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2:00 AM
I like Becks
 
@Dodsy please stop interpreting every sentence maliciously
 
I like german brew
@LeakyNun I'm literally not!
I'm just saying that using the word "crazy" to represent mental illness is offensive.
 
@LeakyNun finally someone who understands me
 
Well, in this case, Twink and I know each other a little, so it's OK.
Don't worry about it.
 
:)
 
2:01 AM
Please cut it out with the flags.
10
 
so @Dodsy you better remove yourself
 
Who is being flagged @bwDraco
 
Yeah, there isn't anything to flag here.
 
When I sense that we're going in the wrong direction, I will try to move the chat in another one.
 
2:02 AM
Sigh. Thanks Draco.
 
By the way, the flagger could be anyone, might not even be in the room.
 
Look, if you insist on flagging non-offensive content simply because you don't like it or you just want attention, then you need to go elsewhere.
3
 
@Jasper enjoy the rest of your night dude. I'm off to do some readings as I do not want to be banned tonight.
talk to you all later.
 
@Dodsy See you in your dreams.
 
:o
that's scary .
 
2:05 AM
goodbye @Dodsy
 
@bwDraco No need to get a mod, there's no problem here, I think.
 
I think it's under control now.
 
But why did you remove that, lol.
 
@Jasper and I are good friends
 
Never mind.
 
2:08 AM
why would you need to get a mod @bwDraco ?
you're just seeing problems where there's no problem
 
Problem solved, it's nothing.
 
Yeah. By the way, someone has been flagging many things here.
We don't really know who.
 
it probably was Dodly
 
Infinities... what am I doing with them
 
but let's not talk about that anymore :)
finally peace in this chat
 
2:10 AM
let's do topology on $\Bbb Q$
 
Bye.
 
goodbye :)
 
Thanks.
 
$\Bbb Q$ with which topology?
 
Maybe to avoid all these flags we can say a bit less in future, things we don't need to (removed)
 
2:12 AM
I like topology
 
And also, to the flagger, please don't flag such things in future.
2
 
*le sigh*
I've summoned the mods.
 
Sure, but I really don't know who flagged what, lol.
And I also don't know what the problem is.
There doesn't seem to be any big problem anywhere.
 
me neither, I don't understand
 
Jun 5 at 5:09, by Shog9
@Avery @bwDraco well, I looked. Turns out @TheGreatDuck and @Dodsy have been trading flags here for a couple of days. Latest volley was courtesy of Dodsy, but TheGreatDuck still leads by a fair margin for raw numbers pointless flags.
 
2:15 AM
LOL
 
hmm... so it was Dodsy
 
That was what happened last time I intervened. I had to escalate it to CMs.
 
We don't know for sure.
 
that's why he left
he was sacred of the mods
 
So... Dodsy might need to be suspended.
 
2:16 AM
I am sorry I hope I didn't cause anything.
 
don't worry Jasper it's not your fault
 
To be honest there has been a fair amount of trolling going on in the chat by multiple users
 
All I did was to go to the calculus class and then come here and then this happens.
 
yeah, this chat is not the same as before
 
Problem is, we won't know for sure until an official intervenes. I'm trying to summon @Shog9 again.
 
2:20 AM
I thought everything was under control now.
 
Yeah I know Shog he's the man.
I think what happened is very simple.
Twink and Dodsy just had a small misunderstanding, that's all.
But there wasn't really anything vicious going on.
As for who flagged what I don't know.
 
I know, he was feeling offended for everything
 
I am going to sleep. I hope that everyone will calm down and let the drama die down. Good night.
 
nite @Jasper
 
No matter what happens... the mystery flagger needs to be held accountable for the abuse.
 
3:06 AM
@TedShifrin done with Latex :)
 
0
Q: Probability Space for rolling two dice - sigma-field

ALannisterI need to come up with the probability space for rolling two dice, which includes $\Omega$ (the sample space), a $\sigma$-field and a probability function. Thus far, I have that $\Omega = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), ...

Anybody up for some probability?
Spock just got hit in the face with some spores.
My flag boy and your flag boy were sitting by the fire. My flag boy said to your flag boy, I'm gonna set your flag on fire.
 
Anyone know what it's called when a symmetric, square matrix composed solely of 1s and 0s is arranged such that all of the 1s are in blocks around the diagonal?
 
Can't say that I do.
 
3:22 AM
Welp. I literally just googled "matrix with blocks of 1s around the diagonal" and the second result was block diagonal matrices.
 
All right then.
You know anything about probability?
 
What I'm really interested in is approximating these. As in, given the adjacency matrix of a graph, shuffle the rows and columns until you get something that's as close to a block diagonal matrix as possible (either entirely or within some "distance" epsilon).
@ALannister Only basic stuff. I can take a look at the question you posted.
 
Thanks @El'endiaStarman
 
Okay, my experience with probability does not include $\Omega$ or $\sigma$-fields. Mind explaining what those are and/or how they're useful?
 
$\Omega$ is the sample space.
A $\sigma$-field $\mathcal{F}$ is a class of observable subsets of the sample space such that the following three properties is satisfied: 1. $\Omega \in \mathcal{F}$, 2. $\forall$ sets $A \in \mathcal{F}$, $A^{c} \in \mathcal{F}$, and 3. $\mathcal{F}$ is closed under countable unions.
 
3:32 AM
What does $A^c$ mean here?
 
A complement.
So, everything in the sample space that is not contained in the set A
 
What's the use of a $\sigma$-field?
 
In this case, it is the set of all possible outcomes.
Every probability space has one
I wish there were more people online at this time of night who were probability specialists. I posted my question 22 minutes ago and only 5 people have even looked at it (2 of those have been me).
 
Yeah, I'm certainly not a probability specialist. However, I have been told in the past that I tend to ask really good questions, so maybe I can help a little in that way. (And besides, I like learning new stuff.)
@ALannister What's a "possible outcome"? The $\sigma$-field you have there for $\Omega = \{1, 2, 3, 4, 5, 6\}$ looks like every possible subset of $\Omega$. What does $\{2, 3, 5, 6\}$ mean as a "possible outcome" for rolling one die?
 
For your second question, it means rolling a 2, a 3, a 5, or a 6.
Omega is rolling a 1, a 2, a 3, a 4, a 5, or a 6.
 
3:41 AM
What's the use of having $\{2, 3, 5, 6\}$ as a separate thing from $\{2\}$?
 
Also, for example, $(\{1\}, \{1,2\})$ corresponds to rolling a 1 on the first die and rolling either a 1 or 2 on the second die.
Because $\{2,3,5,6\}$ also includes the possibility that it could come up 3, 5, or 6. $\{2\}$ includes only the possibility of rollign a 2
 
I think I'm starting to get it.
In any case, my gut is that the $\sigma$-field for the $\Omega$ of rolling two dice is probably (heh) the set of all subsets of $\Omega$.
 
My two reactions: 1) Oh, probability theory! 2) Oh, measure theory...
 
LOL @Semiclassical
The largest $\sigma$-field is @El'endiaStarman and the largest one is the one I'm looking for to characterize the probability space.
 
Hahaha. On a different note, @Semiclassical, thanks for taking the time to explain Lagrangian mechanics a bit to me. That definitely helped, and sorry I didn't get back to you right away.
 
3:47 AM
@Semiclassical can you take a look at the question I posted?
 
@ALannister I'm the largest $\sigma$-field?! That's, like, a step up from a "yo mama is so fat" joke.
2
 
I can look, but the probability that I'll have anything useful to say is almost 0.
Is the
 
@Semiclassical I doubt that's true. I just want to know if I had the right idea for the sigma field for the two dice case.
 
Is the sigma-field supposed to represent the set of all possible 'events' with which you could associate a probability?
 
Yes @Semiclassical
 
3:48 AM
hmm.
 
For two dice, I was thinking you would just have all possible pairs.
 
That is a -big- set.
 
Yeah, it's fricking huge.
 
I suspect the right way to do this is to start with the simplest possible example, namely replacing the two dice with two coins
 
And then, the probability of each individual roll is 1/36
I need to come up with the probability function as well.
 
3:51 AM
eesh
hmm, actually. I wonder if one can think of it like this.
There are, as you say, 36 possible outcomes of the dice roll.
 
So, I'm guessing $P\{(i,j)\} = \frac{1}{36}$ for $i = 1,\cdots, 6$, $j = 1,\cdots , 6$.
 
And the possible events should be combinations of these outcomes.
So that'd suggest 2^36 elements in the sigma-field.
 
Correct.
 
Neat.
 
So, then...would we just pair up what we get for the one-die case, but not pairing up $\emptyset$ at all.
 
3:54 AM
The way I came to that is basically this: Rolling a pair of d6 is equivalent to rolling a single d36. So just do the same construction as you did for the single d6 but now with d36 instead.
 
Except to do that, I applied the binomial theorem.
And I knew what taking 6 possibilities, say 2 at a time looked like
I can't figure out how to visualize it in this case.
 
I'm not sure I know either. But the way I'm thinking is that there's really no difference between 36 events labelled as (1,1), (1,2),...(1,6), (2,1),...,(6,6) and 36 events labelled 1,2,3,...36.
 
Yeah, maybe I can relabel them and try something.
 
@Semiclassical hello semi
 
You can even be pretty literal: (m,n)->6(m-1)+n
hi
 
3:59 AM
Some guy commented on my post and said the same thing.
Cool.
I'm going to go try that. This could take me a while.
But, thanks @Semiclassical
 
I think then the probability you'd associate to the sigma field is just 1/36 times the size of the subset
 
@El'endiaStarman I'm still laughing about that "yo mama so fat" joke.
 
e.g. {(1,1)} would have a probability 1/36, whereas {(1,1),(1,2)} would have probability 2/36
and so forth
 
Yeah, those are both events, though. I think the probability function just assigns probabilities to individual rolls
Which is where the 2/36 comes from
 
right
not sure about how the terminology works tbh
 
4:02 AM
This is going to be more fun than a barrel full of monkeys.
I wish I was a computer.
 
eh, just remember this: a computer is only fast to us because it has a high processing speed
 
On a completely unrelated note, why is a barrel full of monkeys considered fun?
 
from it's perspective, it's still having to do every single manipulation one at a time
 
Seems to me they would probably die in there.
And then when you let them out, they'd scream and throw feces at you.
 
@ALannister :D
 
4:03 AM
Not fun at all.
 
zombie monkeys. disturbing
 
Yes, blue-eyed whitewalker monkeys
 
Well, looks like you're getting some help from Semi. Good. Bedtime for me!
 
I need help understanding centrilizers and normalizers
 
G'night @El'endiaStarman
 
4:04 AM
G'night!
 
well, you'll have to get help from someone else. my knowledge of centralizers and normalizers is confined to what I read on Wikipedia
 
The centralizer is the set of elements that commute with every element of G.
 
the problem is i dont know if we fix some element of g
or take all elements of g
C(A) = g in G such that gag^-1 = a for all a in A
 
WAit a minute. .I think i see.
 
4:06 AM
sigma-field seems basically a synonym for power set in this context @ALannister
 
do you have an easy example just to make this clear?
 
@KasmirKhaan check out this bad boy: mathworld.wolfram.com/Centralizer.html
 
@ALannister am not a bad boy , am decent , and thanks
 
@Semiclassical really, a sigma-field is a sigma-algebra, so any subset of the power set that satisfies the conditions of a sigma-algebra would also work. But for a complete characterization of the probability space, we need the power set.
@Kasmir, no i was saying the thing I was direcxting you to was a bad boy not you.
If I was calling you "bad boy", I would have put a comma after the word "this".
 
gotcha.
 
4:08 AM
Just like "Let's eat Grandma" is not the same as "Let's eat, Grandma".
In the first case, you're probably a member of the Donner Party.
 
"Eats, shoots, and leaves" vs. "Eats shoots and leaves" is the classic version
 
I prefer my shocking, cannibalistic version, @Semiclassical
 
@ALannister In this case you want the largest sigma-field, though, so taking the power set would seem the right choice.
 
Yarp @Semiclassical
 
So that'd seem to resolve the point.
 
4:10 AM
Yarp.
Seems so. Now I've got to go do it.
 
Yarp.
 
Hot Fuzz Hound :)
Good night, @Semiclassical.
 
night
 
4:25 AM
meanwhile |x|-x is injective
 
@LeakyNun Hello :D
 
4:58 AM
PSA: Flags are not "super-downvotes". Don't flag stuff simply because you disagree with it. Abusing flags in an attempt to remove content that isn't spam or offensive can and will get you suspended.
6
(somebody please pin this)
 
5:23 AM
@bwDraco how to pin it?
 
Only a room owner or mod can do that. You can star it, though.
 
@bwDraco thats the best i could do =p room owner might see this now with bigger chance
 
6:12 AM
How do I find what kind of filter from discrete impulse response? And how do I find H(e^jw)? I'm stuck, I can't find any appropriate answer in google
 
 
1 hour later…
7:19 AM
What is the difference between H(w), H(jw), and H(e^jw)?
 
8:06 AM
Oh wow, looks like I slept through a lot of drama
 
@SteamyRoot sup streamy
 
ohi
 
@SteamyRoot do you know about abstract algebra?
group actions to be specific
=p
 
I do, I guess
 
what does that mean?
I read the book did not get it
I mean what is the diffence in the associative law
or the multiplication for that matter
@LeakyNun Sup leaky =p
 
8:21 AM
group action is on set
regular associativity is abou group elements
 
how does it differ ( its operation i mean)
 
how does what differ?
 
Can you say more stuff , because I really dont know what to ask
 
then what can i answer
 
let me find few questions one second
g_1 (g_2 a ) = (g_1 g_2)a
why is this not called associativty ?
 
8:24 AM
what is it called?
 
1a =a , 1 here is the identity
action of a group
I dont know what they mean by that
 
I mean what the axiom is callrd
 
nothing they just wrote that
 
call it associativity then
 
1) g_1 (g_2 a ) = (g_1 g_2)a
2) 1a=a
what is the difference betwen the product of 1)
LH and RH
 
8:27 AM
a is an element of a set
not a group
 
hmm okay
@LeakyNun am gonna go to library and study there >< is it ok to send you the exercices that I do from dummut and foote?
 
group action is a function that takes element of a group and of a set and gives an element of said set
ok
 
I will write them in a clean way and take a picture
then ill email you that , btw you need link to the book ?
its avalble as pdf
 
@KasmirKhaan hopefully.
 
@LeakyNun Ill try to do chapter 1 and 2 =P
as a second read, hope things goes smothly :D
 
8:31 AM
ok
 
Hey everyone!
 
hi
 
anyways wish you have a continued good day , and ill be back in like 4-6 hours
 
Dumb question, if $(X, \epsilon)$ is a CW Complex and $e \in \epsilon$ is an open cell, then is $\text{cl}_X(e) = \bar{e}$ (i.e. is the closure of $e$ in $X$ equal to the closed cell $\bar{e}$)? I think so, because $e$ is homeomorphic to $\mathbb{B}^n \subseteq \mathbb{R}^n$, and $\text{cl}_{\mathbb{R}^n}(\mathbb{B}^n)$ is $\bar{\mathbb{B}^n}$. Since homeomorphisms preserve closures $\text{cl}_X(e) = \bar{e}$
 
8:49 AM
is there a sequence of positive numbers e_n such that there are no sequence of real intervals i_n that covers the cantor set with |i_n| <= e_n for all n?
 
9:01 AM
What's $\epsilon$? @perturbative
 
Heya! @AlessandroCodenotti
$\epsilon$ is a cell decomposition of $X$
 
what's |i_n|? @Leaky
 
@Perturbative Closure of a cell is not in general homeomorphic to a closed disk.
But it is a union of cells.
 
@AlessandroCodenotti length of interval
 
(That's what one of the words in "CW" stands for; C = Closure-finite, closure of any cell is a finite union of cells)
 
9:04 AM
@BalarkaSen So the closure of a cell, doesn't actually equal a closed cell?
 
Well, what do you mean by a closed cell? A closed disk (of some dimension)? No.
Eg consider the standard CW structure on the torus with one 0-cell, two 1-cells and one 2-cell.
The closure of the 2-cell is all of the torus.
 
Ohh, by closed cell I mean some topological space homeomorphic to $\mathbb{B}^n$ (the closed unit ball in $\mathbb{R}^n$)
 
That's what I said. The answer to your question is no.
 
Okay thanks
 
@LeakyNun then the answer is no
You can even cover the Cantor set with intervals such that $\sum\limits_{n\in\Bbb N} |I_n|<e_1$ for every $e_1>0$
 
9:19 AM
but it's supposed to be yes... @AlessandroCodenotti
 
I don't think what you said is stronger
the borel conjecture is supposed to be undecidable
independent*
 
Hm, apparently measure zero isn't enough to imply that property
the fact that it's undecidable doesn't mean that specific instances (such as the Cantor set) can't be decided
 
but it must be yes
or else the conjecture would be provably false
 
I think there's a mistake in my homework: I'm asked for p>=1 to find a in 0;pi included such that cos(2^n * a) is p-periodic
That's impossible right?
 
9:23 AM
with the Cantor set being the counter example @AlessandroCodenotti
 
I don't know what you expect to obtain by asking conjectures you already know the answer to without stating that by the way
 
I don't know the sequence that makes Cantor fail
 
I know, I can read wikipedia, I was mistaken since measure zero doesn't imply that property as I wrote above
 
@JohnDo 0 :)
 
@LeakyNun exactly, but it makes it p-periodic for p=1... The question seems ridiculously easy if you take it as 1-periodic implies p-periodic
so I guess p must be the smallest period
But that seems impossible...
 
9:25 AM
@AlessandroCodenotti I thought that is the definition of measure zero
that the sum can be arbitrarily small
but the sum has nothing to do with the behaviour of the sequence
so re my question: there must be such a sequence, but what is it...
 
9:38 AM
5
A: The Cantor set is not strong measure zero

palladiumtelemannFor a more direct approach: Suppose that $I_1 , I_2 , \ldots$ are open intervals in $\mathbb{R}$ such that $I_n$ has length $3^{-n}$. Note that $I_1$ must be disjoint from either $[ 0 , \frac 1 3 ]$ or $[ \frac 2 3 , 1 ]$ (or both, I guess, if $I_1$ was chosen particularly badly). Let us label ...

found it lol
 
9:49 AM
Hi
The goal of the formula

$distance = \abs{w^T(x_n-x)} $
is to calculate the distance from x_n to the blue plane
Yet it is unclear to me why those people are multiplying with $\hat{w^T}$ rather than using the Pythagorean theorem
Could someone explain?
$distance = |\hat{w^T}(x_n-x) |$
 
10:10 AM
Hi
Any online Crank-Nicolson solver online?
of heat equation
 
user308168
10:59 AM
Can you please come to Math Mods' Office and help me?
 
@bwDraco Well, actually, we don't know for sure now who flagged what and why, so let's just wait and see what Shogun says, since we already called him.
No, I am not going to those other chats now. There's enough drama here already, don't want any more.
 
12:08 PM
Hello
I have a question
 
@Evinda just ask
 
How do we show that a group is cyclic? We guess a generator $\alpha$. If we are able to show that $\alpha^n \neq 1$ for any $n<N-1$ we are done. Note that $\alpha^{N-1}=1$ anyway. Therefore, we just need to check that $\alpha^{\frac{N-1}{p_i}} \neq 1$ for every prime divisor $p_i$ of $N-1$.
I have a question about the above
Why having checked that $\alpha^{\frac{N-1}{p_i}} \neq 1$ for every prime divisor $p_i$ of $N-1$ do we know that there is no $d \in \{1, \dots, N-2 \}$ such that $a^d \equiv 1 \pmod{N}$ ?
 
@Evinda what is $N$?
 
@LeakyNun We pick a prime N in order to show that then $(\mathbb{Z}/N\mathbb{Z})^{\ast}$ is cyclic... @LeakyNun
 
Then why don't you include that?
@Evinda yes, by Lagrange's theorem.
 
12:12 PM
Sorry
From Lagrange's theorem, we have that if $a^d \equiv 1 \pmod{N}$ then $d$ divides $N-1$, or not? @LeakyNun
 
@Evinda yes
although phrasing it in terms of number theory isn't the right way
$a^d \equiv 1 \pmod N$ is number theory; $a^d = 1$ is group theory.
 
Yes, but then why don't we check if $\alpha^{p_i} \not\equiv 1 \pmod{N}$ but if $\alpha^{\frac{N-1}{p_i}} \not\equiv 1 \pmod{N}$ for every prime divisor $p_i$ of $N-1$ ? @LeakyNun
 
Just use $N=37$ as an example to see.
@Evinda when approaching a problem, always use examples.
 
@LeakyNun In this case, we would check the values $a^{2 \cdot 3^2}, a^{2^2 \cdot 3}$.

For example $2^{2 \cdot 3^2}\equiv -1$ and $2^{2^2 \cdot 3} \equiv 26$.

But why does it suffice to check if these 2 values are not equivalent to 1 modulo 37 in order to deduce that 37 is a prime?
 
@Evinda to deduce that 37 is a prime???
no, we're not deducing the primality of 37 at all in this exercise.
 
12:24 PM
@LeakyNun To deduce that $(\mathbb{Z}/37 \mathbb{Z})^{\ast}$ is a cyclic group. And this is equivalent to the fact that 37 is a prime, Isn't it?
 
@Evinda of course not
$(\Bbb Z/10\Bbb Z)^*$ is also cyclic with generator $3$
but prime does imply cyclic.
 
A ok. I see... But then how do we come like that to the conclusion that $(\mathbb{Z}/37 \mathbb{Z})^{\ast}$ is a cyclic group? @LeakyNun
 
that $2$ generates said group.
remember to use $=$ here not $\equiv$ :)
 
So in order to show that $(\mathbb{Z}/N\mathbb{Z})^{\ast}$ is a cyclic group we want to find an $\alpha$ such that $\alpha^{\frac{N-1}{p}} \neq 1$ for every prime divisor $p$ of $N-1$ ? Why does it suffice to check these inequalities? @LeakyNun
 
@Evinda that's what I'm asking you to show us by exploring $N=37$.
@Evinda done?
 
12:38 PM
@LeakyNun That's what I thought:

Suppose that there is some $d \neq 2 \cdot 3^2, 2^2 \cdot 3$ such that $d<2^2 \cdot 3^2$ with $2^d= 1$.

Then we have that $d \mid 2^2 \cdot 3^2 \Rightarrow 2^2 \cdot 3^2=kd$, for some $k \in \mathbb{N}$. So, $2 \cdot 3^2=\frac{kd}{2}$.

So we have $2^{2 \cdot 3^2}=-1 \Rightarrow 2^{\frac{kd}{2}}=-1 \Rightarrow (2^d)^{\frac{k}{2}}=-1 \Rightarrow 1=-1 $, contradiction.
Is this right?
 
$k$ might not be even
$(2^d)^{\frac k2}$ may not be defined
but you got the general idea
 
Oh yes, that's right
 
basically you expand your counterexample to one of the checked numbers
so it suffices to check those numbers alone
 
What do you mean?
Do we pick all the possible values of d? @LeakyNun
 
let $a^d=1$ with $d \ne N-1$. Then, $\dfrac{N-1}d$ is an integer greater than $1$, so there is a prime that divides it, say $p$. Then, $\dfrac{N-1}{dp}$ is an integer, and $(a^d)^{(N-1)/(dp)}=1$, contradicting the cases we have checked.
Therefore, it suffices to check $a^{(N-1)/p}$ for all $p \mid N-1$.
 
12:47 PM
@Semi! Do you have time?
 
@ShaVuklia could I help you?
 
if you're familiar with 1-dimensional waves (physics), then yes
 
Let's see.
 
@LeakyNun Which cases does $(a^d)^{(N-1)/(dp)}=1$ contradicts?
 
@Evinda the case where $a^{(N-1)/p} \ne 1$
 
12:49 PM
Well this is my question:
I don’t understand one thing about the superposition of waves. My book says the following: Say we have $f_1(t)=A_1\cos\omega_1t+A_2\cos\omega_2t$ and $f_2=0$. Then the infinite wave that travels in the positive $z$-direction is given by: $y(z,t)=A_1\cos(\omega_1t-k_1z)+A_2\cos(\omega_2t-k_2z)$. In the same way, when $f_1(t)=f_2(t)=A\cos\omega t$, then we get $y(z,t)=A\cos(\omega t-kz)+A\cos(\omega t+kz)$.

I can see the pattern here; so we write $\pm k_iz$ in the argument with $\omega_it$, and apparently we then take the sum, with equal amplitude, that travel in the opposite direction. My q
 
hey, you said 1-dimensional! :c
 
that.... is 1-dimensional
in space
 
lol
@ShaVuklia I like to think of it as $\cos(kz-\omega t)$ ($\cos$ is even), in which case the explanation is trivial: a wave by itself is $\cos(kz)$, and since it travels in the positive z-direction, the input of $\cos$ should be decreasing (so that it shifts to the right), which gives us $\cos(kz-\omega t)$.
 
@LeakyNun Ah I got it now.... Thanks a lot!!! :)
 
@Evinda no problem
 
1:01 PM
@LeakyNun I don't know what a "wave by itself" would be. I'm guessing you mean the shape of the wave. In either case, that doesn't really help, because you start out with $z$, and then add the time-dependence, while they do the opposite. Anyhow, I'm still confused. I asked a classmate, but thanks for you help.
 
user308168
1:11 PM
2 hours ago, by WDNWBM
Can you please come to Math Mods' Office and help me?
 
@WDNWBM It appears that you are mathematicsaminphysics, right?
 
user308168
@Jasper Yes.
 
@WDNWBM Thank you for identifying yourself.
But I don't want any more drama. I am very tired.
 
user308168
@Jasper You are welcome.
 
@WDNWBM Although I did not agree with your idea of the MSE University, Heather has a much scaled down version which you can see on meta, and now SBA is holding a calculus class there as the first such class in the calculus and analysis room, just to share with you.
As to whether this project will succeed or not, only time will tell.
 
user308168
1:18 PM
@Jasper I think I should first know the reason of my suspension, then I will try to develop that idea.
 
@WDNWBM Which idea? The MSE University idea is doomed, finished.
 
user308168
@Jasper I do not think so.
 
@WDNWBM Why don't you go and take a look at how the calculus class is doing first? It is already very hard to conduct even one class.
Anyway, I am done with this.
 
user308168
@Jasper I have some good ideas, but I should first get rid of this suspension.
 
@WDNWBM It's up to you what you want to do, but I think this idea is finished.
 
user308168
1:29 PM
@Jasper I will not give up easily.
 
@WDNWBM No one can "help" you with that. Suspensions are entirely at the discretion of the math.SE moderator team. Stop trying to drag other users into your problems.
 
user308168
@ACuriousMind I want to hear the main reason from them (mods), but I have not heard it yet.
 
@WDNWBM If you do not understand the reason that was given to you in the suspension message, the proper course of action is to either reply to that message requesting clarification or to use the "contact" form for SE if you think you have fallen victim to an error so that a Community Manager can look into it. Let me say though that complaints about moderator action received that way almost never have any merit.
The proper course of action is not through chat, in any case
 
@WDNWBM Yes, you've said that many times now. Bothering your fellow chat users, who can't know what you're asking for, violates SE's Be Nice policy. Please stop talking about your suspension in chat. As ACuriousMind suggested, there are other ways to contact the Math.SE and SE mods.
 
user308168
@ACuriousMind @El'endiaStarman Thanks. Ok. I stop it.
 

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