If anyone here can explain fermions that would be appreciated!

@robjohn Are you around? I sent you something.

@beginner fermions and bosons differ in spins

hi everybody

looking for some resources

@Chris'ssis I will look

@robjohn Thank you.

@beginner The only time I took the best mark at Physics in high school was at the theorey of relativity, I'll never forget that hour when the professor wanted to give me $9$ instead of $10$ and all kids began to shout "10, 10, 10!!!"

@ADG I know, bosons have integer spin, but fermions are the weird ones with half integer spin

@ADG I am trying to understand how half integer spin makes sense, does it change state when it is rotated once?

recently found in a question about a book on complex analysis, so i studied complex variables by brown and churchill but i see it doen't has evalusting sums using contour integration, and I have completed it (not exactly leaved the irrelevant topics) can now someone suggest a book

@Chris I hope that happens to me!

@robjohn or @Venus would you hel?

*help

@beginner Then the professor asked me another question to show to the kids that $9$ is the best mark to me, but I answered it instantenously giving the correct answer. There was no choice anymore and I received the best mark, 10.

@robjohn hapy new year sorry to distub you: but can you tol me what to do to prove that $\alpha(|x|+|y|)\leq |x-y|$ for $x\neq y$ $\alpha$ is a positive constant

@Chris I correct the teacher heaps of times in math hehe

and then she picks on me in English to make me not do it

how to show that for $\Bbb F_p$ where $p$ is an odd prime that exactly half the non-zero elements are squares?

@balarka

@ADG I'm still in a state of euphoria of New Year :-)

venus and chris always come together and talk the same

they are sisters

or maybe she has two accounts hehe, I like them both

@beginner: I'd really be interested in some evidence about your statement of how to interpret spin.

@beginner: And by evidence I don't mean "a friend of my father told me"

How you doing, @Venus

look in a physics book @huy

I did today and it is the exact definition @huy

@Huy spin is rotational symmetry of fermions and bosons

@robjohn I wanna show you something very weird, at least to me, a thing I discovered these days. It's about a finite summation in terms of Stirling's numbers of the first kind multiplied by something that by addition give $0$. I'll check that a couple of times before to make sure I do not miss a simple thing there. I mean that identity is simply too nice not to be known. Do you have in mind anything like that? Maybe you met something like that somewhere.

@Huy I'm still high. Haha. How's New Year there?

@Venus: Same old. Studying. Don't do drugs.

I often feel being ignored in chat

@beginner do you wanna know about djvu?

@Huy Particles with integer spin are bosons, and with half integer spin are fermions, half integer spins mean that two entire rotations are required to return to the same appearance

@Huy Studying in New Year's eve? Kidding, right? Why didn't you hang out with your GF?

@MikeMiller Itry to do $|x|+|y|=|x-y+y|+|y|$ and i use triangular inequality but a i don't find !!! have you an idea ?please

@ADG yeah I do, I can only download dummit at 8mb in DJVU

@Venus: She's in a different country, at her grandparents' place.

@Huy How about your friends?

@Huy The example the book had used cards, the ace of spades requires a full rotation to return to the same, so it has spin 1, whereas the queen is symmetrical across the center, so it has spin 2

@Venus: What about them?

having spin 1/2 is equivalent to turning it 360 and finding the card upsidedown and having to spin in 360 again to reflip it

make sense @huy?

@beginner like a C_2 axis

what is C_2?

@Huy Why didn't you hang out with them?

Complex plane two tuple?

Venus can you ask @huy if he ignored me? he could learn a bunch from above

@Venus: Most of them aren't in town and the ones that were in town either studied too or wanted to go clubbing, which I didn't think was a good idea on New Years Eve because of the too many people and the prices being more expensive.

@Venus please :)

@beginner see this

@Chris'ssis I thought we worked on those about a year ago. I'd have to search since it was quite a while ago.

one common thing I see is girls in minority in mostly maths or any engineering site

Can I get viruses through djvu?

no it's like .doc .xml .exe etc

@robjohn do you know any general relativity?

mostly it's resembles .pdf

@Huy Haha. We all have friends like that. You seemed to have a conversation with beginner. Don't ignore him.

@ADG can you ask huy if he is ignoring me? Thank you for your help with DJVU

@beginner try grabbing a book "university physics" or "feymann lectures on physics"

@Venus: I won't undo it. What have you been up to? Nice party?

@robjohn I answered you privately. OK, no hurry.

thank you @Venus you are nice

@beginner not your problem most people keep doing this to mee too most of the time in chat,

@robjohn please have you seen my question ?

That is sad @ADG, @Huy should want to learn more :(

What is your question @Vrou?

that mod alpha and x,y? @beginner

what @ADG?

@Huy Party all night & had fun with my family while seeing fireworks in the sky

@beginner no nothing leave it

@ADG nooo tell me

@beginner Have fun with him :-)

@Venus: Cool. I saw the fireworks too. Wasn't as impressive as I hoped though. :-(

@beginner a very very simple question but i don't see what is the right methods prove that $\alpha(|x|+|y|)\leq |x-y|$ for $x\neq y$ $\alpha$ is a positive constant thatis we must find an $\alpha>0$ who give us the inequality

@beginner what is your age

I am $13\pm 2$ :)

@beginner too young maybe! keep it on!

Apparently you have to be 13 to be a member on a chat page :(

so my 11 year old friend has to pretend to be 13

@Huy Maybe because you spent the day alone?

okay @vrou I will solve this

@Venus: I don't think that's the reason.

@Chris'ssis why delete the statements of something that is so old? If anyone is interested, they would have seen it. It is saved in net archives. There is no proof given, only the end result in a couple.

@Huy Then?

@Venus: The fireworks was simply a lot smaller and less impressive than I always pictured it. ^^

@robjohn Yeah, I think in the same terms. I only wanted to be more cautious, but maybe it's not needed.

So this is like the triangular inequality @Vrou

@robjohn At that moment I didn't know I'm going to find some more general results. But, yeah, it's only the end result.

$|x-y|\le|x|+|-y|=|x|+|y|$ so apparently $\alpha<1$

@Huy Jeez! My fireworks is probably even smaller than you saw because I live in small town but we still enjoyed the night. We played firecrackers too, haha.

@Vrouvrou there is no positive constant that will make that true.

Besides, not everyone here celebrated New Year

I know this is a waste of time here, everyone is so ignorant!! huh!

BYE :(:(

@Venus: Do some celebrate Chinese New Year?

@ADG that isn't right because you divided |X|+|Y| by |X|+|Y| not term wise

cya @ADG

@Vrouvrou Try $x=\frac1n$ and $y=\frac1{n+1}$

@robjohn I sent you that weird result (it looks incredibly nice). :-)

@Huy My community yes, but not rest of the town. Chinese is minority here

@vrou yeah it doesn't work I think I got $(\alpha - 1 )(|x|-|y|)\leq 0 $ which is clearly not possible

@Chris'ssis Those must be the signed Stirling numbers

wait no it works, just set $\alpha \lt 1$

wait no it doesn't

no it doesn't work

it would work for $\alpha\lt 1$ if it were $|x-y|$ up there

@robjohn Yeah.

iI need this inequality for $y\neq x$

@robjohn I discovered that identity in a very weird way. Only let me know if you saw it before.

doesn't work @vrou

where did you get it? plot it and you will see

@beginner even for $x\neq y$ and $\alpha=\frac12$

@Huy I gotta go. I have to prepare dinner with my sis & bro for the party with my big family

one told me that it works for $\alpha=\frac12$ when $x\neq y$ @beginner

even then $\frac{.11+.1}{2}\not\leq .01

why didn't you try that and see?

@Venus: Enjoy your meal.

@Huy hi! can we start again and be friends please please??

@venus can you give huy my biggest sorry ever and ask him if we can be friends again??

@venus please!

@Vrouvrou did you try the example I gave? as $n\to\infty$, you will see that there is no positive constant.

@Venus how is it going?

@robjohn how do I prove that for $\Bbb F_p$ where $p$ is an odd prime, exactly half the non-zero elements are squares? I can see it is true for the first few things I trialed

@chris's can you tell Huy I am really sorry and that I want to be friends again?\

@beginner I don't know the story between you and Huy. What did you do?

@chris just copy and paste that please because he ignored me and he cant see

@beginner consider the equation $x^{(p-1)/2}-1=0$

I told him about fermions and it embarred him because he didn't know stuff and I am so young I think

@chris's can you tell Huy I am really sorry and that I want to be friends again?\ (by beginner)

@robjohn I will consider this thank you for the hint

@robjohn wait just to clarify this is for a field with $p$ elements?

@beginner Since $\mathbb{Z}_p$ is a field, there can be at most $(p-1)/2$ roots

Really? I have never heard of that

@beginner a degree $n$ polynomial can have at most $n$ roots.

Can I just clarify what you mean by roots here for $Z_p$ @robjohn?

Usually it means hitting $0$ I thought, but I don't know how that comes in for the integers?

@beginner There are at most $(p-1)/2$ elements of $\mathbb{Z}_p$ so that $x^{(p-1)/2}=1$

@beginner and at most $(p-1)/2$ elements of $\mathbb{Z}_p$ so that $x^{(p-1)/2}=-1$

Oh so this is inverting it by squaring it, that is called primitive roots isn't it?

Okay I have it now thank you robjohn

@beginner Since there are exactly $p-1$ elements so that $x^{p-1}=1$, you know that there must be exactly $(p-1)/2$ elements so that $x^{(p-1)/2}=1$ and exactly $(p-1)/2$ elements so that $x^{(p-1)/2}=-1$

@robjohn (over a field!)

@PedroTamaroff we mentioned that just a few lines up

@PedroTamaroff I believe that that shows there are at most $\frac{p-1}2$ squares. It does not show that there are exactly $\frac{p-1}2$ non-zero squares.

@robjohn The third line shows we have at least $\frac{p-1}2$ squares.

@PedroTamaroff you have to use that $x^2=y$ has a most two solutions to get that. I would mention that, at least.

I received a chat message saying " Potet i halsen" , does anybody know what does it mean

I don't understand what language is this

@DanishALI: According to google translate, it means "potato throat".

@robjohn do you see any obvious way of proving that result involving Stirling's numbers?

Any suggestions??

Lol windows users

@AlecTeal ??

Full drive encryption FTW!

I've had a netbook stolen before, joke was on them, there's a reason the thing was in storage!

I wanted my diary to be on paper though @alec

Also what is $w\in\Bbb C, |w|$?

keeping diary is silly.

I have trouble making decisions unless I write out my options, and I want my decisions to be private

@beginner $w$ is a complex number, $|w|$ is the absolute value of $w$...

I know, but what is the absolute value of a complex number I meant

$|a + ib|$ by definition is $a^2 + b^2$

you can think about $a+ib$ as the point $(a, b)$ in the caretesian coordinate. then the absolute value can be thought of as distance of $a + ib$ from the origin.

So the absolute value function necessarily $f:\Bbb C \to \Bbb R$?

uh, yes.

Oh okay, I haven't thought of it as a plane, but the absolute value is the magnitude of the vector?

I haven't done much with vectors yet hehe

forget magnitudes and vectors. plain old line kiddo

length of a line joining $(0, 0)$ and $(a, b)$

I know now, I just didn't think of it like that

well, now you do

so $|w|=1, w\in\Bbb C\implies$ Re(w)$^2$ + Im(w)$^2=1$ is correct notation? (sorry if latex is wrong, I can't render)

yes. that is right.

in fact collection of all the points with $|w| = 1$ is a unit circle.

do you see why?

Because we have all points at distance 1 from the origin, which traces a circle? I think that is on the metric space line of thinking

right. and forget about metric space

Are metric spaces important?

you throw names too much. :)

yes, @beginner

very important

so do you hehe @balarka , modules and stuff :P

well i used to throw names, yes

but when i say modules, it's on context

metric spaces aren't really relevant at this point.

ok

;)

:)

Can I ask something that is probably wrong now and you won't smack me :)?

I don't smack. Ted does.

Ask away.

Wait it is definitely wrong, but I am trying to show that there is some $z\in\Bbb C$ for $z=rw$ with $w\in\Bbb C$, $|w|=1$ and $r\in\Bbb R, r\geq 0$

I think you have messed up some quantifiers there. Do you mean you want to prove that for all $z \in \Bbb C$, there is a $r \in \Bbb R$ such that $z = rw$ where $|w| = 1$?

If $z$ is a complex number, prove that there exists an $r\geq 0$ and a complex number $w$ with $|w|=1$ such that $z=rw$

Sorry that is it from Rudin - Principles of Mathematical Analysis page 23

Yes, that's what I wrote above (but not what you wrote).

@beginner Well, prove it.

How do I represent my $|w|$ as a two tuple?

$|w|=1$*

You can do it with algebraic computations, but can you think about it geometrically?

Think of some given complex number $z$ situated in your cartesian plane.

Yes it is a line connecting to a circle of radius $1$

And think of the unit circle somewhere around the origin of your plane.

@beginner wat

$z$ is a complex number

oh that is $w$ sorry

forget about $w$

$|w|$*

forget about $|w|$

Okay let me think

$|w| = 1$ is just a unit circle.

why don't you try to draw it, @beginner. writing down helps.

It is a unit circle with two discontinuous points(from calculus) on the real line, where now we have the points $\pm r$?

ok, you're apparently not used to thinking about complex plane as cartesian plane

do it algebraically, the old boring way :P

$z = a + ib$ be your given complex number

$w = c + id$ is some complex number such that $c^2 + d^2 = 1$

Grr!

okay thank you I will do that

math.stackexchange.com/questions/1087500/… I spent ages typing out a question and suddenly the answer hits me right when I'm nearly done.

I HATE THAT.

you need to find an $r$ and appropriate $c, d$s such that $a + ib = r(c + id)$

ok, i'm outta here.

/me glares at @BalarkaSen

Thank you @balarka I will see you tomorrow

@alec that suck :(.But the silver lining, you know how to solve it

There is no silver lining!

No one upvotes here.

Post with own answer?

Oh you did

Upvotes on most SO sites mean "I agree with you" for answers and "you've explained the question well and it's not a duplicate" for questions (with maths duplicates are harder because different symbols will throw off a search)

@AlecTeal It is so much better to use words than to write "$\forall x\in U\exists B:x\in B and B\subseteq U$."

But here rewards are rarely given!

But if I ask a nooby question like math.stackexchange.com/questions/1087412/what-is-a-function given on page 3 of ANY set theory book - upvotes galore (not my Q but a good example)

Write "For each element of the open set, there is a basic neighborhood of this point contained in the set."

EGH, stop whining Alec. =D

I really wanted to break 2k rep in 2014

@pedro What language is your first?

In my first year - when I asked noob questions, I got like 1.2-1.3k of my current 1.4k rep.

@Alec better than no answer