@Slereah Using goto is horrible style in almost all cases :P

@Huy bah

I'll pick a fight, I don't care.

just listen to some adele

no!

@ACuriousMind There's a few good uses of goto

Mainly getting out of nested loops

Gotos aren't bad really

They were just abused a lot and it spooked people

@Huy rekt

@ACuriousMind Did I get flagged?

Hell all control operations are just gotos deep down

@Slereah Yes, but goto often makes code harder to follow when reading it than using specific loops

Well spaghetti code is best avoided, yes

But I feel people tend to have a knee jerk reaction to gotos

Plus really, gotos are great, compared to what was before!

You don't want to increase the instruction counter by hand

You will punch holes in cards and you will like it!

Still not a lot of situations where goto is really advised outside of assembly, but no need to cry for help if you see one either

I have a friend who is a retroprogrammer

He codes for the Amstrad CPC

It is fun to hear his tales

The tricks people used back then to save one byte of memory

@ACuriousMind dude

My dad had to do that back in the day

at 1 in the morning because that's when undergrads got computer time

Our programming teacher used to describe to us how they did a program back in the days

First, write it down on paper

Make sure it is all well and good

Can't have any bugs

Then give it to the typist to punch the punch cards

Then put them in your little briefcase

Take the train to Paris

oh god

Give the punchcards to the computer people at Orsay

Then they run it

And bam, you have your results

Unless it crashes, of course

then you wait a few more weeks because a spot at the computer takes time

how long ago was this?

my dad's college had a computer back in the mid 60s

60's too, I think

I guess back then a computer was too expensive to have one for every university

should have gone to a better uni then!

what does Hawking & Israel talk about

is it an easy read?

It's a series of chapters by various authors on topics in GR

In between Wald and HE I'd say

I want to reread HE one day

and take detailed notes

the Cauchy problem would be a cool senior thesis

Apparently this is what Orsay had

IBM 360

The Cauchy Problem of General Relativity on the Worst Torus

60's groovy chick included

I think Purdue had a Cray

Apparently it had 2-4 MB of RAM

@DanielSank I got it

I was being stupoid

@0celo7 k

@0celo7 k

By the way

You know what had the best keyboards?

no, but I'm sure you'll tell me

The LISP machines

Look at all those keys

press HYPER + RUB OUT

Good times

wat are you on??

It also has a thumbs up and thumbs down button

For facebook, no doubt

APL keyboards are also pretty impressive

@ACuriousMind What exactly does it mean that $\exp_p$ depends smoothly on $p$?

I had a girlfriend who is an actuary and she bought an apl keyboard

so complex

>actuary

Why APL though

How old are you

APL was vaguely popular in the 70's

Or did she work at a company that had legacy APL

really don't know why but this was in like 2010 ...

Legacy probably then

@0celo7 The map $M\times TM \to M, (p,v)\mapsto \exp_p(v)$ is smooth in $p$.

APL is like FORTRAN and COBOL

The languages that won't die

i thought it was pretty dead language but where she worked required her to write APL

no 2011

Oh, and ADA

@ACuriousMind Exactly, what does that mean :P

Hm, no, that isn't right

C++ should probably also die

wut

And Java

what would we be left with then

C# is fine

Python is okay to code in a hurry

C is god, of course

ofc

Assembly you can't really get rid of!

COBOL

php

get rid of that and most people die

Well no

But people might get poor

(Since it is mostly used in banks nowadays)

Some COBOL

::master baits to GDP::

^some FORTRAN

Can you feel that GDP

oh...baby

@Slereah younger than you :p

How do you know

@ACuriousMind What isn't right

@0celo7: Okay, I think it just means the map $\exp: D\subset TM\to M,v\to \exp(v)$, where $D$ is the set of vectors for which a geodesic $[0,1]\to M$ with them as tangent exists, is smooth. $\exp_p$ is the restriction of that map to $D\cap T_pM$, and this is "smooth in $p$" in the sense that $\exp$ is smooth and so this restriction is also "smooth" in a vague sense. (It would become actual smoothness if the maps $T_pM\cong \mathbb{R}^n\to M$ formed a smooth manifold, I think)

@0celo7 My statement wasn't right, $\exp$ is not a map on $M\times TM$.

@ACuriousMind Yes, I see you have elaborated.

@Slereah it says so on your profile :)

You whippersnapper

@ACuriousMind Uh

I don't get it

but I'll leave you be

In particular, I don't see how that helps with my $\rho$ issue.

Well, that I don't know either, you just asked me what it means for the exponential map to be smooth in $p$ :P

@ACuriousMind I was hoping you would have some epiphany and it would fall out

@ACuriousMind Ok, I think the argument here is that there is an nbd $U$ of $p$ containing $q,r$ and this is all happening inside of $B(\rho,p)$. Then there should be a geodesic connecting $q,r$ and thus the geodesic ball around $q$ and $r$ should have a "similar" radius to $\rho$.

So we make these nbds in which all points have a "similar" geodesic ball radius.

And by compactness we can cover $M$ with a finite number of these.

So we just pick the smallest radius.

wad u trying to prove @0celo7

1+1=2 on manifolds

@Huy You wanna help?

I can give you definitions.

no

whyyyyyy

no one wants to help me

well at every point there exists a nbhd with polars and it's compact so you're done?

@Huy that's what I thought

@ACuriousMind didn't like that

I don't know why

@ACuriousMind why

The statement does not only say you can cover the manifold with polars of constant radius, but that that radius gives a polar at every point

oh

that's true

you can get a nbhd for every single point

and since it's compact you can get finitely many nbhds

and then get a minimal radius

but this doesn't mean that this minimal radius now works for every point

just for those which were associated to the nbhds in the first place

doesn't mean that minimal radius works for all different points in the same nbhd

why do you need such a global radius in the first place, @0celo7?

I don't recall such a corollary

so I'm wondering what it'd be good for

@0celo7: ok we proved the statement from your cor 1.4.4 differently I guess

I'm not 100% sure but almost that we only used normal coordinates

and not polars

Polars aren't used.

so why cor 1.4.3

I think 1.4.3 works for polars and non-polars.

the most delicious experiment

Also he doesn't thaw his peas before boiling them

Savage

what happens if you get an infinite while loop on a punch card machine

I think it only reads the cards once

and store the program in the RAM

So the same thing as a regular program

ok but how do you know

it just chugs?

@Huy stahp helping MSE ppl and help me

I'm a small child, I need more help

@0celo7: I'm trying to fix something with my remote plex server

it's stuttering and it's driving me crazy

can't even watch matrix

so help me and clear your mind

@0celo7 IIRC you can prove that with Captain Picard

Captain Picard and the Lindy Hop

god damn you made me lol in the library

The geodesic equation is a first order ODE so locally you have a unique solution

But how do you justify that there is some $\rho_0$ for which that is always true

That is the "locally" part?

$\forall p \exists \rho \nrightarrow \exists \rho \forall p$

I have a book about mathematical counterexample

They provide one for that in the first chapter

I forget what it is

It's actually the very first one!

They use $x\leq y$

$\forall x \exists y .x < y \nrightarrow \exists y \forall x. x < y$

what

but that's true

let $y=x+1$

The first means that "for all x there's a y such that x < y"

wow @0celo7

I'm done talking with you troll

All x have a number bigger than them

The second means "there's a y such that for all x, x < y"

It means that there's a y bigger than all other numbers

which is not the case in $\mathbb{N}$

@Huy what?

It is a fun book to find weird math things

@Slereah oh, derp

so now we understand why my compactness thing is an issue

You're not compact, you're dense

(Heyoooo)

lol

@ACuriousMind you're supposed to be on my side

Wo ist the German love

@0celo7 Since when? Also, I just love silly math puns

@ACuriousMind MOTHER DUCK

@ACuriousMind @Huy Now, given the thing with the $\rho$s, do you guys see how the corollary follows?

So what he wants is: There is some open $U$ containing $p$ such that for all $q\in U$ $B(q,\rho )$ is a convex neighborhood, where $\rho$ is the radius of the geodesic ball of $p$.

@ACuriousMind puns are great!