@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.
I 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), ...
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?
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.
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.
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?
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.
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.
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.
@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".
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.
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 $\text{cl}_X(e) = \bar{e}$
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?
For 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 ...
@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.
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 $a^d \equiv 1 \pmod{N}$ ?
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
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
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 $p \mid N-1$.
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…
@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 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.
@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 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.
I 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), ...
For 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 ...
