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user218912
12:00 AM
you guys accidentally wasted my time for question 1
 
lol
 
user218912
there is a much easier way to do it without expanding it in terms of step functions and applying the klein-gordon operator.
 
user218912
someone showed me
 
user218912
._.
 
user218912
you just take the derivative of the time ordered product directly inside the bra-ket.
 
user218912
12:02 AM
they didn't show me their work though so I'm still figuring out the steps.
 
12:14 AM
sucks to be you
 
user218912
nah I figured it out.
 
user218912
done question 1
 
12:41 AM
@bloo dude problem 3 on my QM homework is so long
 
12:59 AM
>doing homework
I'm glad to be out of school man
 
hmm
I should check that this is correct
but HOW
crap I lost a $u(0)$
 
user218912
1:45 AM
why are you pinging bernard?
 
user218912
and why is your font size so large
 
user218912
is it a requirement to have it like that?
 
Hello, physics experts.
We got this real-world physics question over at Science Fiction and Fantasy. Would Physics be the best place for the OP to ask it?
 
@bloo that's standard 12 size font
@bloo because he knows QM
 
@AlfredCentauri - You seem like a high-rep user. Would such a question be well-suited for Physics, or should it be sent elsewhere?
 
1:54 AM
Seems like nonsense @Adamant
 
@0celo7 It certainly seems to be starting from a mistaken premise, yes.
 
user218912
@0celo7 not at that level.
 
@bloo they're solving the SE for hydrogen
I think he's at that level
 
@yuggib My prof agreed that a $5\epsilon$ argument was best :)
 
user218912
2:06 AM
@0celo7 maybe question 1
 
user218912
you're taking a graduate qm class, he's in freshman physical chemistry.
 
user218912
big difference.
 
@bloo the material is the same lol
 
user218912
@0celo7 no it's not.
 
user218912
they deal with it in a different way.
 
user218912
2:08 AM
only some of it is the same.
 
@DanielSank I'm confused by what to do in this question
How are the t and r coefficients calculated when there are three regions?
what's the definition
 
@0celo7 It's too late for analysis
it's 3:15 here dood
 
@BernardMeurer sounds like prime analysis time to me.
 
Jeez
@bloo Icelord?
@0celo7 Give me an exercise
 
@BernardMeurer Don't look at your notes
That's the only way to learn
You need to reprove things you did in class without looking at notes.
I will help if you need
@BernardMeurer Show that a convergent sequence has precisely one limit.
 
2:17 AM
Hmm Okay
So this sounds like the kind of prof you begin by assuming:
Suppose the the limit is not unique, and that there are $x,y \in \mathbb K, x\neq y$ s.t. $a\rightarrow x$ and $a\rightarrow y$
 
$\Bbb K$?
 
Some field, $\mathbb R, \mathbb C$
Does it need to be real?
Of course
it's real analysis
 
@BernardMeurer Well hell, might as well work in a metric space
@BernardMeurer No
 
@0celo7 Well then leave it some field then
 
@BernardMeurer it's true for any Hausdorff topology
 
2:20 AM
So not true for me
 
@BernardMeurer No, let's just work on $\Bbb R$
Either you work on $\Bbb R$ or on a metric space/Banach space
but working on $\Bbb K$ is nonsense
 
Sure, say $\mathbb R$ then since idk whats a metric space
Got it
 
@BernardMeurer Basically, you work concretely on $\Bbb R$ or you go full-power abstract
 
so $x,y\in\mathbb R$
 
should that be $a_n\to x$
 
2:21 AM
Yes, thanks
 
you can proceed in two ways now
Either assume $x\ne y$ and get a contradiction
Or prove they're the same
 
I was going to try and go by supremum/infimum
Is that a route?
 
Yikes
Maybe
But not easily generalized, especially because $\Bbb C$ has no sup/inf
 
Hm, should I try coming up with something else then?
I guess
OH
Definition of convergence!
 
that's a good place to start
 
2:24 AM
Well it has no ordering
Hard to do sups without ordering
 
How do I defnine this shit using $V_\epsilon(a_n)$ again?
 
what the hell is that
You don't, why are you looking at a neighborhood of $a_n$
 
There, fixed, neighbourhood of radius $\epsilon$ of $a_n$
Because that's how I learned to define convergence
Can I check abbot?
 
Ok, define convergence
What does $\lim a_n=a$ mean
No looking
 
it means that after a certain p every n>p will be in a neighbourhood epsilon of a
 
2:28 AM
p?
 
It means that for some epsilon, there is some N such that for all n > N, |a_n - a| < epsilon
amirite
 
We use $N$ normally
 
Okay, $N$
 
@Slereah You're very clever.
 
@Slereah mathjax plox
Okay that's what a limit means for me
 
2:30 AM
Wasn't too hard since I've read analysis all day
 
@Slereah You should be able to define a limit anyways
 
Well worst case scenario
I just remember the Tom Lehrer song about limits
There's a delta, for every epsilon
 
$\lim a_n=a\Leftrightarrow (\forall\epsilon>0)(\exists N\in\Bbb N):n\ge N\implies |a_n-a|<\epsilon$.
 
It's a fact that you can always count upon
There's a delta, for every epsilon
And now and again, there's also an N
But one condition I must give
The epsilon must be positive
What a lonely life all the others live
 
@BernardMeurer Clear?
 
2:32 AM
In no theorem, a delta for them
How cruel, how sad, how tragic
 
@0celo7 Yep, absolutely
 
I forget the next line
 
@BernardMeurer Ok I will show you how to do this
 
Wait
Walk me through it
So we have a definition, gut
we want to show that $\exists! \epsilon$
 
user218912
@0celo7 you never show me how to do stuff :(
 
2:35 AM
@bloo You're not his adoptive brother
 
user218912
tbh not including that one time DS helped me on that part of the question in the problem set, I've been figuring out and answering all of this QFT stuff on my own.
 
Let $(a_n)\subset\Bbb R$ be a sequence. Suppose $a_n\to x$ and $a_n\to y$. Let $\epsilon>0$. There is an $N_1\in\Bbb N$ such that $n\ge N_1$ implies $|a_n-x|<\epsilon/2$. Also, there is an $N_2\in\Bbb N$ such that $n\ge N_2$ implies $|a_n-y|<\epsilon/2$. Let $N=\max\{N_1,N_2\}$. Then for $n\ge N$, $$|x-y|\le |x-a_N|+|y-a_N|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$ Thus $|x-y|<\epsilon$ for any $\epsilon>0$, hence $x=y$.
@BernardMeurer what?
@bloo so all that help I gave you with derivatives?
 
user218912
@0celo7 and that.
 
$\epsilon/2$?
 
oh god I broke it
Damn
That should read: Then, since $N\ge N_1$ and $N\ge N_2$, $$|x-y|\le\cdots$$
 
2:38 AM
why over two?
 
@BernardMeurer so when you add your two estimates together you get $\epsilon$ instead of $2\epsilon$
@BernardMeurer Maybe we should start with something simpler.
Since you can't use notes.
 
Didnt I prove this over skype the other day?
 
@BernardMeurer Prove: $x=y\Leftrightarrow (\forall \epsilon>0):|x-y|<\epsilon$.
 
But by trichotomy
 
@BernardMeurer no
that was the uniqueness of the supremum
 
2:40 AM
AH
 
prove what I just said pls
 
it's a good proof
very useful thing to know
 
who are x and y?
 
$x,y\in\Bbb R$.
 
2:41 AM
$\forall x, y \in \mathbb R$
Ah
 
One direction is trivial!
 
user218912
can't we talk about physics here
 
No.
 
user218912
I need exposure
 
Hmm
Well
Every number has a symmetric by definition
so if x=y, -y is x's symmetric
and by definiton x-x=0
since we defined eps as >0 the solution will always be true
 
2:45 AM
Are you trying $\Rightarrow$?
What is "symmetric"
"solution"
 
$\forall x\in \mathbb R, \exists z : x-z=0$
 
But yes, that's the idea. $x=y\Rightarrow x-y=0\Rightarrow |x-y|=0\Rightarrow |x-y|<\epsilon$ for any $\epsilon>0$.
The other direction is harder
I'm going to do laundry, brb
 
That is the symmetric, the notation however is that $\forall x\in \mathbb R, \text{ iff } x+z=0\text{ then } z=-x$
 
You will want a contradiction
 
So the other way around: we know the abs will make it positive so $|a-b| \geq 0$
Since this is for all epsilon we can suppose the lower bound case eps=1
So for that to be true $|x-y|$ need to be such that $|x-y|<1$ always
Wait
these are reals not naturals
fuck
Hm
How do I denote the infinitesimal? The smallest number larger than 0
 
2:53 AM
@Slereah can tell you
but why the hell do you want it
that's not standard analysis
 
@0celo7 You told me to find a solution I'm finding one
 
"infinitesimal" is not a thing in real analysis
 
if we assume the smallest possible value of eps we can show that the solution of x-y must be 0
Tsc, well fuck then
$$x=y\Leftrightarrow (\forall \epsilon>0):|x-y|<\epsilon$$
Dude
 
yeah
 
The reals are infinitely dense
 
2:59 AM
that's not proper terminology
 
How the fuck do I denote it then?
ffs
 
denote what
 
$$\forall x,y \in \mathbb R, \exists z\in \mathbb R : a<z<x$$
 
The reals are a linear continuum.
 
There, it's dense af
Lol, I'll write that on my exams "Since the reals are dense af we can assume..."
But okay, it's a linear continuum
 
3:02 AM
No one says that in analysis though.
I think there's a special name for that property in the real number case
 
If there were x,y such that the inverse implication wasn't true... Fuck I lost my train of thought with this density thing
 
do you want me to tell you?
 
I had it, now idk anymore
Not yet
 
If I were an analysis prof I would give this on every test
even advanced analysis
I find the correct argument really tricky
it's one line
 
since R is a linear continuum the only pair x,y that can exist so that the difference is smaller than every eps in R is (x,x) so then it's 0
 
3:04 AM
but I always have to think about it
@BernardMeurer Proof?
 
Because if it was anything else there would be infinite possible eps smaller than it
Lemme try
 
user116211
Hey, @0celo7, you must know the proof of rational numbers being dense in $\mathbb R\,.$
 
@MAFIA36790 Uh, vaguely
Why?
 
user116211
While the crux of the proof is dependent on Archimedean property and I got most part of the proof, I still couldn't comprehend some thing.
 
ok?
 
user116211
3:10 AM
Wait, ...
 
user116211
 
user116211
I didn't get why $$S = \{n\in \mathbb N| n/q\color{red}{\geq}b\}\,.$$
 
$$\forall x,y \in \mathbb R, \exists z\in \mathbb R : x<z<y \\ (\forall \epsilon>0):|x-y|<\epsilon \\ x\neq y \implies |x-y| > 0 \\ \therefore \epsilon > |x-y| \implies \exists z\in \mathbb R : \text{ fuck this shit}$$
 
user116211
When will be the equality case in $n/q \geq b\,?$
 
@BernardMeurer Want to know it?
You're close.
 
3:13 AM
@0celo7 No, I want to find it
 
@MAFIA36790 Reading.
 
user116211
@0celo7 sure.
 
@MAFIA36790 How have they formulated the Archimedean property?
 
user116211
@0celo7 You mean how they stated the property?
 
$(\forall x\in\Bbb R)(\exists n\in\Bbb N): n>x$?
@MAFIA36790 Why what?
 
user116211
3:14 AM
For each pair of real numbers $a$ and $b,$ there is a natural number n such that $na\gt b\,.$
 
@MAFIA36790 ok same thing
 
user116211
@0celo7 yes.
 
So what's your question?
 
user116211
@0celo7 In the set $S,$ why did they use the equality sign in $n/q\geq b\,?$
 
user116211
When will be the equality case?
 
3:16 AM
when $bq$ is an integer.
i.e. if $b$ is an integer
*positive integer
 
user116211
$1/q$ is $b-a$ where $a\gt 0$ and $b\gt a\,.$ So, that means $b-a/q\geq b\,.$
 
$$\forall x,y \in \mathbb R, \exists z\in \mathbb R : x<z<y \\ (\forall \epsilon>0):|x-y|<\epsilon \\ x\neq y \implies |x-y| > 0 \\ \therefore \epsilon > |x-y| \implies \exists z\in \mathbb R : |x-y|<z<\epsilon \\ \therefore \epsilon > 0 \implies |x-y|=0 s.t. \forall \epsilon, \epsilon > |x-y|$$
 
dude
 
Best I can do
 
$q$ is a natural number
 
user116211
3:18 AM
WTH
 
if $b$ is a natural number, then $bq$ is a natural number
 
user116211
checking
 
user116211
I mean unless $q=1$ and $a=0,$ how can it give the equality with $b\,?$ Also, it was stated initially that $a\gt 0\,.$
 
user116211
So, no chance of equality
 
user116211
Or am I missing something?
 
3:21 AM
yes you're missing something
if $b$ is a natural number, then $bq$ is a natural number
do you agree?
 
user116211
@0celo7 Yes, I agree with this statement but I'm unable to see how this statement is related to the equality case.
 
@0celo7 yes it's a closed algebra
 
because then there is an $n\in\Bbb N$ such that $n=bq$
 
@0celo7 Solve for the amplitude of the rightward traveling wave in the rightmost region.
 
@0celo7 DUDE ITS 4:30 AM
WTF
 
3:23 AM
namely $bq$ itself
 
@0celo7 Is my proof right even?
 
That, divided by the incoming wave amplitude, is the transmission coefficient.
 
user116211
@0celo7 Ahhha!!
 
user116211
So, is there some sort of generator that makes an element of natural number product of finite numbers of other natural numbers?
 
user116211
Or is it an axiom?
 
user116211
3:25 AM
@DanielSank o/
 
@BernardMeurer no
I will tell you in one minute.
 
@0celo7 I give up
I'm going to bed
See y'all
 
user116211
@BernardMeurer Yes, go to bed; o/
 
@BernardMeurer WAIT
typing
Assume $x\ne y$. Then $|x-y|=:\epsilon_0>0$.
Then $|x-y|\not <\epsilon_0$, which contradicts $|x-y|<\epsilon$ for all $\epsilon>0$.
QED.
 
user116211
I got the the thing, but @0celo7, the thing you stated, is it an axiom of Natural Numbers?
 
3:32 AM
@MAFIA36790 what?
what did I state?
 
user116211
@0celo7 Am I seeing the application of Trichotomy Law?
 
@DanielSank I see.
@MAFIA36790 Yes.
 
user116211
10 mins ago, by 0celo7
because then there is an $n\in\Bbb N$ such that $n=bq$
 
@MAFIA36790 dude
what's so hard to understand
if $bq$ is a natural number, then of course there's a natural number equal to $bq$
 
user116211
._.
 
3:34 AM
namley $bq$ itself
 
user116211
@0celo7 sure.
 
so there's your equality!
 
user116211
okkkay :)
 
user116211
Also, can I apply the Archimedean Property for natural numbers?
 
user116211
After all it was said that $a$ and $b$ were real numbers.
 
3:35 AM
natural numbers are real numbers...
 
user116211
I can take them as natural numbers too, isn't it?
 
user116211
@0celo7 Yes, that's what I'm saying.
 
user116211
Thanks @0celo7....
 
4:36 AM
0
Q: Can someone explain the voting system for the upcoming moderator election?

NathanielIn the upcoming moderator election we each get three votes - a first, second and third choice. In order to use these effectively, I would like to understand the mechanics of how they will be used to determine the election's outcome. A side panel on the election page says We will calculate th...

 
5:03 AM
@MAFIA36790 \o
 
user228700
5:47 AM
Hi everyone :-)
 
user116211
Hello.
 
@Kaumudi Hi, everybody.
 
user228700
@MAFIA36790 What does the term "score" mean in the language of statistics?
 
user116211
@Kaumudi Depends on the context of the problem.
 
user228700
5:50 AM
@MAFIA36790 It depends? Huh. Well, what about in the following definition:
 
user228700
"Measures of dispersion are descriptive statistics that describe how similar a set of scores are to each other"
 
That's really bad wording.
 
user228700
I understand the definition of measures of dispersion. What is the meaning of the term score here?
 
user116211
Very much bad.
 
user228700
@DanielSank Really? :/
 
5:52 AM
...unless this text is already talking about scores and you've left out context.
@Kaumudi Are you keeping context from us?
If not, then "score" is a dumb and confusing way to say "value".
 
user228700
@DanielSank Uh, it uses the term score multiple times before this. How does that matter? No, there is no other context to be given :/
 
user228700
@DanielSank Then why use the term score? Where is it used correctly, then?
 
@Kaumudi Well, then the writer has chosen to use rather non-standard vocabulary (at least, in my opinion).
 
user116211
They are measures which expresses the spread of observations in terms of the average of deviations from certain (central) values.
 
@Kaumudi To me, "score" is the value associated to how many times you get the ball into the goal in football.
@MAFIA36790 Correct.
 
user228700
5:55 AM
@DanielSank U mean like frequency?
 
@Kaumudi No.
If I kick the ball in the goal twice, and you do so three times, the score is 2-3.
 
user228700
Ohh.
 
user228700
Huh. Okay, then I'll just ignore the word for now.
 
Good idea.
Why are so many books so bad?
 
user228700
@DanielSank I found that on a website.
 
user116211
5:57 AM
@DanielSank They are not standard books, so I'll not make a comment on those books.
 
Oh, I know! Because unlike software, book sources are not made available on hosting services like github which allow people to file issues.
 
user116211
They are just for aggregating formulas and that's it.
 
@MAFIA36790 Not true, sir! Some books explain well. Not many though.
 
user116211
@DanielSank As I said, I'll not make any comment on those high-school books.
 
user116211
I have seen though books use score; but it's not a regular usable terminology.
 
user116211
6:03 AM
It varies from authors and contexts what score actually is.
 
7:21 AM
So here's a funny thing... my numerical integration code is coming up with $\int_0^1 (1 - 2\xi + 2\xi^2)\mathrm{d}\xi = 1$, when it should be $\frac{2}{3}$. If I give it $\int_0^1 \mathrm{d}\xi = 1$, it gives the right answer. If I give it $\int_0^1 (1 - 2\xi)\mathrm{d}\xi$, it gives $-\frac{2}{3}$ whereas the correct answer is $0$. If I give it $\int_0^1 (1 - \xi)\mathrm{d}\xi$ it produces $0$ (and very quickly, too) but the answer should be $\frac{1}{2}$.
Any thoughts from the hive mind on what could be wrong with my code?
oh wait, I figured it out - it was setting $\xi=1$. duh
 
7:48 AM
Felt proud to be able to explain current density going out of slab of thickness 2z but infinite in the xy plane, and how the B field is contributed by each J:
Caption: Consider any $z\neq 0$, then since the B field from each J decays as $\frac{1}{r^2}$ then the field contributed by each J in the slab in a given z will be asymmetric for field lines that are pointing to -y and those pointing to +y. Therefore, there will be some leftovers when all the contributions are summed together.
For $z=0$, however, the contribution is symmetric from top and bottom, thus the field always cancel there leaving no field
As for why there is no y component in the field, because the slab is infinite in the y direction, any point along any fixed y is symmetric left and right, thus loops always cancel, leaving no y component
 
 
1 hour later…
8:53 AM
@DavidZ why are you writing that stuff? Use a library that implements the best algorithms and has been tried, tested, debugged, profiled etc
 
 
1 hour later…
10:09 AM
@innisfree got any suggestions?
 
user116211
@0celo7, You are still awake or your phantom phone is logging you still ;P
 
Group theory is kinda cool, especially when you start turning concepts like cosets, subgroups, torsion, generators etc. into some pictures
The idea of orbits really help in visualising most groups
 
10:37 AM
@DavidZ sure, depends on the language. But generally quadpack a Fortran library, which is part of GNU scientific library (gsl) for c and cpp, and used by scipy quad in python
QUADPACK is a FORTRAN 77 library for numerical integration of one-dimensional functions. It was included in the SLATEC Common Mathematical Library and is therefore in the public domain. The individual subprograms are also available on netlib. The GNU Scientific Library reimplemented the QUADPACK routines in C. SciPy provides a Python interface to QUADPACK. == Routines == The main focus of QUADPACK is on automatic integration routines in which the user inputs the problem and an absolute or relative error tolerance and the routine attempts to perform the integration with an error no larger than that...
 
When will Fortran die
 
@innisfree I'm already using GSL
 
Yeah, I know, I'm quite familiar with much of GSL ;-)
 
Oh sorry I misunderstand you
Uh mistunderstood you, even
The worst thing about my programming education is that we were encouraged to write numerical recipes ourselves (integration, differential equations, eigenvalues of matrices etc)
It's good to understand them, but it's crazy to make your own buggy implementations of such common things
 
10:47 AM
Yeah, agreed
I'm all for using libraries
Whatever my problem is, it's a bug in my "driver" code, not the integration routines themselves.
 
In mathematics, a pro-p group (for some prime number p) is a profinite group G {\displaystyle G} such that for any open normal subgroup N ◃ G {\displaystyle N\triangleleft G} the quotient group G / N {\displaystyle G/N} is a p-group. Note that, as profinite groups are compact, the open subgroups are exactly the closed subgroups of finite index, so that the discrete quotient group is always finite. Alternatively, one...
[Running gag] Slereah, I heard you like p adic numbers
 

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