebraic topology is probably not that complicated but I never see algebraic topology more complicated than some homology groups

So it appears quite horrifying to an outsider like me

Balarka Sen

Oh, it is kind of nontrivial.

It's a hard area of algebraic/differential topology

@ACuriousMind So I guess you probably won't see a lot of places where you will find the calculation done

ACuriousMind

@0celouvsky Well, in this case it turns out that you can define an interesting invariant of a $G_2$-manifold when it is the boundary of a $\mathrm{Spin}(7)$-manifold, and we kinda want that invariant for all $G_2$-manifolds so we need to know that they all are the boundaries of some spin manifold, which is equivalent to the assertion that the cobordism group is trivial.

0celouvsky

15:21

@BalarkaSen I found an alternative proof of that $L^p$ thing. It's actually a really neat thing: if $L^p(\Omega,\mu)\ni f_n\to f$ pointwise, you can't say anything about the $L^p$ limit. But if you know $||f_n||\to ||f||$, then you can say $f\in L^p$ and $f_n\to f$ in $L^p$ norm.

ACuriousMind

@BalarkaSen Well, at least that page you linked gives a general classification that should give the same result if I understood what it is talking about :D

0celouvsky

It's surprising because convergence of norms usually gives you zero information.

ACuriousMind

I guess I'll file this under "somewhat worrisome black box" and move on.

Balarka Sen

Ah, yeah, a lot of manifold invariants are defined by fiddling with a manifold it bounds. (eg, all 3-manifolds bound 4-manifolds)

and showing it's independent of the cobordism

@ACuriousMind Yeah.

0celouvsky

@ACuriousMind Christ

Balarka Sen

15:24

@0celouvsky a'right but i dunno analysis

0celouvsky

@BalarkaSen The idea is that the only way $f_n\not\to f$ in $L^p$ is if "mass gets pushed out to infinity"

Balarka Sen

got that

0celouvsky

So the convergence of norms is the statement that you don't lose mass, basically

For a counterexample, take $\chi_{[0,1]}$ and scoot it off to infinity

Balarka Sen

sure, sure. i'm just sayin', i don't particularly care much

0celouvsky

so @BalarkaSen do you like my new name?

Balarka Sen

15:28

it's neat

0celouvsky

I should add a german twist. @ACuriousMind ?

0ßelouvsky?

Will PSE let me put a ß in my name?

ACuriousMind

I think it allows arbitrary unicode?

So a ß in particular will be fine

0celouvsky

so I can put the Egyptian hieroglyphic ejaculating penis in my name?

I need an answer

Secret

[Caption] I think I need to revise joint probability distributions, cause I need to understand how states go from no correlation to some correlations

ACuriousMind

@0celouvsky You can, in the sense it is technically possible. However, that'd be somewhat distasteful and we'd probably have to reset it.

0celouvsky

15:32

Fair enough

Secret

The only states that are uncorrelated in this diagram are the basis states

So that means, given some vanishingly small $\epsilon > 0$ the following state has a nonzero correlation

0celouvsky

ugh

Secret

$$\frac{\sqrt{\epsilon - 1}}{\sqrt{\epsilon}}\lvert + -\rangle + \frac{1}{\sqrt{\epsilon}}\lvert - +\rangle$$

0celouvsky

@ACuriousMind why is the wavelength of fire ~ tungsten filament ~ sun

why can we see fire?

it has to be more than a coincidence

Secret

This is so weird, yes, geometrically and algebraically it is easy to understand what is happening because you are essentially rotating the state in the hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_2$ thus nothing is really weird. However, the transition from no correlation to some correlation looks so abrupt

ACuriousMind

15:40

@0celouvsky It's a miracle

0celouvsky

why do people know the refraction index of diamond randomly!?

Secret

that is, one only need to rotate the state a little bit off the basis state and suddenly you gain some correlations

0celouvsky

why do people know any random values of things randomly

ACuriousMind

An engineered one in the case of the filament - you make the filament precisely so thick that it reaches the correct temperature at the usual current

0celouvsky

the only reason I know the mass of the electron is because of ~~voldemort~~ you know who

other than that I cannot tell you any physical constants!

Balarka Sen

15:42

voldemort knows the mass of electron?

0celouvsky

but these people know specific heats of iron!?

@BalarkaSen yeah, check the 6th book

Balarka Sen

???

ACuriousMind

Also, I'm not convinced "wavelength of fire" ~ "wavelength of sun" is really accurate. What's true is they're both visible, but consider that blackbody radiation can't really leave the visible spectrum once it has non-neglegible part above infrared

0celouvsky

he uses pair production to make 511keV photons

the magic is not really magic

it's just quantum field theory

ACuriousMind

So all "hot" radiation will be visible

Secret

15:44

The following state, however, has no correlations of any kind because it is a product state
$$\frac{1}{2}(\lvert ++\rangle+\lvert +-\rangle+\lvert -+\rangle+\lvert --\rangle)$$

0celouvsky

@ACuriousMind hmm, I didn't know that

what's the precise statement?

Balarka Sen

@0celouvsky i know the charge of a neutron

ACuriousMind

@0celouvsky The "peak" can move past the visible spectrum, but at that point the blackbody is radiating so much energy that the tail in the visible spectrum is still, well, distinctly visible

Balarka Sen

win win

ACuriousMind

We call that "glowing white-hot", I think

0celouvsky

15:46

ahh

interesting

@BalarkaSen I know c to one decimal place

one significant figure I mean

c=1

ACuriousMind

Also, @0celouvsky, you might be interested to know that the yellow color of the sun is an atmopheric effect

0celouvsky

@ACuriousMind I don't know why people say it looks yellow. It looks white to me

ACuriousMind

Well, but it's yellow-tinted

0celouvsky

I'm color blind

ACuriousMind

In any case, don't look too long into the sun to see if it's yellow or not :D

0celouvsky

15:49

what is the atmospheric effect?

@ACuriousMind So is the majority of radiation from the sun in the visible range?

ACuriousMind

Rayleigh and Mie scattering

0celouvsky

I'm assuming we've evolved to see that part

ACuriousMind

@0celouvsky I believe so, yes.

0celouvsky

I just think it's interesting that lots of other things radiate in the same range

And it's a pretty narrow range of the whole spectrum!

ACuriousMind

@0celouvsky Well, what are those lots of things except for other blackbody radiation?

Balarka Sen

15:50

I am forgetting, is H_alpha in Balmer series the one where electron jumps from n = 6 to n = 2?

0celouvsky

@ACuriousMind spectral lines from atoms are not blackbody, are they?

Balarka Sen

or n =3 to n = 2?

ACuriousMind

@0celouvsky They ain't, but they are also frequently outside the visible range

Also, obligatory Abstruse Goose

Balarka Sen

Ok, I guess the latter

0celouvsky

I was watching the Joe Rogan Experience with that one guy who used to be married to Katy Perry

Pretty sure he was out of his mind high, but he said some interesting things

Like what if we couldn't smell...it would be impossible to comprehend what smell is

So in out understanding of the universe we're limited by things we can comprehend via our senses

So there might be a lot of stuff we're missing because of that

His solution was take LSD and shrooms

ACuriousMind

15:54

@0celouvsky Related, What's it like to be a Bat?. Such impossible-to-explain subjective experiences are often called Qualia

Secret

So... generalising this observation and using this [PSE] (http://physics.stackexchange.com/questions/164700/how-are-superposition-and-entanglement-related/164716#164716), we could have claimed that the linearly polarised state

$$\frac{1}{\sqrt{2}}(\lvert +\rangle+ \lvert -\rangle)$$

is a correlation between the clockwise polarised state or anticlockwise polarised state

One can demonstrate the mathematical similarities by using the up state and down state projectors, in that one of the basis state will be inner product to 0 and vanishes

ACuriousMind

As with so many things, philosophers of mind have so far not even agreed on whether they exist in a meaningful sense :P

Slereah

I was a bat once

It was alright

0celouvsky

> Lewy's example led L. HORMANDER [3] to develop a systematic method of constructing linear partial differential equations without solutions.

lol

Slereah

Sounds rather dickish

What did he do, give them to his students as homework?

skill patrol

15:59

usually

Balarka Sen

@0celouvsky good idea. i also suggest murder and psychosis

0celouvsky

Yosida likes puns

Secret

Well I guess, my semi philosophical question is, for any two particle state, it is just a ray in the hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_2$, but why do they behave very differently, e.g. some of these are product states and some of these are entangled states and some of these are something in between, is it because we are effectively only seeing a shadow of them even if we combine results from both subsystems and establish a correlation ?

0celouvsky

He defines a fundamental solution of a partial differential operator $P(D)$ as a distribution $E$ s.t. $P(D)E=\delta$

PDE!

@BalarkaSen did you check out r/imgoingtohellforthis?

Balarka Sen

yeah i did

Secret

16:02

[Caption] They all look like rays in here...

0celouvsky

Every constant coefficient PDE has a fundamental solution!

Crazy!

Secret

0celouvsky

dear god

Secret

Above is a bad example demonstrating why you should never ever try to draw these spaces. They are huge

0celouvsky

@Secret you should draw L^2[0,1]

Secret

16:06

Well, I don't even know what $L^2$ look like

also I suspect it will be completely covered with ink once you plot all possible functions in

0celouvsky

it is infinite dimensional

Slereah

aren't we all infinite dimensional, in the end

0celouvsky

no

@ACuriousMind is 3-dimensional by AdS/CTF

he's an AI on the boundary of spacetime

Secret

PS, don't try to draw the rational line, you will get rekt very quickly by its density

0celouvsky

lol

@Secret try to draw a non-Hausdorff manifold

Balarka Sen

16:17

you can easily do that modulo enlarging the notion of drawing

0celouvsky

:P

Slereah

The split real line is pretty easy to draw

0celouvsky

that's Hausdorff if you draw it

you need to draw the topology somehow

Slereah

that's what the little cup is for

0celouvsky

what?

Balarka Sen

16:23

It's a sketch not an actual picture is what 0celo7's saying. but that shouldn't be a problem

0celouvsky

@BalarkaSen the problem is that a subspace of a Hausdorff space (the page) is Hausdorff

Balarka Sen

go away

0celouvsky

so you need to indicate the topology somehow

Balarka Sen

that's not a problem

Slereah

You draw a little $\supset$ on the end of $(0, \infty)$

and a little ball on the $0$

to signify the topology

Balarka Sen

16:25

i like the line with two origins better

Slereah

The best 1D non-hausdorff manifold is the complete feather

Secret

[Caption] Cofinite topology wrapped around a half infinite cylinder:

Balarka Sen

@Slereah nah, leaf space of reeb

Secret

A few open sets illustrated in gray

Slereah

The complete feather is like

Take $\Bbb R$

Split it at every point $x \in \Bbb R$

Repeat the process

forever

it's a very fluffy manifold

Secret

16:29

Is $\Bbb{Q} \times S^1$ a manifold?

Slereah

What's a bijection between $\Bbb Q$ and $\Bbb R$

Secret

There isn't one since $\Bbb{Q}$ is countable while $\Bbb{R}$ is not

Slereah

there's your answer

0celouvsky

@Slereah that does not look like a manifold to me...

Slereah

@0celouvsky And yet it is

0celouvsky

16:32

proof?

Balarka Sen

It is not a manifold. Any nbhd of any point is disconnected. In the product topology anyway.

0celouvsky

I'm assuming it doesn't have the subspace topology from $\Bbb R^2$?

Slereah

Ah

Apparently it doesn't branch at every point

Only at rational points

books.google.fr/…

0celouvsky

Even if it branches at one point it's not a manifold...

Balarka Sen

@0celouvsky The topology is different.

It's like $\Bbb R \times \{0, 1\}$ with negative real axis squished at both. The neighborhood at the branch point belongs entirely in 1 of the 2 branches, not a tripod

Slereah

16:34

Only if you assume hausdorfness is part of the manifold definition

Balarka Sen

Locally modeled on stuff like that

0celouvsky

christ, what is that?

Bernardo Meurer

I'm a non-hausdorff manifold

Balarka Sen

I wonder how leaf space of the Reeb foliation looks like. Never gave it a thought.

0celouvsky

@BalarkaSen Every time I try to learn about foliations I am turned off by the really quite technical definitions

Slereah

16:36

I'm guessing it looks like a thing that can't be represented in 2 dimensions

0celouvsky

Plaque charts and whatnot

Various submanifold topologies

Slereah

I want to achieve satori so I can picture what a principal bundle looks like

Balarka Sen

@0celouvsky Yeah, it looks quite technical because it's really hard to put things in words. Once you get them it's not very hard.

0celouvsky

@Slereah just picture a trivial bundle with your favorite 1-d base space and 1-d Lie group

Slereah

Yeah but

How do u do the structure group

I mean I can picture $(R, S, Z_2)$

But how do you picture $(R, S, S)$

0celouvsky

16:45

what is $(R,S,S)$ supposed to be?

Slereah

Base space $\Bbb R$, fiber $U(1)$ with structure group $U(1)$

or something

Wait, is the structure group of principal bundles supposed to be the same as the fiber itself?

Or am I mistaken

0celouvsky

I don't think you can really `see' how the structure group acts

it should be the same picture regardless of the structure group

@BalarkaSen is that right?

Slereah

Isn't the mobius strip specifically one with $\Bbb Z_2$ as the structure group?

Balarka Sen

@Slereah Leaf space of a Reeb foliation? I think it's like a line where every point neighbors two points far away somewhere

(the boundary leaves)

@Slereah @0celo I think you can probably see the action of the structure group using monodromy?

Slereah

IIRC the twisted torus also has a non-trivial group

0celouvsky

16:52

@BalarkaSen No, I mean if you just draw the bundle space you wouldn't be able to tell what the structure group is

Slereah

wouldn't they just all look like the product space if that was true

0celouvsky

locally a product space

and that's exactly what they are

Slereah

Well yeah but the global part comes from the structure group

So

0celouvsky

that's a regularity condition, not a pictorial condition...

ACuriousMind

@Slereah The edges give a $\mathbb{Z}/2\mathbb{Z}$-bundle over $S^1$

But the strip itself is a line bundle

@Slereah yes, in a principal bundle the fiber is isomorphic to the structure group

The crux of the whole construction being that the fiber lacks a distinguished identity element, though, so the isomorphism is not "natural"

0celouvsky

16:59

@ACuriousMind what is natural, anyway

OBE

@0celouvsky why'd you drop the 7?

ACuriousMind

@0celouvsky I don't feel like giving the category-theoretic construction that you wouldn't appreciate, anyway :P

0celouvsky

because it didn't make sense any more

@ACuriousMind do you mean a natural transformation?

ACuriousMind

@0celouvsky Not exactly

0celouvsky

there's something more natural??

skill patrol

17:03

Without the 7 it's not cool anymore.

0celouvsky

I'm not having an old woman tell me what's cool or not.

skill patrol

-_-

::changes username to 777::

0celouvsky

you won't do it

no balls

OBE

@0celouvsky who are you? XD

0celouvsky

what kind of question is that?

ACuriousMind

17:06

@0celouvsky Ah, sorry, it does rather straightforwardly involve natural transformation. For instance, the natural isomorphism to the double-dual of finite-dimensional vector spaces is just the natural transformation between the identity functor and the double-dual functor evaluated on a certain space.

Balarka Sen

a guy from the internet

skill patrol

^

OBE

@0celouvsky idk who 0celouvsky is but I know 0celo7

0celouvsky

@ACuriousMind we've been over this before, so I don't know why you don't think I'll understand it

OBE

where is he btw? haven't seen him around lately.

0celouvsky

17:07

@OBE then don't talk to me I guess

ACuriousMind

@0celouvsky I didn't say you wouldn't understand it! I said you would not appreciate it :P

0celouvsky

@ACuriousMind category theory is a big part of analysis on Banach manifolds, I guarantee we'll talk about it more one day

Bernardo Meurer

17:38

@dmckee Are you around?

dmckee

Sorta.

I'm allegedly working.

5

Bernardo Meurer

@dmckee Quick C99 question

If I have a 2-D array foo[MAX][MAX]

dmckee

I'm a native speaker of the 89 standard, but OK.

Bernardo Meurer

and then some function that takes in a 1-D array bar(foo[MAX])

I can pass the "2nd dimension" of the array to the function easily with bar(foo[x]) where x is some int

Is there a similarly easy way to pass a "1st dimension" of the array?

dmckee

A "2D array" in c is more properly conceived of as an array of arrays.

Bernardo Meurer

17:41

::nods::

dmckee

What you do when you call bar(foo[0]) is formally passing the first element of bar.

Which is usually written as a row.

As I understand the question, you are asking about passing a column?

If so, then no. There is no easy way.

Bernardo Meurer

Yep, that's what I was asking

Sigh

Here I go allocating more memory

dmckee

If you have to have that feature than you may want to implement the "2D array" in some other data structure.

Bernardo Meurer

I want to change this thing to some nice struct eventually

but I'm a lazy bastard right now

dmckee

A 2D linked structure or a big hash-map where the index is formed by, say, concatenating the row and column values.

What's best depends on what other featuers you need the thing to have.

O(1) random access lookup or O(N) walking and so on.

Bernardo Meurer

17:45

@dmckee Have you played the game 2048?

dmckee

@BernardoMeurer If you can afford the memory then storing the 2D array in both row-major and column-major is an option.

@BernardoMeurer Yeah.

Bernardo Meurer

That's what I'm doing, in C with SDL

dmckee

OK, so there is a trick code-golfers use.

Bernardo Meurer

the 2-D array in question hold the 5x5 board, and each cell has the index of 2^n of the square in there

where 0 means no square

dmckee

Rather than passing a row or a column, do this....

* Store the cells in a 1D array.

* You'll have to do the row-column math yourself, but that is easy.

* Pass an element and a skip-distance.

Bernardo Meurer

17:47

    for (int row = 0; row < MAX_BOARD_POS; ++row){
        int buf[MAX_BOARD_POS] = {0};
        for(int i = 0; i < MAX_BOARD_POS; ++i) buf[i] = _board[i][row];
        collapse(buf, false);
        for(int i = 0; i < MAX_BOARD_POS; ++i) _board[i][row] = buf[i];
    }

dmckee

* Make the skip distance 1 to walk a row.

Bernardo Meurer

I did that

dmckee

* Make it row-length (i.e. 5) to walk a column.

Bernardo Meurer

Not exactly a good solution lol

JR told me to use a 1-D array too, but it seemed unnecessary

dmckee

There is, of course, some fussing stuff around the limits.

And you have to decide if you want to walk it circularly (use modular arithmetic) or not.

@BernardoMeurer Well, it's mess, but you should encapsulate all that in a set of access functions to that you just present a opaque object to the rest of the program.

Bernardo Meurer

17:51

@dmckee Oh, that is inside a function

If you have a GH account I can give you access to the repo

dmckee

Much as I'd like to play with this notion, I'm afraid I have too much on my plate right now.

Bernardo Meurer

@dmckee LAAAAAAME

:)

Thanks for the help, I'll use the crappy for loops for now, probably move onto something better once basic functionality works

dmckee

^ That's the way to "just get it working".

Bernardo Meurer

Is that a good thing or a bad thing?

DanielSank

Hi, everybody.

Bernardo Meurer

17:58

@DanielSank <3

DanielSank

<3

@0celouvsky </3

dmckee

@BernardoMeurer It's a good thing for turning in assignments, and may be a good or bad thing for maintenance depending on how messy the code is. It's a bad thing from the point of view of elegance and is often less than optimal from a performance point of view.

Bernardo Meurer

@dmckee You had me at the first sentence

0celouvsky

@DanielSank :( what

DanielSank

@0celouvsky I like teasing you.

Bernardo Meurer

18:11

@dmckee What naming convention people use in C?

for like functions and variables and so on

dmckee

All of them. Depends who you ask.

Bernardo Meurer

I'm thinking camelCase for functions and under_scores for variables

0celouvsky

@DanielSank hmm

dmckee

I loathe underscores.

Bernardo Meurer

@dmckee :(

I like it

They're nice to read

dmckee

18:13

I also find that I rarely need multi-word variable names if (a) I've partitioned the code into enough functions and (b) I think carefully before choosing them.

So I use all-lower for variables and camelCase for functions.

Bernardo Meurer

@dmckee I have board_size and square_size for example

dmckee

I also use a prefixedCamelCase when I'm writing an encapsulated objects to make up for the lack of namespaces.

Bernardo Meurer

and some #defines like MAX_BOARD_POS

dmckee

I have to say that namespace encapsulation is something that c++ does much better than c.

Bernardo Meurer

::nods::

0celouvsky

18:21

@DanielSank as a show of good faith, I will impart upon you the knowledge that any PDE with constant coefficients has a green's function

Sir Cumference

Howdy

Ben Niehoff

any (linear) PDE? really?

ah, never mind

a Green's function, certainly

but there might be many :D

Bernardo Meurer

18:39

@0celouvsky Your name should be 0velouFREEvsky

To show that you like to be Free as in Freedom

Emilio Pisanty

naw, mate

next in line should be 0celoñez

since it appears that the theme is patronyms

after which probably 0celoupolous

whatcha think @0celouvsky?

Ben Niehoff

@EmilioPisanty do you know much about the etymology of -ez? I'm curious

was there a genitive in Old Spanish?

the oldest Spanish I know anything about is Quijote's Spanish, which is not that much different from modern, and certainly didn't have cases

ah, I just found it on Wiktionary

"From the Latin genitive suffix -icī, from -icus, originating as a calque of surname-formation conventions of the Visigoths."

dmckee

18:57

O'0celo7?

3

Sir Cumference

@dmckee ...-san

dmckee

Mac0celo7?

0celouvsky

@dmckee oooo

@BenNiehoff certainly? For a physicist I'm sure it's trivial :D

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