Mathematics - 2016-05-23

2:44 AM

from f(x):R to (-1,1) and let the function defined by g(x)=cos x is it surjective?
I suppose it must be surjective since it maps exactly to the range of a cosine function. How do you guys see this?

Eric Stucky

You probably meant [-1,1]

Neman Nasawa

yes.

Eric Stucky

If you want to show something is surjective, you better know what it means to be surjective, in a very technical sense. But your idea is the right one. If the codomain is equal to the range, then the function is surjective.

Neman Nasawa

well how can I say that using the range or an example?

Eric Stucky

?

say what?

g is surjective?

Neman Nasawa

2:49 AM

yes

Eric Stucky

What does 'surjective' mean?

As I said, you can't avoid this.

Neman Nasawa

A function f from A to Bis called onto, or surjective, if and only if every element b∈B there is an element a∈A with f(a) = b. A function f is called a surjection if it is onto

so you mean I say g is surjective using the definition?

Eric Stucky

Yeah, you'll need to use the definition.

Semiclassical

evening @mike

Eric Stucky

But unless you have a rather unusual definition for cos(x), it's not going to be so easy to find the a for every b.

Neman Nasawa

2:55 AM

what if i just use the elements [-1,1] for b?

Eric Stucky

well, of course, you can only use elements in the codomain

But what I mean is

If $b=2\sqrt 2 - \pi^1.001$, then what is $a$?

It's not so easy to answer this question, you can see.

I think I would use the intermediate value theorem.

Neman Nasawa

well then How?

by the way, what does the IVT says?

Eric Stucky

If you don't know the intermediate value theorem, then you probably weren't expected to think about the question so hard ;)

but I'll tell you

If $f:I\to \Bbb R$ is continuous, where $I$ is an interval, and if $f(x)=y$ and $f(a)=b$ and $a<x$, then

for every $q$ such that $b<q<y$, there exists a $p$ such that $a<p<x$ and $f(p)=q$.

Neman Nasawa

well i could not get that since the message is difficult to read.

Eric Stucky

It's a very technical result.

That's why I said

if you don't know it, you were probably not supposed to think about the question so hard :P

Neman Nasawa

3:06 AM

ok so I just use the surjection definition or what?

Eric Stucky

I honestly don't know how to do it without the IVT, besides just saying "everything in the codomain is in the range". Perhaps there is something clever you can do with the exponential function.

The problem is that the usual definitions of 'cos' as a function are obnoxious to work with.

Neman Nasawa

if that is the case, I suppose you upload a file or notes on IVT about this particular subject then I will work on it.

Mike Miller

Hi @SemiC

Eric Stucky

Neman: This reference is pretty good

Neman Nasawa

ok thanks.

Balarka Sen

8:10 AM

@MikeMiller No, they do not prove the Morse lemma.

Also they seem to be proving something weaker than density of Morse function. Namely, given a function $f : M \to \Bbb R$, $f_a = f + a_1 x_1 + \cdots a_n x_n$ is Morse for almost every $a = (a_1, \cdots, a_n) \in \Bbb R^n$.

Is that equivalent to density?

sasha

9:50 AM

0

Q: Computing losing positions in modified Wythoff's game efficiently

Wythoff's game is as follows: there are two players $A$ and $B$ ( $A$ being the first player ) and there are $2$ piles of stones. When his turn a player can remove one or more stones from anyone pile or same number of stones from both the piles. A player unable to make a move loses. Also it is...

Ell

10:20 AM

Hi folks

N3buchadnezzar

10:33 AM

heya

Ell

$ x^3 = 1\mod 7 \implies x = 2\mod 7$

but why?

Tobias Kildetoft

@Ell Well, it is not true

Akiva Weinberger

$1$, $2$, and $4\mod7$ all work

Ell

hmm

Akiva Weinberger

$1^3=1\pmod7\\2^3=8\equiv1\pmod7\\4^3=64\equiv1\pmod7$

Ell

10:44 AM

My mark scheme says this:

$x^3 = 1\mod 7 \implies x\times(x\times x) = 1\mod 7 \implies x = 2\mod 7$

and I'm trying to understand why it would say that

the full system of equations is:

$x = 3\mod 5$
$x^3 = 1\mod 7$

Kaustabha Ray

Does the run time complexity of Linear Programming or non linear programming depend upon the number of number of constraints? If yes does it increase or decrese?

Akiva Weinberger

10:59 AM

@Ell Yeah, $x$ does not need to be $2\pmod7$. For example, $x=8$ (the number, not modulo anything) satisfies $\begin{cases}\phantom{{}^3}x\equiv 3\pmod5\\x^3\equiv1\pmod7\end{cases}$ and isn't $2\pmod7$.

N3buchadnezzar

Heya

Tobias Kildetoft

11:29 AM

If I already use [,] and $\langle,\rangle$ for other things and I prefer not to use just (,), what would be a good choice to denote a bilinear form?

The reason I prefer not to use (,) is that there are a lot of other such parenthesis floating around, so it gets a bit too hard to distinguish them

s.harp

11:51 AM

$\beta(,)$

Tobias Kildetoft

That still puts the same number of additional regular parentheses

s.harp

$xBy$ :)

if you are very daring you can use $\cdot$ also

Tobias Kildetoft

Heh, would probably look terrible unfortunately. The thing I that I need to apply this bilinear form to some fairly long expressions

So I was mainly looking for some nice alternative types of brackets

s.harp

If you are writing on paper, you can do large double brackets from mathematica, 〚〛

kind of hard to see, but the doublestroke version of "[", $\LARGE 〚$

Tobias Kildetoft

Yeah, I need them in LaTeX as I am writing a paper

(I don't think journals accept papers with parts written by hand any longer)

s.harp

12:07 PM

can you read the character that I put in? I'm sure there is a unicode version of it

Tobias Kildetoft

That might be an option, yes

Hmm, it also seems that it exists in some TeX package

s.harp

oh yeah, I totally forgot, if you can think of some type-setting / font thing then there exists a tex package for it :)

Tobias Kildetoft

time to get playing with detexify I guess

quite possibly I should also use | instead of , in the middle to distinguish it better

s.harp

$\llbracket$ $\large \llbracket$

ah you need package stmaryrd

Tobias Kildetoft

thanks

s.harp

12:11 PM

sunsite.informatik.rwth-aachen.de/ftp/pub/mirror/ctan/fonts/… look at all those symbols O:

Tobias Kildetoft

awesome

Mike Miller

1:35 PM

@Balarka In particular, you can choose such an $(a_i)$ arbitrarily small. That's stronger than denseness: it gives some slight control kver the perturbations you need to make.

Semiclassical

morning chat

Mats Granvik

Can anyone tell me if my question will be deleted by the Community robot within a week now that the question has one downvote? The votes are at -1 and there are comments on the question. Will the comments protect it from the bot?

Semiclassical

don't know. which question, out of curiousity?

Mats Granvik

-1

Q: Find a polynomial such that this proposed root finding algorithm fails.

Is this polynomial root finding algorithm below known, and under what conditions for the choice of polynomial coefficients does it find at least one root? Description of the algorithm: Consider the polynomial: $$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 = 0$$ Replace $x^n$...

polynomials reference-request roots recursive-algorithms

Mike Miller

@Mats: The community bot only deletes closed questions, I believe.

And in any case doesn't go near questions until a month after posting.

Mats Granvik

1:51 PM

@MikeMiller I once was deleted by the Community bot, with the vote score -1. That other question had no comments.

Mike Miller

Yes, comments are unrelated to the algoeithm. But either your quesrion was a year old or it was a month old and closed.

Mats Granvik

@MikeMiller ok. And thanks for the up vote whoever it was.

Balarka Sen

2:13 PM

@MikeMiller Ah, interesting.

ramsay

Can anyone tell me what does the person who answered this question mean by saying $a$ cannot be in the image of $g(x)$

Balarka Sen

Right, I was probably not awake when I said that. I have no idea why I thought it was weaker than denseness.

Choosing $a$ to be very small perturbs $f$ to a Morse function. That's precisely denseness, duh.

I'll try out some more exercises from chapter 1 today.

Secret

Is there exists a multiplicavie analogue of the modulus function

e.g. usual modulus function inolves elements of the form $x+nb$ where n is an integer and b is a constant

is there exists some kind of finite group like $x*b^n$ ?

Mike Miller

2:29 PM

@Balarka They shouldn't be too hard. You know most of this stuff.

Secret

for example, we have $\mathbb{Z}/5\mathbb{Z}$ and it involves classes with elements of the form $a+5b$
Is there exists finite sets which has elements of the form $a*b^5$? What are the names of these sets (if any)?

Mike Miller

2:46 PM

@BalarkaSen There are a few general function space results you should know, all in Hirsch's differential topology book, which I'll say a couple of. 1) $C^\infty(M,N)$ is dense in $C^0(M,N)$. This is also true relatively: if a map is smooth on a closed subset, you can homotope it to be smooth everywhere while fixing the map in the subset. This is also true relative to the boundary of non-closed manifolds.

(This implies that $C^\infty(M,N) \hookrightarrow C^0(M,N)$ is a homotopy equivalence.)

You can also smooth sections of a bundle through sections, or smooth maps from a fiber bundle so that if it's already fiberwise smooth, each $f_t$ is still fiberwise smooth. That sort of thing.

Oh, I think I told you the embedding result already, about denseness in function spaces. To repeat in case I didn't: if $\dim N > 2 \dim M$, then $\text{Emb}(M,N)$ is dense in $C^\infty(M,N)$.

Balarka Sen

OK, (1) is somewhat interesting. Is it hard to prove?

Mike Miller

Not really. Don't bother, though.

Balarka Sen

@MikeMiller Right, you mentioned this.

Obliv

how does it make sense that the order of $sr$ and $sr^2$ is $2$? These are elements of the dihedral group $D_{6}$ I get that $|r| = 3$ and $|r^2| = 3$ since $r^3 = 1$ and $r^6 = 1$ but $|sr| = 2$ doesn't make sense. $s^2 r^2 = r^2$ not $r^3 = 1$.

Mike Miller

This should still apply if the codomain is a Frechet manifold, though Hirsch doesn't prove this.

Obliv

2:52 PM

ah nvm I see. $sr^2$ is not $s^2 r^2$ done independently. It's $(sr)^2$ so it's a reflection then rotation twice.

Anubhav.K

3:05 PM

@BalarkaSen I've started reading algebraic geometry, and so far I proved that two affine plane curve intersect only finitely many points. Now I am thinking, if we consider the base field as $\mathbb C$ , then can we say that they intersect transversally or not?

Semiclassical

A phrase I tend to use in answers lately: "All that remains is X, a task I leave to the interested reader."

translation: the rest of this isn't interesting to me. do it yourself.

Balarka Sen

@Anubhav.K No. $y = x^2$ and $y = -x^2$.

Anubhav.K

I mean upto constant (unit) multiplication

Mike Miller

I have no idea what up to constant means

Balarka Sen

No idea what you mean.

Anubhav.K

3:08 PM

let me state it

Let $f,g$ be two irreducible polynomial in $C[x,y]$

Then $f(x,y)=g(x,y)=0$ has only finitely many intersection point.

Can we say or give any condition so that they intersect transversally?

Balarka Sen

Yes.

Anubhav.K

How do I prove?

Balarka Sen

For one, note that the zero sets has to be smooth manifolds to actually say transversality.

Once you have that, you just write down the conditions for transversality of smooth manifolds.

Anubhav.K

hmm

Mike Miller

It's just going to be literally "the intersection only occurs in the smooth locals and the tangent spaces sum to the tangent space of the ambient variety"

Balarka Sen

3:11 PM

That.

Mike Miller

this makes perfect sense over any field tho I don't know what good it does yoh

Anubhav.K

DO you think Shafarevich is a good text to begin with?

Balarka Sen

I learnt from it.

I think it's good.

Anubhav.K

How far did you read from here?

Balarka Sen

Chapter 2.

Anubhav.K

3:13 PM

I am reading it, I'll come up with doubts

I hope you can help

Balarka Sen

You should start from chapter 1.

Anubhav.K

I am doing that

Balarka Sen

Sure, I'll help whenever I can.

Mike Miller

Ah, all of my friends abandon topology.

Anubhav.K

:)

Balarka Sen

3:16 PM

I still like topology over algebraic geometry, and only want to study the topological kind of algebraic geometry. But I guess I am not literally your friend.

Mike Miller

You don't watch TV and eat pizza with me but I was referring to the people in this room.

Balarka Sen

@Anubhav.K In one sense, transversality in algebraic geometry is subtle. The thing is that algebraic varieties can be singular, and one can still ponder on intersection of singular algebraic varieties. In that case making them intersect in "general position" becomes a lot harder, unlike for smooth manifolds where transversality theorem says on can always make smooth submanifolds intersect transversally upto a homotopy. Just sayin'.

By which I remember I still have to learn transversality theorem.

So, back to work...

Mike Miller

by a small homotopy

if it was by some necessarily large homotopy the result would be a lot less useful

Balarka Sen

Hmm, you're right, but I am not sure if I know what "small homotopy" is yet.

Lembik

what is -1 mod 3?

Mike Miller

3:25 PM

What could it possibly mean?

Lembik

or.. it must be 2?

because 1+2 == 0

Semiclassical

if "a mod b = c" is intended as "the least nonnegative integer c such that a-c is divisible by b"

then sure

Balarka Sen

@MikeMiller I don't know, that the homotopy can be chosen to deform the inclusion map to some other map which is arbitrarily close to the inclusion map?

In some appropriate norm on the function space (uniform norm, most likely).

Mike Miller

How bizarrely convoluted. It just means $f_t$ is close to $f_0$ for all $t$.

Balarka Sen

Whoops, right, I should also care about the intermediates to be arbitrarily close to the original map too. Meh. Otherwise it can go all over the place.

Mike Miller

3:34 PM

aka, this is anothe denseness result, provided you know the space of maps is locally path-connected.

Balarka Sen

Right, makes sense.

Secret

@MikeMiller are there multiplicative analogues to a mod b, where there is some function S such that $a S b$ has elements $a*b^n$ n is integer? What are the names of these structures in general?

Balarka Sen

We are saying that the subspace of $C^\infty(M, N)$ consisting of maps $M \to N$ transverse to some submanifold $W$ of $N$ is dense.

Mike Miller

one also has relative etc versions of this theorem which proves that $C^{\infty,W}(M,N)$ is homotopy equivalent to the original space of smooth maps

could improve this to "transverse to a countable set of submanifolds" if so desired

Balarka Sen

Cute.

Mike Miller

3:40 PM

useful!

Balarka Sen

I have never thought about all this from the function space POV. Very interesting.

Mike Miller

@Secret Seems unlikely to be of much use and whence probably not named. Why ping me?

Secret

I saw algebra in your profile, and the rest are afking, thus figure that I can ask you

what property that makes it differ from the mod that makes it less useful?

Mike Miller

If you're interested, you should explore its properties yourself to see why it doesn't do very much.

@BalarkaSen: For instance, you can use this to set up a homology theory for which the intersection product is visible at the chain level, or the ability to take fiber products at the chain level.

Balarka Sen

wow, I am curious what the homology theory is.

or rather, how to build a homology theory out of that

Mike Miller

3:49 PM

It's a tiny bit technical, give me a bit.

Balarka Sen

Sure, thanks for taking the time to type it out.

Secret

ok, i'll see what I can learn from it

Mike Miller

Let $P$ be a smooth manifold with corners. Say a geometric chain is a smooth map $\sigma: P \to M$; and two geometric chains are isomorphic of there's a diffeomorphism $P \to P'$ taking $\sigma$ to $\sigma'$. Say a geometric chain is small if $\sigma(P)$ is in the image of a smooth map from a manifold of smaller dimension. Say a geometric chain is negligible if both $[P]$ and $[\partial P]$ are small.

OK, now define $C_*(M)$ to be the $\Bbb Z/2$-vector space generated by isomorphism classes of geometric chains, modulo $\sigma \sim 0$ if $\sigma$ is negligible, and modulo the relation $[\sigma] + [\sigma'] = [\sigma \sqcup \sigma']$.

$P$ is supposed to be compact up there, sorry.

Balarka Sen

Understood. Please go on.

Mike Miller

The boundary operator $\partial$ makes this into a chain complex. The homology of this chain complex is naturally isomorphic to singular homology. This is not too hard to prove. We needed to quotient out by negligibility or else $H_*(pt)$ would be wrong (and we needed to pass to isomorphism classes lest we not have a set of generators).

this notion of negligibility is old; it's been around since people tried to define singular homology via cubes, and in that setting one needed to add a notion of nondegeneracy for the same reason

we could in addition define the chain complex $C_*^U(M)$ to be the geometric chains transverse to a submanifold $U$. The transversality-denseness result from above more or less straightforwardly implies that $C_*^U(M) \to C_*(M)$ is a chain homotopy equivalence, whence this also defines singular homology.

We thus have a literal intersection product $C_k^U(M) \to C_{k-\dim U}(M)$ given by intersecting with $U$. Passing to homology we get intersection product.

Balarka Sen

4:03 PM

@MikeMiller Interesting. I was wondering what negligibility means.

Mike Miller

In the simplicial setting, a small chain is automatically zero in homology, iirc.

Balarka Sen

Right.

Mike Miller

if $M = U \cup V$, pick a function $f$ such that $f|_{U - V} = 0$, $f|_{V - U} = 1$. pick a regular value of $f$; say $1/2$. define $W = f^{-1}(1/2)$. then the mayer-vietoris boundary map is given on the chain level by $C_k^W(M) \to C_{k-1}(U \cap V)$.

similarly making a transversality assumption one can make chain-level fiber products. these are nice tools.

Balarka Sen

@MikeMiller Ah, now I see. This is very fun.

Steve

Given the Fourier series:

$F(x)=\sin{x}+\sum_{n=1}^\infty \frac{1}{5^n} \cos{nx}$

How do I find $a_0$?

Mike Miller

4:11 PM

what is $a_0$

anyway it's nontrivially more work to work oriented so I won't do that but the construction is nice. i learned this from (and think this is due to) max lipyanskiy, geometric homology.

Balarka Sen

Mhm, ok. This is a nice idea.

Steve

$a_0=\frac{1}{\pi} \int_{-\pi}^\pi f(x) dx$

Mike Miller

you integrate everything in sight

(or note that you already have a fourier series and it doesn't include a constant term so $a_0 = 0$)

Steve

Do I integrate everything in $F(x)$?

Mike Miller

@BalarkaSen this works for Frechet manifold codomain. in particular, because every countable CW complex is homotopy equivalent to a Hilbert manifold, you can define homology for any space you think about that way.

Steve

4:20 PM

Or is it because $\frac{1}{5^n}$ doesn't contain any x's?

that $a_0=0$

Mike Miller

presumably if you allow your Hilbert space to be non-separable you can encode the homotopy type of any cw complex

Balarka Sen

I didn't know that every CW complex is homotopy equivalent to a Hilbert manifold.

Steve

@MikeMiller I'm not sure what to make of what you wrote

Mike Miller

embed it cell-by-cell and take a regular neighborhood

anyway that's all i've got for you, @BalarkaSen. back to the grindstone and whatnot

EinsL

I was checking an answer in a post from 2013, I saw a mistake in the answer (there was a sec instead of a csc). I edited the answer and waited for the approval of a moderator. I return one hour later and a user with lots of points and I guess some mod privileges edited the answer. That is just hilarious.

Mike Miller

4:34 PM

@EinsL Any user with over 2000 reputation can edit a post. When you suggest an edit, it goes into a queue that everyone with that reputation can see. What presumably happened is someone opened the queue, saw your edit, and saw further improvements they could make.

Semiclassical

4:46 PM

I find it funny how the answers i spend the least effort on are the ones which tend to get more upvotes

Balarka Sen

@MikeMiller right, that's what i am doing

Mike Miller

yeah.

Lord_Farin

@Semiclassical Funny/disheartening/annoying/pleasing.

I've been through all of those.

s.harp

@Semiclassical yesterday I wrote up a trivial "no" in two lines, got 3 votes, today I wrote up about 1 1/2 pages, get 1 vote lol

Semiclassical

not that it's entirely surprising: If I don't have to spend a lot of effort on it, then 1) it's probably on a problem that's lower level and of wider interest, and 2) I can get it out faster and therefore avoid having someone else post first

Lord_Farin

4:49 PM

@Semiclassical At least it demonstrates the relativity of reputation.

Mike Miller

my best answers tend to do alright, my good answers tend to do terribly, and my trivial answers tend to attract upvotes

3

Lord_Farin

^

Mike Miller

ugh this reminds me of the answer i spent all day on recently that nobody paid attention to that day

it ended up with three upvotes but still that was a salty day

Balarka Sen

which one is that?

Mike Miller

7

Q: $S^3\times \Bbb CP^\infty$ is not homotopy equivalent to $\left(S^1\times \Bbb CP^\infty\right)\big/\left(S^1\times \{x_0\}\right)$

Both $S^3\times \Bbb CP^\infty$ and $\left(S^1\times \Bbb CP^\infty\right)\big/\left(S^1\times \{x_0\}\right)$ have cohomology ring isomorphic to $\Bbb Z[a]\otimes \Lambda[b]$ with $|a|=2$ and $|b|=3$, as can be seen from Künneth and cellular cohomology. Thus, the cohomology ring structure can't ...

algebraic-topology

double salt because the question was highly upvoted

it's like, man, did you like the idea of the question but not want to actually see an answer?

Semiclassical

4:52 PM

my issue with that answer is that i only upvote questions/answers I understand :p

but then, i wouldn't upvote the question either

Balarka Sen

oh no i don't understand the answer.

Semiclassical

on the principle that, if i can't understand what's going on, i shouldn't be giving an opinion on it

Mike Miller

sure, that's fine. it's just annoying to give a detailed, correct answer to a highly upvoted question and have nobody pay attention

Semiclassical

yeah

another pet peeve: people who keep asking questions in comments without changing their actual question in any meaningful way

cough cough

Lord_Farin

5:12 PM

@Semi Solved.

Akiva Weinberger

@MikeMiller The upvotes are people's way of saying "Yeah, I was just about to say that" sometimes

Balarka Sen

5:31 PM

People can also downvote to say that

Mike Miller

@Akiva So what does refusing to accept an answer mean? ;)

Akiva Weinberger

Maybe the poster couldn't really understand the answer, and later forgot about it

or maybe the poster wasn't really sure how it answered the question

Balarka Sen

I wonder if you got the reference.

Akiva Weinberger

I did.

Mike Miller

That's usually what clarifying comments are for - if a poster isn't quite sure about something, they can always ask further questions.

Abhishek Bhatia

5:55 PM

Hi guys!

I have a silly question.

Why is Riemann sum equal to the definite integral?

Or why is area under a curve equal to the anti-derivative?

Mike Miller

one is the definition and the other is the fundamental theorem of calculus.

Abhishek Bhatia

Riemann sum is equal to the definite integral by definition?

Mike Miller

yes

Abhishek Bhatia

how?

user19405892

hi

what does it mean if a number is congruent to its inverse modulo $n$?

Mike Miller

6:10 PM

@AbhishekBhatia It's literally the definition of the definite integral. The limit of Riemann sums.

Abhishek Bhatia

@Mike okay, I need to re-read FTC.

also, is there any rigid rule to which theorem is FTC 1 and which is FTC 2?

I found them mixed up in 2 different books.

Mike Miller

They're more or less the same theorem. I don't know why people bother to label them as two different things.

Owatch

7:01 PM

Hello all.

Is there a word for an expression that has no variables?

Eg: 1+2

Akiva Weinberger

@user19405892 It means that $x\equiv x^{-1}\pmod n$, or $x^2\equiv1\pmod n$, or $x^2-1$ is a multiple of $n$.

Owatch

7:13 PM

I guess a constant expression.

LifeOfPai

Hello, someone can tell me what the derivative of:

$xy^2+yz^2+xyz+x^2y^2z^2 = 5 $
I need to solve the drivative of $\frac{d^2z}{dy^2}$

Maybe it`s good for the first derivative?
$\frac{dz}{dy}= (2yx+z^2z'+xzz'+2x^2yz^2*z')$

user19405892

8:51 PM

hi

Karl Kronenfeld

9:33 PM

oh hi

Hi @Karl.

Hello @Balarka.

What's up?

9:58 PM

morning

Karl Kronenfeld

Ever try to wait out a salesperson, i.e. maintain a conversation until they eventually contradict themselves?

Mike Miller

no

Karl Kronenfeld

Definitely a worthwhile experience.

Mike Miller

eh

Karl Kronenfeld

You sound skeptical, well everybody is skeptical at first.

Forever Mozart

10:03 PM

oh my

take the derivative and do it

just do itttt

Balarka Sen

@KarlKronenfeld lol

sounds like the kind of thing you will do, indeed

Karl Kronenfeld

@ForeverMozart You can't do it for him/her?

Forever Mozart

if i am paid only

Karl Kronenfeld

@BalarkaSen a troll's life

Mike Miller

@BalarkaSen did you learn anything

Neman Nasawa

10:27 PM

can I get an answer by clicking this: math.stackexchange.com/questions/1797286/…

Akiva Weinberger

11:02 PM

@NemanNasawa The parenthesis in (-1,1] is almost certainly a typo. It should be [-1,1].

Semiclassical

11:20 PM

evening chat

Forever Mozart

i just ran a mile and I'm about to die

who wants to join the mile a day club

TheGreatDuck

guys

is it alright or generally accepted for someone to write a paper-sized answer to a question that is broad, incredibly complex, or simply inspiring to the answerer?

I've noticed this in a few answers here and there and I'm not sure whether or not it's the accepted norm or if those people are doing something not normally approved of.

Of course, this is a question for opinion, not actual policy.

Forever Mozart

11:36 PM

it's not the norm, but it's fine. sometimes the problem requires a long answer, and other times there may be some insight gained by a long circuitous argument

Steven Stewart-Gallus

Gausse's Law is $ \phi = \frac{Q}{\epsilon_0} = \oint_{S} \vec{E} \cdot \mathrm{d} \vec{A} $. The surface area of a sphere is $ A = 4 \pi r^2 $. So for a point charge $ \phi = \frac{Q}{\epsilon_0} = \oint_{S} \vec{E} \cdot \mathrm{d} \left(4\pi r^2 \hat{r} \right) $ and $ \phi = \frac{Q}{\epsilon_0} = 4 \pi r^2 \oint_{S} \vec{E} \cdot \mathrm{d} \hat{r} $. But how do I get from Coulomb's law from here?

TheGreatDuck

@ForeverMozart ok. thanks

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$Mathematics$ Mathematics