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3:00 AM
And that means you really need to know the difference between pointer and arrays. Which is obscured a bit by the interchangability of the access semantics and a lot more by people claiming that they are the same.
 
and then read the cells linearly self->size*self-size at a time
 
For your current purpose a fixed size history declared as
int history[UNDO_DEPTH][NUMCELLS*NUMCELLS]; or similar will work.
But then you will have no choice but to copy data around.
 
It would be a little more clever to use a single array for both the active board and the history and treat it as a circular buffer with indexes holding the location of the active state, and the last item in the history.
No copying, no memory management hassle, automatic overwriting of stale history. It's cute.
 
@TanMath Uhhhh, wut?
 
3:04 AM
@dmckee Woah
That is a good idea
 
@TanMath I really, really think you should review the standard random walk problem before you worry about the quantum case.
 
so we were looking at this article, and thinking it was kind of sketchy...
 
Another one that I stole. I have decades of experience in stealing good ideas.
 
@BernardoMeurer My god you're young. This is standard C trickery.
 
@DanielSank any good resources?
 
3:05 AM
@dmckee Why circular though?
 
(For anyone who thinks I'm being mean to Bernardo: We're buddies. It's a joke and I expect he'll fire back)
@TanMath Hmmmmmm. The first chapter of Reif's statistical mechanics book does a good job.
 
@DanielSank I am blessed with youth you decrepit old man
 
I keep meaning to write a blog post about it...
(See?)
 
@DanielSank about random walk?
 
@TanMath I don't have a particularly amazing example in mind. Try Googling around.
 
3:06 AM
(To anyone who thinks I'm being mean to Daniel: I am, suck it)
 
@TanMath Yeah.
 
When you first start you put the initial state of the board at index 0. Then the next state at index 1 and so on. When your ready to do the next one are index BUFFER_SIZE-1 you're out of space, so you go back and use index 0 again...
 
@DanielSank just the mathematical basis of it?
 
The modulus operator % is going to be your friend in this.
 
@dmckee This project will never ship, lol
 
3:07 AM
What I just described will be faster to code and debug than the pointer-to-pointers-to-arrays scheme you exhibit above.
I know. I've done both of them at times.
 
@dmckee Sweet then
I'm glad I spent about 10 hours today reworking this code to make it struct-oriented
Now I can build on this and do nicer things
Anyway, 4AM time to go sleep
if I code now I will blow shit up
 
3 mins ago, by DanielSank
(For anyone who thinks I'm being mean to Bernardo: We're buddies. It's a joke and I expect he'll fire back)
2 mins ago, by Bernardo Meurer
(To anyone who thinks I'm being mean to Daniel: I am, suck it)
lol
 
@TanMath Well, based on your question about the 1/N probability in the FMO complex, I think you could benefit from thinking about the difference between probability distributions of a random process and particular realizations of that process.
Or perhaps I misunderstood the question.
 
i just feel, if the probability is 1/N, how is it directed in any way.. how can it "solve" a shortest-path problem
it would just be a random walk, nothing more, nothing less...
 
Well ok let's back up.
The probability is 1/N when?
 
3:11 AM
Okay, heading to bed
 
@BernardoMeurer <3
 
@dmckee Thanks for the help as always!
@DanielSank <3
Goodnight folks!
 
@DanielSank second page of the article:
". At the optimal level of phase breaking the system relaxes
quickly to the uniform probability distribution rho_nn = 1/N"
 
I'm going to see Ghost in the Shell.
Wish me luck because if it's bad, I'll have to go through a ritual forty day mourning period.
 
enjoy!
@DanielSank lol
 
3:12 AM
@TanMath Not yet, in about half an hour.
 
oh...
 
Then you should watch something which has to be good instead
 
@BalarkaSen It's worth the risk.
The Ghost in the Shell animated TV series is one of my favorite TV series ever.
 
Fair :)
I've never seen it.
 
It is truly excellent.
It's on Hulu, I think.
This is one of the greatest tools of all time.
It's a ruler, made of circuit board, with all kinds of useful stuff including surface mount footprints.
 
user228700
3:40 AM
@DanielSank: Hi. Seeing as you work at Google, I'd like to file a complaint against the system for a stupid error that it is making (:-P)
 
user228700
 
user228700
 
user228700
17?!
 
make that 1998 and you're done
ezpz
 
user228700
I did, but why the error?!
 
user228700
3:50 AM
@BalarkaSen x'D This may just be my favorite "word".
 
who's to blame google when there exists people incompetent at subtracting numbers
it's totally fine
 
user228700
Dude, it's Google.
 
's only human
 
user228700
...except it's not :-P
 
hola amigos
 
4:00 AM
mystical greetings
 
@BalarkaSen functional analysis is terrible
I thought I understood spectral theory, but this guy wants it in so much generality that the operators are not densely defined
maybe I should be an algebraist
 
@Kaumudi.H heehee
File a ticket
I recently filed a bug related to to time zones in Gmail.
 
@0celouvsky analysis is too hard
 
there's three types of eigenvalues
it's too hard to remember
@BalarkaSen algebraic topology is too hard
 
@0celouvsky The hardcore stuff (homotopy theory etc) are. The stuff in Hatcher is essentially easy if you're willing to think in terms of pictures, which is what I am good at
I basically can't do math, I can draw nice pictures
 
4:07 AM
Huh. The position operator in quantum mechanics has no classical eigenvalues, but it does have so-called "discontinuous eigenvalues"
@BalarkaSen I just don't know how to trust pictures
The fact that analysts don't need pictures beyond intro analysis is quite appealing
 
if it feels good you trust it
 
user228700
@DanielSank ._. Wokay.
 
I am drawing lots of pictures because learning foliations
beautiful stuff
 
pics?
 
can't upload mine. here's a friendly dude from candel-conlon:
those 1st level pipes all sit inside one another. 2nd level pipes do so too
 
4:15 AM
sit inside one another?
 
like the reeb foliation yeah
 
so this is like a pulled apart view?
 
Yup
The geometry of the leaf is actually complicated.
 
I see
So if I have the operator $X$ on $L^2(\Bbb R)$ which gives $(Xf)(x)=xf(x)$ whenver that makes sense in $L^2$, am I crazy to think that for $\lambda\in\Bbb R$, I should have $(\lambda -X)^{-1}$ be represented by $(\lambda-x)^{-1}$
Seems reasonable
But why shouldn't this be unbounded is the 1E6 rupee question
Maybe...stretch out some characteristic functions
 
hey
 
4:19 AM
hola
 
@Kaumudi.H So your mum and sister weren't just teasing you then? :-)
 
ugh integrals
what's $\int x^{-2}\, dx$?
 
user228700
@JohnRennie Eyy, you're up earlier than usual! :-)
 
user228700
No, they were not! :-o
 
$-\frac{1}{3}x^{-3}$
no
 
4:22 AM
@Kaumudi.H Couldn't sleep for some reason.
 
$-x^{-1}$?
 
user228700
@JohnRennie Oh. That's a bummer :-|
 
@JohnRennie Because you know you don't understand the general topology of manifolds
 
@Kaumudi.H It's not a problem. I just got up a bit earlier.
 
user228700
4:23 AM
Riight.
 
user228700
My mind was blown because of all the shtuff!!
 
I have a fair bit to do today anyway, so it's not a bad thing to get a head start :-)
 
user228700
:-) Wokay.
 
So ...
 
i didn't sleep all night so i feel ya
 
user228700
4:24 AM
@BalarkaSen Bro, you have a problem.
 
@BalarkaSen You're worried about point set topology as well then? :-)
 
@Kaumudi Yeah... naw
 
@Slereah ugh
 
wot
 
user228700
@JohnRennie Right. So their original plan was to buy me 18 small gifts but they were quick to drop that although I did end up getting a lot of crap.
 
4:26 AM
How should I prove that the $1/x$ operator is unbounded
it doesn't really seem that bad
why should it be unbounded
 
user228700
Hang on, I will show you two of the best ones:
 
isn't $1/x$ unbounded basically the archimedean property?
 
user228700
 
no, I mean the operator $1/X$
 
oh
then who knows
 
4:27 AM
unbounded on $L^2$
 
user228700
 
user228700
(^ Coldplay mug)
 
user228700
This, BTW, is how the script reads, @JohnR:
 
user228700
 
It's turned out to be a good 18th birthday then :-)
Are you going out for a meal tonight?
 
4:32 AM
I should get a T-shirt with "Times are a-chaingin' a-tellin', and the changin' ain't free"
But my Karl Marx is my homeboy T-shirt never materialized into being, so...
 
user228700
@JohnRennie So good! :-o They gave me lots more! At 12, they brought out the cake they had hidden and I cut it. They also gave me a silver bracelet, a framed photo of me and my two sisters, a beautiful dress to wear, a bottle of perfume, a bag of sorts and my mum even wrote me a letter!
 
Cool :-)
 
user228700
:-D It's been awesome and my mind has been thoroughly blown.
 
the usual trick for $X$ doesn't work
 
user228700
And no, we'll eat outside tomorrow perhaps, after my exam.
 
4:34 AM
rip
 
what are bivectors?
hey . . . . . .
 
@Kaumudi.H that makes sense. Help you chill down again after the stress.
 
user228700
Yeah :-) Do u have any of the delicious food you made yesterday left?
 
@Kaumudi.H Are you really asking if I didn't EAT THE LOT! ?
Me? :-)
 
user228700
:-) So no fancy lunch today?
 
4:40 AM
Was yesterday a holiday?
 
@Kaumudi.H I quite fancy some sausage rolls. It's a plain meal but nice. With some jerk sauce.
@Cows Every day is a holiday when you're retired :-)
 
@JohnRennie nice!!
:D
 
user228700
Is it just me or does "jerk sauce" sound wildly inappropriate? (:-P)
 
user228700
Jerk is a style of cooking native to Jamaica in which meat is dry-rubbed or wet marinated with a very hot spice mixture called Jamaican jerk spice. == Etymology == The term jerk is said to come from the word charqui, a Spanish term of Quechua origin for jerked or dried meat, which eventually became the word jerky in English. Jerk is also derived from the action of jerking, which referred to poking meat with holes so that flavor could more easily be absorbed. The term jerk spice (also commonly known as Jamaican jerk spice) refers to a spice rub. The word jerk refers to the spice rub, wet marinade...
 
user228700
Ah, I see.
 
4:42 AM
Indeed. What with faggots and jerking my eating is wildly inappropriate.
 
user228700
x'D True!
 
Now I think about about you could probably make jerked faggots.
 
Please don't
 
Though I'm not sure how well jerk seasoning goes with liver.
Actually I doubt it goes at all well with liver.
 
user228700
...I have no inputs to provide in this discussion.
 
4:45 AM
You might find it interesting to try jerk sauce. I think the nearest in Indian cooking would be a very hot sweet chutney.
 
user228700
Heh. I doubt if it's available here.
 
The sauce itself is vegetarian so you'd be able to eat it.
It works very well on pizza. I've eaten plain cheese pizza with jerk sauce and it's very nice.
 
user228700
Ooh, sounds great! I'll be on the lookout for it :-)
 
...wtf
 
4:49 AM
do you want to get banned?
 
user228700
@JohnRennie Hmm, wokay, thanks!
 
user228700
@0celouvsky Food. We're talking about food. Don't blame us for the actions of those who named this stuff in the first place.
 
@0celouvsky The last time you flagged me for mentioning the traditional English meal of faggots the ban was immediately rescinded. So let's not go there again eh?
 
user228700
What's nice is that most everybody remembers my birthday as it is on April fools' day and all and I bring this up because a random dude from high school has messaged me ._.
 
Hmm, neither Amazon.in nor eBay.in have the jerk sauce. You might be able to get it froma specialist supplier, but it would probably be very expensive.
 
4:53 AM
I forgot it's April fools day
 
user228700
@JohnRennie Hmm. I'll try and find it somewhere. There are some shops that might have it.
 
It's flavoured with a spice called allspice.
 
What I like best is jerk barbecue sauce. It's the sauce I said was a bit like a hot, sweet chutney. Jerk spice is just a powder, or sometimes a paste made with oil, and it doesn't contain any sugar.
 
Interesting how the comma got in there!
 
user228700
5:00 AM
@JohnRennie Hmm, I see.
 
@JohnRennie I flagged you?
Ah, so with $X^{-1}$ I have to look at functions that get close to $0$
good idea @Slereah
What about $f_n=\chi_{[1/n,1]}$...
 
I think @0celouvsky is talking to a fictional version of me
 
5:15 AM
alternate universe overlap
 
So I have $||f_n||=\sqrt{1-1/n}=\sqrt{(n-1)/n}$ and $||X^{-1}f_n||=\sqrt{n-1}$
 
(in that universe I'm a retried quantum field theorist, Slereah is a topologist and 0celo7 is still a functional analyst because he sucks)
 
What topology am I
Am I the Alexandrov topology
 
So, letting $\hat f_n=f_n/||f_n||$, we have: $$||X^{-1}f_n||=\frac{||X^{-1}f_n||}{||f_n||}=\frac{\sqrt{n-1}}{\sqrt{n-1}/\sqrt n}=\sqrt n\to\infty$$
yass
@Slereah you're the $\sigma(E'',E')$ topology
the weak topology on dual of the strong dual
I've got to sleep lads
cheerio
 
5:36 AM
@Sebastiano $A$ the EM four-potential not the four-acceleration.
If you look at the Wikipedia article the EM four-potential is written in terms of $E$ and $B$ so taking the derivatives to calculate $F$ is straightforward (but tedious).
 
Hello @JohnRennie
 
Then when you have $F$ you just need to calculate the trace of $F^2$. The easiest way to do this is take the Hadamard product of $F$ and the transpose of $F$ and add all the entries in the product.
That gives you the trace of $F^2$. This is easier than it sounds, and you'll get the result in your post.
@Ramanujan Morning
 
Good morning @JohnRennie. Could you give me some idea on modular arithmetic?
 
Maths isn't my strong point, but I'll help if I can. What do you need to know?
 
@JohnRennie, Actually I want to know what does it mean, where do we apply and how does it work?
 
5:43 AM
That's an awfully broad question. You really need to do some reading around to get an idea of what modular arithmetic is. There must be plenty of introductory articles out there in Googlespace.
I don't think modular arithmetic is used much in physics if at all. It's mostly used in pure maths in number theory.
 
I tried everything found in Google but could not understand
@JohnRennie, what is the meaning of mod?
 
Suppose you're doing integer arithmetic and you divide $x$ by $y$ i.e. $x/y$.
 
Yeah @JohnRennie
 
Unless $x$ is an exact multiple of $y$ there will be a remainder. For example 8/3 is 2 with remainder 2 because 8 = 2*3 + 2. Does that make sense so far?
 
Yeah. I am getting a bit. @JohnRennie
But what if -7 is divided by 5 @JohnRennie
 
5:49 AM
The mod function gives you the remainder, so 8 mod 3 = 2.
 
@JohnRennie, so what for -7 divided by 5
 
@Ramanujan to answer this note that $x mod y = (x+y) mod y$ because adding multiples of $y$ doesn't change the remainder when you do the division. Is that OK?
 
-7 mod 5=-2 mod 5?@JohnRennie, do you mean same?
 
Yes, and now add 5 again ...
 
To what?
@JohnRennie
 
user228700
5:52 AM
@JohnR: Math?! :-o
 
@Ramanujan -2 mod 5 = (-2+5) mod 5 = 3 mod 5
And 3 mod 5 you can do ...
 
@JohnRennie, could you please elaborate how is it uses here?
3
Q: Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$

RamanujanProve that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ My Attempt: $$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$ $$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$ $$=\dfrac {\cos^2 40°.\cos^2 80°+\cos^2 20°.\cos^2 80°+\cos^2 20°.\co...

 
@Ramanujan At a quick glance I can't see how that has anything to do with modular arithmetic ...
 
@JohnRennie, did you notice Lab's answer?
 
@Ramanujan hmm, I don't see why he needs $m = -1,0,1 \pmod 3$
He shows that $x=120^\circ m+20^\circ$ where $m$ can be any integer.
 
6:03 AM
@JohnRennie, he says that a,b,c are roots of the equation. But how?
 
He starts with the identity: $$ \cos 3x =4\cos^3x-3\cos x $$ This is proved here if you're unsure how that identity is obtained.
 
Yeah. @JohnRennie, but how did he conclude that a,b c are roots of the equation
 
@Ramanujan by direct substitution i.e. plug in $a=\cos20^\circ$ etc and show it works.
 
O.@JohnRennie, A pair of quantities having same dimensional formula are. ;;1). Momentum, impulse. 2). Momentum, energy. 3). Energy, pressure. 4). Force, power
 
I suspect that what is puzzling you is that his proof doesn't seem to have a nice clear chain of reasoning i.e. start at the beginning and work through, logical step by logical step, to get to the conclusion.
And that's because it doesn't. He made an intuitive leap, a guess basically, to get a polynomial that linked $a$, $b$ and $c$. Then he was able to use Vieta's formula to derive links between $a$, $b$ and $c$ that could be used to answer the original question.
A lot of maths proofs are like this. You manage to find some unexpected property that you can use to solve the problem.
 
6:19 AM
Ok..
@JohnRennie, A pair of quantities having same dimensional formula are. ;;1). Momentum, impulse. 2). Momentum, energy. 3). Energy, pressure. 4). Force, power
 
You can do that. Just write down the equations for each quantity.
 
@JohnRennie, impulse, energy
??
 
Well what is the equation for impulse?
 
@JohnRennie, I don't know?
 
In classical mechanics, impulse (symbolized by J or Imp) is the integral of a force, F, over the time interval, t, for which it acts. Since force is a vector quantity, impulse is also a vector in the same direction. Impulse applied to an object produces an equivalent vector change in its linear momentum, also in the same direction. The SI unit of impulse is the newton second (N·s), and the dimensionally equivalent unit of momentum is the kilogram meter per second (kg·m/s). The corresponding English engineering units are the pound-second (lbf·s) and the slug-foot per second (slug·ft/s). A resultant...
 
6:38 AM
(planning something in the background...)
 
user228700
@JohnR: When everything is awesome and my best friend hasn't called yet -_-
 
Everything is awesome - enjoy it while it lasts! :-)
Real life has a habit of sneaking back in :-)
 
user228700
:-) Yeah. My cousins (with whom I speak once in two years eek!) called! :-o
 
user228700
This is so strange; I didn't expect for more than 2 people to call :-|
 
It's nice to know you haven't been completely forgotten by the rest of the world. Even if you have been locked in a cell revising for the last year and a bit.
 
user228700
6:45 AM
Exactly! :-) All but my stupid best friend -_-
 
user228700
Oh, so u think u'll read the FBaWTFT screenplay?
 
@Kaumudi.H No. I'll wait until the novel is published. I forget when it's due, but I don't think it's too far off.
 
user228700
The novel? What novel?!
 
Isn't Fabulous Beasts being rewritten by JKR as a novel?
 
user228700
*Fantastic Beasts and no, no novel.
 
6:54 AM
Hmm, I wonder where I got that idea from.
 
user228700
Err, I suspect it was me :-| I remember now that for awhile, I thought it was being novelized.
 
Ah, OK :-)
 
Into a novelty?
:-)
 
7:18 AM
@JohnRennie Just make it.
 
7:35 AM
@JohnRennie You have been very kind and I am very happy that there is someone who has listened to me. At least for now, the system does not make me ask more questions. I greet you with so much cordiality.
 
99.999% of my friendships are long distance, including those that live in the same city as I am
We basically maintain that via social network and more importantly, highly resilient memories of us being together
These memories makes our friendships almost spacetime independent like entanglement, but also very robust unlike entanglement that is fragile
 
8:08 AM
@ACuriousMind I knew you'd disapprove!
@Sebastiano I must admit it hadn't occurred to me that you thought $A$ was the four-acceleration, though to be fair we do use the same symbol for both the four-acceleration and the four-potential.
I now understand why you were getting confused!
 
@JohnRennie I do, in particular because that's not a layman's question. The asker has a tendency to be very literal, and I'd be surprised if, after reading your answer, they'd not come back with "Why is X not calculable with scattering amplitudes?" after they encounter the first QFT phenomenon that's not about scattering.
 
@ACuriousMind if you think it's unhelpful I'll delete it. My aim is only ever to make things clearer not murkier.
 
I also know that many non-hep QFT practitioners disapprove of hep theorists acting as if their scattering amplitudes and whatnot are the be-all and end-all of QFT, so I'm also trying to be a bit more inclusive there ;)
 
8:25 AM
0
Q: Trying to understand why the answer was deleted

akhmeteliMy answer to What is special about Maxwell's equations? was deleted by a moderator. The OP asked what is special about the Maxwell equations, which, according to the OP, are basically a combination of "4 equations that were already formulated by other physicists". It looked like the OP was surpr...

 
And that is why I am not understanding anything about the 4-acceleration and 4- potential. Right now I'm at school and I'm trying to understand at least something more. I had to cancel my request but the questions they would like to make a lot. $\textbf{E}\bullet \textbf{B}=\textbf{E}’\bullet \textbf{B}’$


There are distinct possibilities:

$\textbf{E}=\textbf{B}=0$

$\textbf{E}\bullet \textbf{B}\ne 0$

$\textbf{E}\bullet \textbf{B}\ne 0$

$\textbf{E}\bullet \textbf{B}= 0 \longrightarrow \textbf{E}’\bullet \textbf{B}’= 0 $ with $\textbf{E}, \textbf{B}\ne 0$.
 
8:41 AM
@Sebastiano What is $\bullet$? I'm not sure what exactly your question is.
 
9:08 AM
@ACuriousMind Hi and good morning. symbol of scalar product (in LaTeX).
 
The scalar product is usually written $\cdot$, not $\bullet$. Anyway, what exactly is the question? $E\cdot B = 0$ means that the electric and magnetic fields are orthogonal, $E\cdot B \neq 0$ means they aren't.
 
@ACuriousMind I know this. It is not the problem knowing the physical considerations. i use $\cdot$ for usual multiplication
 
You know you are sleepy when you need to zoom in to read a journal article
 
9:49 AM
@AccidentalFourierTransform Sadly, that seems to not be true.
 
Walter Lewin might join PSE :D
 
10:12 AM
Hi guys! I have a question about the ADS super potential. What does it mean exactly that gaugino condensation generates the ADS superpotential for F < N-1 where F is the number of flavours and N the number of colours of the super gauge theory? Physically, I have a gaugino condensation, then.. how can this condensate generate a super potential which influence the dynamics of the system?
 
Says who? @YashasSamaga
 
10:38 AM
@YashasSamaga That Walter Lewin? He just went away from Quora
I was introduced to physics with WL's videos
 
11:32 AM
In quantum computing and specifically the quantum circuit model of computation, a quantum gate (or quantum logic gate) is a basic quantum circuit operating on a small number of qubits. They are the building blocks of quantum circuits, like classical logic gates are for conventional digital circuits. Unlike many classical logic gates, quantum logic gates are reversible. However, it is possible to perform classical computing using only reversible gates. For example, the reversible Toffoli gate can implement all Boolean functions. This gate has a direct quantum equivalent, showing that quantum circuits...
Now learning some quantum circuitory
 
11:43 AM
@ACuriousMind :-(
can we just pretend that it is true?
 
@skillpatrol @JaimeGallego youtube.com/…
 
The account was created just after the comment in YouTube, so I think that's him
 
I think I have seen that profile; I thought it was an impersonator.
Well, anyone could have made an account after seeing that comment to troll around.
 
11:50 AM
Might be.
 
"Here follows a note from my former colleague Professor Barton Zwieback at MIT."
:O
:O IT IS HIM!
5
A: Will a current-carrying conductor be affected by its intrinsic magnetic field?

Walter LewinYour question is a classic question. Yes the 2 wires will repel each other. The classic explanation is that this is the result of the $\vec{v}\times\vec{B}$ Lorentz force. Here $\vec{B}$ is the magnetic field that is produce by the current (Ampere's Law). This force is in opposite direction in th...

Quora did a good job.
 
12:10 PM
is it only me or sci-hub is down?
 
@AccidentalFourierTransform "Search temporarily unavailable"
 
oh no
:-(
I cant even.
 
12:24 PM
@JohnRennie Are you around?
 
12:44 PM
or @dmckee
 
@ACuriousMind lol
 
12:59 PM
what's even funnier is that he actually investigated it :-)
 
1:14 PM
@YashasSamaga halp
 
user228700
1:28 PM
 
user228700
1:52 PM
However, this is something to consider:
 
2:11 PM
I think April Fools is just an excuse to disappoint people
 
Ok I obviously DON'T UNDERSTAND how to choose detector settings for a CHSH computation
o wait a sec, typo...
 
@Sir My favorite joke is to tell and convince someone on April 1st that they will be fooled and not doing it
 
@BalarkaSen Imma use that :)
 
$$\mathcal{B}=\{\lvert++\rangle, \lvert+-\rangle, \lvert-+\rangle, \lvert--\rangle\}$$

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

$$\lvert \updownarrow -\rangle=\lvert \updownarrow\rangle\otimes \lvert - \rangle$$

$$R (\theta)=\begin{pmatrix}1 & 0& 0& 0 \\ 0 & 1& 0& 0 \\ 0 & 0& \cos \theta& -\sin \theta\\ 0 & 0& \sin \theta& \cos \theta\end{pmatrix}$$

$$\lvert m\rangle=R(\theta)\lvert \updownarrow -\rangle=\frac{1}{\sqrt{2}}(\lvert +-\rangle-\sin \theta \lvert -+\rangle+\cos \theta \lvert --\rangle)$$
nope still not working, choosing $\theta = \frac{3\pi}{2}$ and then the Tsirelson bound will be violated, but that cannot be true since the state that is obtained when setting that angle of rotation is one of the triplet bell states which the correlation function should give $2 \sqrt {2}$
 
user228700
@SirCumference I find myself so lucky in this regard; nobody is mean enough to terribly fool the birthday gal :-D
 
2:19 PM
@Kaumudi.H Brilliant
 
user228700
I am still a little bit weary when opening presents for fear that they might be empty but that hasn't happened in the 18 years I've been alive so yay! :-P
 
no sorry, I am stupid again: $1+\sqrt{2}=2.41421356237...<2\sqrt{2}$ thus all is well
 
@Kaumudi.H Oh yeah, and: Happy Birthday! :)
 
user228700
Thanks, man :-)
 
"One morning Gregor Samsa woke up from uneasy dreams and found himself turned into a giant beetle"
This almost sounds like a birthday prank.
(credits to Franz Kafka)
 
2:23 PM
but still, shouldn't I should get $2\sqrt{2}$ since $\theta =\frac{3\pi}{2}$ is a bell state...?
 
@BalarkaSen Family leaves him to starve, it was all a prank man!
Can't you see the camera?
 
user228700
@SirC: My exam is tomorrow, BTW :-|
 
@Kaumudi.H The JEE?
 
user228700
Yep.
 
@Kaumudi.H Dang, good luck, but you'll probably do fine
Keep confident
 
user228700
2:27 PM
Thank you :-) How are your exams going? (Told they were coming up IIRC(?))
 
@Kaumudi.H Well I'm studying with my friends. Some of the profs are actually surprisingly nice, giving us formulae in some cases
 
Good luck with JEE. That stuff's too hard.
 
user228700
@SirCumference Nice!
 
user228700
@BalarkaSen Gulp. Hopefully not way too hard. Thanks.
 
user228700
Dammit, didn't he change his username to "2017" at one point? Wokay, those writing JEE tomorrow: All the best :-) (:-P)
 
2:32 PM
I have to decide what I want to do.
The hardest part of doing something
 
@BernardoMeurer I will help all day tmrow. I need to sleep now.
I have some problem sleeping before the exam day :p
 
2:52 PM
I would appreciate it if someone could explain-to-OP/mediate/step-in/vote-to-reopen/vote-to-close here.
 
why would angular velocity be relevant in pulley free body diagrams?
 
@Qmechanic I was actually typing a comment right now...
give me a sec
 
@YashasSamaga That's alright man, good luck on your exam!
 

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