2:59 AM
@Kenshin Something seems kind of fishy here

3:53 AM
hello?

hi

Do you want an analysis problem?

nty. I'm not focusing on analysis specifically
i'm focusing on EVERYTHING

...
Do you want an everything problem?

okay

4:00 AM
@OBE Let $M$ be a compact $3$-manifold (without boundary). If $\pi_1(M)=1$, show that $M\approx S^3$.

what is $\pi_1$?
some topology thing?

homotopy group

right I heard of that before

Oh, $M$ is assumed to be connected.

I s'ppose
Otherwise u can just pick $S^3 \sqcup S^3$

4:07 AM
is the homotopy group with $1$ label called the fundamental group?

yes

it just means every curve is contractible to a point

Hmm
Am I banned?

I'm reading about it right now. then I'll try to solve the problem.
why?

Does anyone know what that frak letter is?

4:09 AM
it's because it ends in ?1
remove that part

It might be E but that doesn't make sense
@OBE I had to edit the image in Imgur

oh

@Slereah Wait, what's smooth in French?

does that squiggle mean isomorphic

It means homeomorphic.

4:11 AM
$\cong$ is isomorphic?

Yes.
What's the motivation for calling a sigma algebra $\mathscr B$
After Borel I guess

@0celo7 lisse

@Slereah that's what you'd call $C^\infty$?

$\mathfrak{ABCDEFGHIJKLMNOPQRSTUVXYZ}$
Looks like E yeah
En mathématiques et en analyse, les classes de régularité des fonctions numériques constituent un catalogue fragmentaire appuyé sur l’existence et la continuité des dérivées itérées, sans se préoccuper de la forme ou de l’allure de la fonction (monotonie, convexité, zéros, etc). Toutefois, les classes de régularité ne reflètent en aucun cas un type exhaustif des fonctions : en particulier, les critères portent sur la globalité du domaine de définition. == Domaine en dimension n = 1 == Si J {\displaystyle J} est un intervalle de ...
Fonctions lisses ou régulières

weird @0celo7 every book I'm looking at uses $\approx$ for isomorphic and ~ for homeomorphic
they are physics books btw

4:17 AM
$\sim$ is homotopic or homotopy equivalent
or is that $\simeq$

idk

Just accept that I mean homeomorphic. What is the issue?
Who gives a shit?

nothing.
just wanted to know which is used more.

depends

i need to finish this chap on homotopy and also manifolds after.

4:20 AM
Howdy

[Me in the maths chat] Attempting to derive some super high valued iterated integral formulae

@SirCumference hey

i.e. things of the form $\int ^{(m)}=\int \cdots \int$

What language should I learn?

Kikongo
@OBE What book?

4:23 AM
icelandic is pretty cool but I dropped it

@0celo7 ...uh, why?

So the vocal cord parasites take you out

So far I've got English and (mostly) Spanish under my belt
@0celo7 thanks
tho seriously, I'm considering Russian or German

4:24 AM
@0celo7 teach me a word
or saying

I really like this book because it tells you why all the concepts are relevant to physics immediately after describing them.

where do you even find these books
@SirCumference Ich bin nicht ein Idiot.

@0celo7 I somehow don't trust that...

Just a feeling

4:26 AM

it shows you all the books that have all those words on the same page

OK, lemme try from memory
Ich bein nein Idiot
I'm way off...

that's like bad grammar isn't it?
kind of

0

The title pretty much sums it up. Should I change the accepted answer if someone answers the question "better" or more extensively ? Even though the accepted answer is right ?

4:29 AM
By tomorrow I'll read the papers Einstein wrote in German

you'll be like a purist JD.

oh no
@OBE do you think you can solve this problem?

@ODE I'd be JD^2

@0celo7 the one you gave me?

yes

4:32 AM
give me a few days

@OBE So...why's your name capitalized now?

It was unsolved for 100 years, you need to read more

wow you spoiled it

Does it stand for ordinary...bifferential equation?

what

4:33 AM
dude how come every problem you give me is some super duper problem you can't solve yourself. (70% of the time)

why do you take problems from him
are you his mommy

wut

When's the last time I gave you a problem I couldn't solve?

what

Do you want a problem I have solved?

4:34 AM
not right now.

I don't know of any cross-disciplinary problems that are reasonable

I'm trying my best
well I'm trying to try my best at least

here is a problem

omg

4:37 AM
@0celo7 LOOK I found it
a formula for what I wanted

What?

@OBE That makes one of us... ;-;

$\delta(\mathrm d^4x) = \partial_\mu(\delta x^\mu)\mathrm d^4x$
what's the proof for that

Jesus christ
what the hell is that
that's even worse than what I was doing earlier

variation of spacetime measure

4:39 AM
with respect to what?

what do you mean wrt. to what?
the fields?

???
what are you trying to do, boy?

when deriving the euler-lagrange equations for field theory
you vary the action right?
but don't you also need to technically vary the measure?
also

Nope.
Why should you?

so you just vary the lagrangian only?
idk

4:41 AM
correct

then what's this
my derivation of E-L for field theory was much simpler

Wrong? I don't know what you want me to say.

what's this book trying to do
oh
why's it varying the lagrangian and the measure?

Beats me.

@Slereah any idea?

4:43 AM
I don't have the book.

I'm pretty sure this is wrong
the e-l equations it derives at the end don't look like what I'm familiar with

wat

oh well better scrap this book
¯_(ツ)_/¯

Pic

4:46 AM
Lol
What does that delta thingie mean

functional derivative
wait no
it's some new notation
one sec
it's the "lagrange derivative"

@Slereah Aha, the only distributions that have support = one point are delta functions or derivatives
@OBE better throw this book away

and the lagrange derivative of that lagrangian is the e-l equations

hope you didn't spend money on it

if that's non standard then yea
i didn't

4:49 AM
@0celo7 even worse they are SINGULAR SUPPORTS
dun dun duuun

@0celo7 what has to be satisfied to have a chain rule for functional derivatives?
since you can't compose functionals

beats me
be more specific

you can't compose functionals but you can compose a functional with an operator that has functional components
therefore you can make a chain rule
you do that by taking the $n \to \infty$ limit of a multivariable function to turn it into a functional and doing the same for an $R^m$ function you compose it with which will look like a function with functional components.
i think...
this is for field theory btw

can you give an example?
did you check wiki?

yeah
but some of it I made up myself but it should work (he didn't explain all the details so I had to figure it out myself)
@0celo7 i don't have a real life example though because Idk of any kind of application for this

5:00 AM
Details of what? I'm still confused by what you want to do
Composing functionals won't work well, so idk what chain rule you want

I don't know. apparently it's used in condensed matter (functional chain rule) but I don't have any examples right now.
what I wanted to do was construct a general one

Think some more and ask a better question
I'm off to bed, cheerio

same
night
i'll think a bit more

1 hour later…
6:10 AM
@AccidentalFourierTransform I don't like rabbit. We used to eat it a lot when I was a child because the local countryside was overrun with rabbits and my brother and I used to hunt them for the pot. However I find rabbit meat rather soft and textureless and generally unpleasant.

user228700
6:32 AM
@JohnR: Morning :-)

Morning :-)

user228700
@JohnRennie Yep :-) How did u know about that? You weren't around when I deleted my messages...

I'm a room owner so I see deleted messages.
Along with the mods.

user228700
Oh wow, I didn't know about that.

YOU CAN RUN BUT YOU CAN'T HIDE!! :-)

user228700
6:35 AM
Geez :-|

user228700
Say, dyou happen to know anything about the monotonicity of functions?

Err ... that depends. What did you want to know?

user228700
I'm trying to understand a problem to check the monotonicity of some functions about a given point.

Monotonicity of $f$ just means $f' \ne 0$ doesn't it?

user228700
Erm, not exactly.

6:39 AM
Ah, OK, it could have a point of inflection

user228700
Yes. In fact, it could have several points of inflection, as long as these points aren't contained in an interval (which has infinite points, essentially)

OK it means $f'$ doesn't change sign

user228700
Yep.

user228700
The graphs for all functions have been given and you'd think that this would make it ridiculously easy for me to comment on their monotonicity but these functions aren't exactly continous about the given point.

user228700
6:41 AM
Image. Let me take a photo...

user228700

user228700
I am to ascertain the monotonicity of these functions about the point $x=a$

The second one is monotonic....
The first one is not

Monotonicity means that if $x \le y$ then $f(x) \le f(y)$

user228700
...and I find myself a bit confused because of the discontinuity of the graph about $a$

6:45 AM
@JohnRennie Not really, it could be decreasing also...

user228700
^

user228700
@2017 Right. Why not? Like I said before, all this discontinuity is a bit confusing...

In the first graph it is possible to find values of $x$ and $y$ for which $x \lt y$ but $f(x) \gt f(y)$.
@2017 no, what I've said is correct.

@Kaumudi.H For (i), just before a, the function is increasing while just after it is decreasing...

Monotonicity means the function preserves order

6:47 AM
@JohnRennie What you said is correct only if the function is an increasing function....
For a decreasing function if x<y then f(x)>f(y)

user228700
@JohnR: Even monotonically decreasing functions are termed "monotonic". Lookie here:

user228700

Oh OK, just tack on a and vice versa to my statement.

@JohnRennie :)
@Kaumudi.H So did you get it now?

user228700
@2017 Riight. And how to address the discontinuity about $a$?

6:49 AM
@Kaumudi.H What do you mean by address?

@Kaumudi.H: take the left part of the first graph. It's obvous that for this part if $x \le y$ then $f(x) \le f(y)$

It doesn't even matter
if there is a discontinuity or not
the definition remains same

Now take $x = a - \epsilon$ and $y =$ the extreme right of the graph

user228700
Riight.

$x \lt y$ but $f(x) \gt f(y)$
Which contradicts our earlier conclusion. So the function is not monotonic.

user228700
6:50 AM
@JohnR: I was having trouble with the fact that the function is discontinuous at $a$. I understand it now :-) Thanks, guys.

Whee, I got a maths question right! :-)

user228700
:-)

7:52 AM
@Kaumudi.H Are you appearing for BITSAT?

user228700
Yep.

@Kaumudi.H So which date slot are you selecting, after advanced or before advanced?

user228700

@Kaumudi.H I'm confused because I haven't prepared for the english test
in bitsat
so I think i need one extra week for that
is the level of test same on all dates ?
or are the later tests tougher?

user228700
Ohh, I see. Well, I've been chatting with you for a few weeks now and it seems that u are fluent in English; I don't see why you'd need another week. Besides, IIRC, there are only a few slots left after advanced; certainly not one more week and all.

user228700
7:58 AM
Also, this "level of tests" is debatable; some have told that the test gets easier toward the end and some others have told that it's easiest in the first few days; it's entirely unpredictable.

@Kaumudi.H Ah, thanks for the suggestion. I will talk to my parents about this. BTW the slot booking begins on 20th march, right?

user228700
Sure. Yep, 20th March.

Okay, thanks :)

user228700
I have another quick question about monotonicity; if, for a given point, the derivative changes from positive to zero, why isn't it monotonic about that point?

@Kaumudi.H Who said it is not monotonic about that point? In such a situation it can be called monotonic (non-decreasing function). However, it cannot be called "strictly increasing".

user228700
8:07 AM
> Who said

user228700
My textbook .__.

As wikipedia frames it "In calculus, a function {\displaystyle f} f defined on a subset of the real numbers with real values is called monotonic if and only if it is either entirely non-(increasing or decreasing). That is, as per Fig. 1, a function that increases monotonically does not exclusively have to increase, it simply must not decrease."
In mathematics, a monotonic function (or monotone function) is a function between ordered sets that preserves or reverses the given order. This concept first arose in calculus, and was later generalized to the more abstract setting of order theory. == Monotonicity in calculus and analysis == In calculus, a function f {\displaystyle f} defined on a subset of the real numbers with real values is called monotonic if and only if it is either entirely non-(increasing or decreasing). That is, as per Fig. 1, a function that increases monotonically does ...

user228700
I'm well aware of the definition; I'm confused because my crappy textbook disagrees :-/

You know your textbook is crappy :-P

user228700
Uhhhhh.

user228700
8:09 AM
Thanks anyway :-) I'm going to try to figure out if I've made a mistake.

user228700
Is it possible to comment on the monotonicity of a function at a point that is not in its range? (End-point)

Is the function defined for points not in its range?
I would have thought not, so it's impossible to comment on its monotonicity or otherwise.

user228700
Uh, no, it's not. And yet, this problem...

user228700
Alright, so the function is given by-

user228700
1) $x$ ; $0\le x\le1$

user228700
8:15 AM
2) $[x]$ ; $1\le x\le 2$

user228700
Where $[x]$ represents the Greatest Integer Function (given by the greatest integer less than or equal to a given element in the domain)

Infinity is the greatest integer

user228700
Ohhhhhhh. Dang it, I missed that the value of the function at 2 is 1.

user228700
Alright nope, I'm a bigger idiot than I was before; the value of the function at 2 is 2, not 1.

user228700
In any case, the graph would look something like this, no? :

8:21 AM
@2017 thanks

user228700

@Kaumudi.H So what is the question?

user228700
Does that look OK? If my textbook is correct, then nope, it would turn out that I've made a mistake...

The graph seems ok

user228700
Right. Well, now u see what I was talking about; my textbook says that the function is non- monotonic at 1.

8:26 AM
How does your textbook define monotonicity ?

user228700
In the proper way; the same way u defined it a few messages ago.

Hmm, the function is definitely monotonic at x=1 according to the definition...

user228700
Would u say that it's monotonically increasing or decreasing at 1? The former, yes?

monotonically non-decreasing

user228700
Any by "non-decreasing", u mean to highlight the fact that it's not strictly increasing, yeah?

8:30 AM
@Kaumudi.H right

@2017 why we cannot say , it is monotonically increasing

sorry , monotonically non increasing

It is just a semantics problem...
Different books have different conventions...

what we will write in exam

user228700
8:33 AM
Heh. Well, my book says "It's neither M.I, nor M.D"

board and jee

@Kaumudi.H That's right
It is neither

i think both answers are correct

@Koolman They won't ask such stupid questions in exam :P
And if they ask both of them should be correct

oh yes

user228700
8:35 AM
How is it neither? It's certainly monotonically increasing, just that it's not strictly increasing.

user228700
The way they've defined this about given points is kinda vague...

@Kaumudi.H Monotonically Increasing=Strictly Increasing (according to what I learnt)

user228700
And this is what I've learned:

user228700
2

I'm used to the following: $f$ is increasing iff $x\le y \Rightarrow f(x)\le f(y)$ strictly increasing iff $x< y \Rightarrow f(x)< f(y)$ decreasing and strictly decreasing: similar, with the inequalities for $f$ reversed. Monotonic: either increasing or decreasing strictly monotonic: either...

user228700
Anyway, I'm glad to know that it just comes down to the semantics.

8:36 AM
@Kaumudi.H I told you that it is a semantics problem :P

user228700
In any case, @BalarkaSen I could use ur help with this :-P

"(In general there is always some freedom when it comes to definitions, this is why I wrote 'I'm used to'. In case you are reading a textbook on analyisis the author should define these terms and then stick to them)."

user228700
And clearly, my textbook hasn't stuck with the definitions they introduced before.

just move on

[What is not art] For me, it is anything that makes me treat them as daily life items and the notion of art never came to mind
For example, with no thinking, I literally almost sat down on the middle column thinking they are just stone benches

9:34 AM
i think its really weird to consider that all the stars were once touching each other..
is this fact true?

stars did not exist back in that era

this guys is saying that when the universe was still pretty small compared to today's size
and stars were there
wait
lemme put it again in a different way

@MartianCactus That would be a child's perspective :) In reality things are far more complicated.

he is saying that when stars were formed, universe as still pretty small and stars were touching each other
yes thats why
i think he is wrong here , right?

"Stars were touching" doesn't make much sense.
@MartianCactus As I said, the author is explaining the facts in a child's perspective. If you want to get into the technicalities then start with the Wiki page (en.wikipedia.org/wiki/Big_Bang).

9:41 AM
oh
so what he means is they were closer together than today right?

@MartianCactus First go and read the Wiki page. Otherwise whatever I say won't make sense.
This is a really good answer :) physics.stackexchange.com/a/136861/102705
232

The simple answer is that no, the Big Bang did not happen at a point. Instead it happened everywhere in the universe at the same time. Consequences of this include: The universe doesn't have a centre: the Big Bang didn't happen at a point so there is no central point in the inverse that it is e...

Key Point: The universe didn't shrink down to a point at the Big Bang, it's just that the spacing between any two randomly selected spacetime points shrank down to zero.

oh!!
but isn't that
the same thing?
ohh

@MartianCactus What makes you think that "stars" were touching?

so the space between the galaxies is expanding

Did you read what Slereah said ?

9:50 AM
not the galaxies moving away from each other!
i will in a second once I finish what I am currently reading :D
and about the star touching stuff
i knew it was wrong
it would be too unstable
but the article was saying it
so i came here to verify