Mathematics - 2017-09-17 (page 3 of 6)

Dodsy

2:00 AM

I like Becks

Leaky Nun

@Dodsy please stop interpreting every sentence maliciously

Dodsy

I like german brew

@LeakyNun I'm literally not!

I'm just saying that using the word "crazy" to represent mental illness is offensive.

Twink

@LeakyNun finally someone who understands me

Jasper

Well, in this case, Twink and I know each other a little, so it's OK.

Don't worry about it.

Twink

:)

bwDraco

2:01 AM

Please cut it out with the flags.

10

Twink

so @Dodsy you better remove yourself

Dodsy

Who is being flagged @bwDraco

Jasper

Yeah, there isn't anything to flag here.

bwDraco

Someone flagged chat.stackexchange.com/transcript/message/40049832#40049832

Jasper

When I sense that we're going in the wrong direction, I will try to move the chat in another one.

Dodsy

2:02 AM

Sigh. Thanks Draco.

Jasper

By the way, the flagger could be anyone, might not even be in the room.

bwDraco

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

Dodsy

@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.

Jasper

@Dodsy See you in your dreams.

Dodsy

:o

that's scary .

Twink

2:05 AM

goodbye @Dodsy

Jasper

@bwDraco No need to get a mod, there's no problem here, I think.

bwDraco

I think it's under control now.

Jasper

But why did you remove that, lol.

Twink

@Jasper and I are good friends

Jasper

Never mind.

Twink

2:08 AM

why would you need to get a mod @bwDraco ?

you're just seeing problems where there's no problem

bwDraco

Problem solved, it's nothing.

Jasper

Yeah. By the way, someone has been flagging many things here.

We don't really know who.

Twink

it probably was Dodly

Leaky Nun

Infinities... what am I doing with them

Twink

but let's not talk about that anymore :)

finally peace in this chat

Leaky Nun

2:10 AM

let's do topology on $\Bbb Q$

bwDraco

Bye.

Twink

goodbye :)

Jasper

Thanks.

Twink

$\Bbb Q$ with which topology?

Jasper

Maybe to avoid all these flags we can say a bit less in future, things we don't need to (removed)

Twink

2:12 AM

I like topology

Jasper

And also, to the flagger, please don't flag such things in future.

2

bwDraco

*le sigh*

I've summoned the mods.

Jasper

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.

Twink

me neither, I don't understand

bwDraco

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.

Jasper

2:15 AM

LOL

Twink

hmm... so it was Dodsy

bwDraco

That was what happened last time I intervened. I had to escalate it to CMs.

Jasper

We don't know for sure.

Twink

that's why he left

he was sacred of the mods

bwDraco

So... Dodsy might need to be suspended.

Jasper

2:16 AM

I am sorry I hope I didn't cause anything.

Twink

don't worry Jasper it's not your fault

Balarka Sen

To be honest there has been a fair amount of trolling going on in the chat by multiple users

Jasper

All I did was to go to the calculus class and then come here and then this happens.

Twink

yeah, this chat is not the same as before

bwDraco

Problem is, we won't know for sure until an official intervenes. I'm trying to summon @Shog9 again.

Twink

2:20 AM

I thought everything was under control now.

Jasper

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.

Twink

I know, he was feeling offended for everything

Jasper

I am going to sleep. I hope that everyone will calm down and let the drama die down. Good night.

Twink

nite @Jasper

bwDraco

No matter what happens... the mystery flagger needs to be held accountable for the abuse.

Twink

3:06 AM

@TedShifrin done with Latex :)

ALannister

0

Q: Probability Space for rolling two dice - sigma-field

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), ...

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.

El'endia Starman

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?

ALannister

Can't say that I do.

El'endia Starman

3:22 AM

Welp. I literally just googled "matrix with blocks of 1s around the diagonal" and the second result was block diagonal matrices.

ALannister

All right then.

You know anything about probability?

El'endia Starman

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.

ALannister

Thanks @El'endiaStarman

El'endia Starman

Okay, my experience with probability does not include $\Omega$ or $\sigma$-fields. Mind explaining what those are and/or how they're useful?

ALannister

$\Omega$ 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.

El'endia Starman

3:32 AM

What does $A^c$ mean here?

ALannister

A complement.

So, everything in the sample space that is not contained in the set A

El'endia Starman

What's the use of a $\sigma$-field?

ALannister

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).

El'endia Starman

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?

ALannister

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.

El'endia Starman

3:41 AM

What's the use of having $\{2, 3, 5, 6\}$ as a separate thing from $\{2\}$?

ALannister

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

El'endia Starman

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 $\Omega$.

Semiclassical

My two reactions: 1) Oh, probability theory! 2) Oh, measure theory...

ALannister

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.

El'endia Starman

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.

ALannister

3:47 AM

@Semiclassical can you take a look at the question I posted?

El'endia Starman

@ALannister I'm the largest $\sigma$-field?! That's, like, a step up from a "yo mama is so fat" joke.

2

Semiclassical

I can look, but the probability that I'll have anything useful to say is almost 0.

Is the

ALannister

@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.

Semiclassical

Is the sigma-field supposed to represent the set of all possible 'events' with which you could associate a probability?

ALannister

Yes @Semiclassical

Semiclassical

3:48 AM

hmm.

ALannister

For two dice, I was thinking you would just have all possible pairs.

Semiclassical

That is a -big- set.

ALannister

Yeah, it's fricking huge.

Semiclassical

I suspect the right way to do this is to start with the simplest possible example, namely replacing the two dice with two coins

ALannister

And then, the probability of each individual roll is 1/36

I need to come up with the probability function as well.

Semiclassical

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.

ALannister

So, I'm guessing $P\{(i,j)\} = \frac{1}{36}$ for $i = 1,\cdots, 6$, $j = 1,\cdots , 6$.

Semiclassical

And the possible events should be combinations of these outcomes.

So that'd suggest 2^36 elements in the sigma-field.

ALannister

Correct.

Semiclassical

Neat.

ALannister

So, then...would we just pair up what we get for the one-die case, but not pairing up $\emptyset$ at all.

Semiclassical

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.

ALannister

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.

Semiclassical

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.

ALannister

Yeah, maybe I can relabel them and try something.

Kasmir Khaan

@Semiclassical hello semi

Semiclassical

You can even be pretty literal: (m,n)->6(m-1)+n

hi

ALannister

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

Semiclassical

I think then the probability you'd associate to the sigma field is just 1/36 times the size of the subset

ALannister

@El'endiaStarman I'm still laughing about that "yo mama so fat" joke.

Semiclassical

e.g. {(1,1)} would have a probability 1/36, whereas {(1,1),(1,2)} would have probability 2/36

and so forth

ALannister

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

Semiclassical

right

not sure about how the terminology works tbh

ALannister

4:02 AM

This is going to be more fun than a barrel full of monkeys.

I wish I was a computer.

Semiclassical

eh, just remember this: a computer is only fast to us because it has a high processing speed

ALannister

On a completely unrelated note, why is a barrel full of monkeys considered fun?

Semiclassical

from it's perspective, it's still having to do every single manipulation one at a time

ALannister

Seems to me they would probably die in there.

And then when you let them out, they'd scream and throw feces at you.

El'endia Starman

@ALannister :D

ALannister

4:03 AM

Not fun at all.

Semiclassical

zombie monkeys. disturbing

ALannister

Yes, blue-eyed whitewalker monkeys

El'endia Starman

Well, looks like you're getting some help from Semi. Good. Bedtime for me!

Kasmir Khaan

I need help understanding centrilizers and normalizers

ALannister

G'night @El'endiaStarman

El'endia Starman

4:04 AM

G'night!

Semiclassical

well, you'll have to get help from someone else. my knowledge of centralizers and normalizers is confined to what I read on Wikipedia

Kasmir Khaan

:D

ALannister

The centralizer is the set of elements that commute with every element of G.

Kasmir Khaan

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

ALannister

WAit a minute. .I think i see.

Semiclassical

4:06 AM

sigma-field seems basically a synonym for power set in this context @ALannister

Kasmir Khaan

do you have an easy example just to make this clear?

ALannister

@KasmirKhaan check out this bad boy: mathworld.wolfram.com/Centralizer.html

Kasmir Khaan

@ALannister am not a bad boy , am decent , and thanks

ALannister

@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".

Semiclassical

gotcha.

ALannister

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.

Semiclassical

"Eats, shoots, and leaves" vs. "Eats shoots and leaves" is the classic version

ALannister

I prefer my shocking, cannibalistic version, @Semiclassical

Semiclassical

@ALannister In this case you want the largest sigma-field, though, so taking the power set would seem the right choice.

ALannister

Yarp @Semiclassical

Semiclassical

So that'd seem to resolve the point.

ALannister

4:10 AM

Yarp.

Seems so. Now I've got to go do it.

Semiclassical

Yarp.

ALannister

Hot Fuzz Hound :)

Good night, @Semiclassical.

Semiclassical

night

Leaky Nun

4:25 AM

meanwhile |x|-x is injective

Kasmir Khaan

@LeakyNun Hello :D

bwDraco

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)

Kasmir Khaan

5:23 AM

@bwDraco how to pin it?

bwDraco

Only a room owner or mod can do that. You can star it, though.

Kasmir Khaan

@bwDraco thats the best i could do =p room owner might see this now with bigger chance

Unknown123

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

Unknown123

7:19 AM

What is the difference between H(w), H(jw), and H(e^jw)?

SteamyRoot

8:06 AM

Oh wow, looks like I slept through a lot of drama

Kasmir Khaan

@SteamyRoot sup streamy

SteamyRoot

ohi

Kasmir Khaan

@SteamyRoot do you know about abstract algebra?

group actions to be specific

=p

SteamyRoot

I do, I guess

Kasmir Khaan

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

Leaky Nun

8:21 AM

group action is on set

regular associativity is abou group elements

Kasmir Khaan

how does it differ ( its operation i mean)

Leaky Nun

how does what differ?

Kasmir Khaan

Can you say more stuff , because I really dont know what to ask

Leaky Nun

then what can i answer

Kasmir Khaan

let me find few questions one second

g_1 (g_2 a ) = (g_1 g_2)a

why is this not called associativty ?

Leaky Nun

8:24 AM

what is it called?

Kasmir Khaan

1a =a , 1 here is the identity

action of a group

I dont know what they mean by that

Leaky Nun

I mean what the axiom is callrd

Kasmir Khaan

nothing they just wrote that

Leaky Nun

call it associativity then

Kasmir Khaan

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

Leaky Nun

8:27 AM

a is an element of a set

not a group

Kasmir Khaan

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?

Leaky Nun

group action is a function that takes element of a group and of a set and gives an element of said set

ok

Kasmir Khaan

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

Leaky Nun

@KasmirKhaan hopefully.

Kasmir Khaan

@LeakyNun scribd.com/doc/310760343/Abstract-algebra-Dummit-and-Foote-pdf

@LeakyNun Ill try to do chapter 1 and 2 =P

as a second read, hope things goes smothly :D

Leaky Nun

8:31 AM

ok

Perturbative

Hey everyone!

Leaky Nun

hi

Kasmir Khaan

anyways wish you have a continued good day , and ill be back in like 4-6 hours

Perturbative

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 $\text{cl}_X(e) = \bar{e}$

Leaky Nun

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?

Alessandro Codenotti

9:01 AM

What's $\epsilon$? @perturbative

Perturbative

Heya! @AlessandroCodenotti

$\epsilon$ is a cell decomposition of $X$

Alessandro Codenotti

what's |i_n|? @Leaky

Balarka Sen

@Perturbative Closure of a cell is not in general homeomorphic to a closed disk.

But it is a union of cells.

Leaky Nun

@AlessandroCodenotti length of interval

Balarka Sen

(That's what one of the words in "CW" stands for; C = Closure-finite, closure of any cell is a finite union of cells)

Perturbative

9:04 AM

@BalarkaSen So the closure of a cell, doesn't actually equal a closed cell?

Balarka Sen

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.

Perturbative

Ohh, by closed cell I mean some topological space homeomorphic to $\mathbb{B}^n$ (the closed unit ball in $\mathbb{R}^n$)

Balarka Sen

That's what I said. The answer to your question is no.

Perturbative

Okay thanks

Alessandro Codenotti

@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 $e_1>0$

Leaky Nun

9:19 AM

but it's supposed to be yes... @AlessandroCodenotti

Alessandro Codenotti

why?

Leaky Nun

I don't think what you said is stronger

the borel conjecture is supposed to be undecidable

independent*

Alessandro Codenotti

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

Leaky Nun

but it must be yes

or else the conjecture would be provably false

John Do

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?

Leaky Nun

9:23 AM

with the Cantor set being the counter example @AlessandroCodenotti

Alessandro Codenotti

I don't know what you expect to obtain by asking conjectures you already know the answer to without stating that by the way

Leaky Nun

I don't know the sequence that makes Cantor fail

Alessandro Codenotti

I know, I can read wikipedia, I was mistaken since measure zero doesn't imply that property as I wrote above

Leaky Nun

@JohnDo 0 :)

John Do

@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...

Leaky Nun

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...

Leaky Nun

9:38 AM

5

A: The Cantor set is not strong measure zero

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 ...

found it lol

privetDruzia

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?

$distance = |\hat{w^T}(x_n-x) |$

BAYMAX

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?

Jasper

@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.

Evinda

12:08 PM

Hello

I have a question

Leaky Nun

@Evinda just ask

Evinda

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 $a^d \equiv 1 \pmod{N}$ ?

Leaky Nun

@Evinda what is $N$?

Evinda

@LeakyNun We pick a prime N in order to show that then $(\mathbb{Z}/N\mathbb{Z})^{\ast}$ is cyclic... @LeakyNun

Leaky Nun

Then why don't you include that?

@Evinda yes, by Lagrange's theorem.

Evinda

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

Leaky Nun

@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.

Evinda

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

Leaky Nun

Just use $N=37$ as an example to see.

@Evinda when approaching a problem, always use examples.

Evinda

@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?

Leaky Nun

@Evinda to deduce that 37 is a prime???

no, we're not deducing the primality of 37 at all in this exercise.

Evinda

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?

Leaky Nun

@Evinda of course not

$(\Bbb Z/10\Bbb Z)^*$ is also cyclic with generator $3$

but prime does imply cyclic.

Evinda

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

Leaky Nun

that $2$ generates said group.

remember to use $=$ here not $\equiv$ :)

Evinda

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

Leaky Nun

@Evinda that's what I'm asking you to show us by exploring $N=37$.

@Evinda done?

Evinda

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?

Leaky Nun

$k$ might not be even

$(2^d)^{\frac k2}$ may not be defined

but you got the general idea

Evinda

Oh yes, that's right

Leaky Nun

basically you expand your counterexample to one of the checked numbers

so it suffices to check those numbers alone

Evinda

What do you mean?

Do we pick all the possible values of d? @LeakyNun

Leaky Nun

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 $p \mid N-1$.

Sha Vuklia

12:47 PM

@Semi! Do you have time?

Leaky Nun

@ShaVuklia could I help you?

Sha Vuklia

if you're familiar with 1-dimensional waves (physics), then yes

Leaky Nun

Let's see.

Evinda

@LeakyNun Which cases does $(a^d)^{(N-1)/(dp)}=1$ contradicts?

Leaky Nun

@Evinda the case where $a^{(N-1)/p} \ne 1$

Sha Vuklia

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…

Leaky Nun

hey, you said 1-dimensional! :c

Sha Vuklia

that.... is 1-dimensional

in space

Leaky Nun

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 $\cos(kz-\omega t)$.

Evinda

@LeakyNun Ah I got it now.... Thanks a lot!!! :)

Leaky Nun

@Evinda no problem

Sha Vuklia

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?

Jasper

@WDNWBM It appears that you are mathematicsaminphysics, right?

user308168

@Jasper Yes.

Jasper

@WDNWBM Thank you for identifying yourself.

But I don't want any more drama. I am very tired.

user308168

@Jasper You are welcome.

Jasper

@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.

Jasper

@WDNWBM Which idea? The MSE University idea is doomed, finished.

user308168

@Jasper I do not think so.

Jasper

@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.

Jasper

@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.

ACuriousMind

@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.

ACuriousMind

@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

El'endia Starman

@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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$Mathematics$ Mathematics