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12:06 AM
Blah, I quit
@Studentmath What is the matter?
I can't get anything solid in here.. won't manage to prove it elegantly
And I've been working on this for over a week now
It's saddening.. but ach so
Gonna catch some sleep
@Pedro: Be careful. Essential sing must be isolated.
12:26 AM
@Ben Did you resolve your problem?
12:44 AM
@Alizter No, I'm stuck between choice B and C. The correct answer should have the same remainder as when 2014/4. but choice B is an improper fraction when I divide by 4.
see the comment I gave to the second anwser
@TedShifrin Why? It is a matter of convention is it not?
@Huy My challenge and progress is in my user profile description(It isn't going well because I am so busy)
@PedroTamaroff I've seen it both ways, but I'm fairly certain it's more common that essential singularities must be isolated.
What's the context of theq uestion?
@MikeMiller Right. Well, we say $c$ is a singularity of some $f$ if it is holomorphic in $D-c$.
So we always take isolated singulatrities, I guess.
12:48 AM
So what about say $1/\sin (1/z)$?
What do you call $0$?
Did you get my ping about Conway?
I will look at it.
Again, I have bored with Remmert's book.
I was bored at first then got into now I'm meh/10 again.
Conway is not very prosaic in his writing, but he suffices, and has exercises.
and as Daniel said Ahlfors is standard for areason
For what reason?
The material is cleanly presented, covering a wide range of important topics?
12:54 AM
Oh, OK. I will look at Alforhshrhs.
@MikeMiller Actually, the presentation is not as clean as I had thought. user1337 had several questions exposing not-so-good presentation by Ahlfors. I've never noticed that when I read the book, since I had read another book before and auto-corrected the glitches.
@DanielFischer I do that quite often, hehehe.
1:17 AM
@DanielFischer I've never read it start-to-finish, but the various sections I read for more background were always nice. Can you give an example?
I looked it up, and found a couple of them. They don't look too terrible...
1:42 AM
@Pedro: The point is that there's no Laurent expansion at non-isolated singularities. Other books: Lang (quite good), Narasimhan (more sophisticated), and Hille (2 vols. I actually don't know it personally). I'm not fond of Stein-Stakarchi. I like Ahlfors, dislike Conway.
1:54 AM
@PedroTamaroff essential singularity?
@TedShifrin: Hey, Ted!
2:05 AM
@robjohn How do you format a proof? Can you link me to what you consider to be really good proof formatting? I had done a proof here, but it ended up looking really ugly. math.stackexchange.com/questions/988631/…
@TedShifrin OK.
@Committingtoachallenge Why not look at some proofs on the site? I usually keep it to one equation per line, break thoughts up into paragraphs to prevent monolithic blocks of text, which are hard to read.
Hi @robjohn: Any further thoughts on my anti-stalker? :)
@TedShifrin I've never read Stein-Shakarchi, but as Stein was my advisor, I usually like his books.
@TedShifrin lemme look at something
I can't find any nice proofs of a similar style to the one I did, e.g. proving numerous statements to get the final result
2:14 AM
I taught out of it and was disappointed. Stein's brilliant, but I don't agree with some of his writing decisions (like log was way way late).
@TedShifrin I see a couple more downvotes since I last looked.
Yup, no epidemic, but IMHO targeted
@TedShifrin however, he is not downvoting enough to get community manager attention.
@TedShifrin I can have a CM look at things to see if the downvoter is just downvoting lot of people, or if he is simply targeting you in a careful manner.
I won't be surprised if it gets worse in 6 days when Rene is off suspension.
Thanks, @robjohn, you're very kind.
The range from algebraic geometry to elementary series and diff geo makes me suspicious ...
2:29 AM
Evenin' foiks
no comments/votes .. :( @robjohn is there anything wrong with my proof here ?
any one got a moment to help a noob with some maths
@DavidH morning sir :-)
@r9m It is likely too difficult for most to understand
Hi all
2:34 AM
@r9m I've only read the first part, but I would separate the sections and perhaps title each as to what each does (e.g. A Second Approach)
@r9m Separating sections with *** is useful
@r9m I very much agree with Robjohn, I didn't read much of it, and not enough to even realise it was two approaches(I would have read it if I had known that)
When people anticipate there is a great lot of reading to go, they are more likely to lose interest
@robjohn okay ,, thanks ! I edited it :-)
@Committingtoachallenge ah you are right !! that does happen .. thanks !
@r9m See what I did? It separates the section a bit more, visually.
2:40 AM
@robjohn That looks much better !! thanks !!!
Is this true? "If the domain of $f$ has more than one element, $f$ is bijective iff it has a single right inverse." where right inverse refers to $fg=1_s$(or in normal notation $g\circ f = 1_s$)
It sounds true to me, but the question ends with "Hint: To get counter examples, look at mappings from $\mathbb{N} \to \mathbb{N}$", what am I giving the counter example for?
3:00 AM
@Committingtoachallenge seems to me the relation $g\circ f={\rm id}$ means $g$ is a left inverse of $f$
you already know the claim you're supposed to be giving a counterexample to: it's the only claim being discussed!
Yo @anon
that's my greeting yo
Yo have a copyright? :)
this forum
anyone can help me to run fmincon in a loop?
3:18 AM
@robjohn :P
@r9m what?
@robjohn yo© copyright protected .. pun
3:58 AM
"Yo yo" is copyright protected :-)
Shameless plug for bounty i just added
Q: Fourier serie of $\sqrt{1 - k^2 \sin^2{t}}$

AfreekaI'm struggling with a Fourier serie. I need to find the Fourier series of the following function. That's the function under study: $f(t)=\left[\sqrt{1-k^2\sin^2t}\,\right]$. The function is even and $\pi$-periodic. The Fourier series should be in this form: $f(t)=\frac{a_0}2+\s...

4:22 AM
@anon What...? If I am to give a counterexample, that implies I am trying to disprove it? But apparently it is true. Also $(g\circ f)(x)=g(f(x))=xfg$ so $fg=1_s == g\circ f = 1_s$
writing functions on the right = heresy
@anon writing functions on the right =Cohn's Classic algebra
What is your favourite series for algebra @Anon?
I don't have a favorite algebra book.
4:26 AM
What is your speciality?
just look at my tags
I did, it says algebra
Where is the love for the texts?
Do you have any favourite textbooks at all @Anon?
not really
I think you're right, that f is bijective iff it has a single postcompositional inverse
4:29 AM
Yeah, I don't understand why it suggests a counter example for something
Perhaps to show that if it has one element it isn't true
it also seems true for precompositional inverses too
Yeah it appears so
Where did you learn all your algebra from @Anon? Only lectures?
stuff here and there, mostly internet
4:52 AM
hi, I'm trying to get a UDP receive set up in Simulink and I'm having trouble getting the data through correctly. Can anyone help?
Is this an ongoing Contest in M.SE ? :D COOL !!!!! @Anastasiya-Romanova .. the very first problem is an A-Bomb ! :-) and if the difficulty of the problems is going uphill from this point let God have mercy on our souls !! (our = {M.Se users}\{Integration Gurus}) ..
@Chris'ssis ^ have you seen this Contest ? :D looks exciting ..
5:17 AM
Hello, I'm new to chat rooms—are there timestamps? Can I tell how old each thing posted here is?
@columbus8myhw Yes, put your mouse over a post, and then go to the left most point of the post, click the down arrow
Google this:
(My friend gave me this one, which also works:)
sin(1/55555555555 degrees)
(Hint as to why it works: It says "degrees" there.)
There's some sort of conspiracy to use "Beautiful" in question titles tonight.
@Committing sorry couldn't see that .. I was away from chat window ..
@Committingtoachallenge not likely .. the way they frame sentences are radically different .. ! (referring to (removed) no. 1) :P but I'm starting to sense a common pattern in tunk-fey and cleo's responses !! :P
Who's Tunk-Fey? Does he give nondescriptive answers, like Cleo does?
5:24 AM
@columbus8myhw no .. his solutions are very detailed and elaborate !
No, he just gives the same answers with all the working
Including once he gave the same answer with a simple error near the start, and got the same answer ;)
He posted after Cleo on everything until someone pointed out that they thought he was Cleo
Hm... I think I remember seeing that guy around.
@KajHansen hhehhe :P lol
Cleo recently joined Integrals and Series. If Tunk joins, I'm subscribing to this "conspiracy theory."
Does Latex work on here? $$\psi\left(\frac ab\right)=\sum_{\substack{\rho^b=1\\\rho\ne1}}(\rho^a-1)\ln(1-\bar\rho)-\gamma$$
5:26 AM
It does with a little help of a bookmark
That's the one
Ah, got it!
(What I posted is—I think—equivalent to Gauss's digamma theorem. I discovered that formulation of it by my self, and I think that it's more elegant.)
@columbus8myhw Nice one !!
@r9m I think that you can prove that $\psi(1-s)-\psi(s)=\pi\cot(\pi s)$ using this... but I need to go to bed. Bye!
@columbus8myhw ic :) Nice !!
6:00 AM
@AnthonyCarapetis Why do you come to the chat if you never talk?
just to annoy you I guess
You broke your 103 days of no talking within 10 seconds of it being mentioned
Are you always here? Or have you just not joined for a long time?
The latter
6:06 AM
Unfortunate, that makes it far less exciting
6:36 AM
This chat is so quiet now days :'(
can any matlab people give me a hand with: stackoverflow.com/questions/26570439/…
2 hours later…
8:19 AM
If we define exponential function in Banach algebra $A$ as $e^a=\sum\limits_{n=0}^{+\infty} \frac{a^n}{n!}$ and if we define $\Delta:A \to A:x \mapsto ax-xa$ is there some nice closed form for $(e^{\Delta})(x)$? I suppose yes, because $\Delta$ looks like some commutator.
@r9m It's an easy question
8:38 AM
@r9m HINT: start by letting $$-\log(\sin(x))=y$$
Then split the integral you get into 2 integrals, the first one is very easy, and the second one is just a bit less easy. Hence, Q.E.D.
What is the first integral? Well, it's $$\int_0^{\infty} \log(y) \log(1-e^{-2y}) \ dy $$ that flows naturally (no effort needed).
The second one is $$\int_0^{\infty} \log(y^2+\pi^2) \log(1-e^{-2y}) \ dy$$.
Keep in mind that after some manipulation we get a form of the type $\frac{x}{a^2+x^2}$ in our integrand, but we know what kind of integral to use here, and furthermore this means that we're going to use double integrals to compute it.
@Chris'ssis Do you ever use phase plane analysis or laplace, fourier?
@Committingtoachallenge Yeah...
hmmm, actually I remember a very nice question where I used Laplace transform ...
Even phase plane analysis?
@Committingtoachallenge I'm not used to this description in English.
8:54 AM
Solving $\begin{pmatrix} y_1' \\ y_2' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$
For the eigen values of that $A$ (above is y'=Ay)
@Chris'ssis Just wondering if your hardcore maths uses any qualitative methods for finding information without solving the system
As W|A shows, proceeding as before, one can also get some series in terms of $\operatorname{Ci}$ function, but $$\LARGE \text{HAPPILY}$$ these series are known.
Well, wait a second ...
Do you see this question here?
Q: Computing in closed form $\sum_{n=1}^{\infty}\frac{\operatorname{Ci}\left(\frac{3}{4}\zeta(2) \space n\right)}{n^2}$

Chris's sisWhat tools would you recommend me for computing the series below? $$\sum_{n=1}^{\infty}\frac{\operatorname{\displaystyle Ci\left(\frac{3}{4}\zeta(2) \space n\right)}}{n^2}$$ I lack the starting ideas, I need some. Thanks.

in a similar approach we can do that series we need to do. These series were already studied and there is a paper available for it below
To conclude, all you need is some patience, write the proof nicely, arrange it and if you want, post it.
@Jasper I got another 5 votes again today :/
@KajHansen Exactly, straight four 'beautiful' question in geometry now.. :/
9:14 AM
@Sawarnik 5 downvotes?
@Committingtoachallenge ... (see below the 2 pictures)
(since you were talking about Laplace ...)
Hope you got them.
Looks very nice @Chris'ssis
@Committingtoachallenge Thank you. I worked on it for a while.
9:23 AM
So are Riemann zeta and Xi functions under complex analysis? I thought I remember reading you saying that you are using no complex analysis in your book
@Committingtoachallenge I mean (largely speaking) I don't use contour integration ... (things might change a bit during the time)
I'm preparing to watch the game between SIMONA HALEP - SERENA WILLIAMS :D (AGAIN)
It starts here in one hour and half (it's the final).
9:56 AM
do you talk about math today?
or about women again :P?
Math @teddy, but do you know non-linear system??
Q: Can someone explain linearisation on nonlinear systems to me?

CalculusI want to find all critical points of the following nonlinear system: $$\def\b{\begin{pmatrix}}\def\e{\end{pmatrix}}$$ $$\b y_1' \\ y_2'\e = \b 5y_2 -15 \\y_2^2 - y_1 ^2\e$$ Then use linearisation to find the type and stability of the critical points. So first of all, finding the critical po...

@Calculus. Ok. no, sorry. I know nothing about that
@Calculus are you from germany?
10:20 AM
@Committingtoachallenge No up :D, we and Jasper had got it yesterday as well :/
@Teddy Hi!
@Teddy Nono sorry not from germany
@Sawarnik Why :\ face then? 5 upvotes is a :)
A: Integral Contest

Chris's sisFirst, let $-\log(\sin(x))=y$ that yields $$\underbrace{\int_0^1 \log(y) \log(1-e^{-2 y}) \ dy}_{\displaystyle \pi^2 \log(A)-\frac{1}{12}\pi^2 \log(A)}-\frac{1}{2}\underbrace{\int_0^1 \log(\pi^2+y^2) \log(1-e^{-2 y}) \ dy}_{\displaystyle\sum_{k=1}^{\infty} \frac{\operatorname{Ci}(2k \pi)}{k^2} -...

@Committingtoachallenge because getting 5 upvotes in one moment for two days straight is freaky :/ ... but then :D :) :)
@Sawarnik Can you help me with nonlinear system problem?
10:27 AM
@Sawarnik Is it probably one person?
@Calculus Aw, I know absolutely nothing about it :D :P
@Committingtoachallenge Yes, I think.
@Sawarnik Okay, that is fine, is that your real name?
@Calculus Yeup :D
@Sawarnik I wish I had someone giving me $5$ upvotes a day xD
@Committingtoachallenge If you help my question, I might give you 5 :))
10:28 AM
@Committingtoachallenge Sure, you ll get today :D
@Sawarnik Very nice name :)
@Calculus Hmm, why? :D
@Sawarnik It sounds awesome
Maybe I am pronouncing it wrong though, Sa-wah-nick
@Calculus Oh, wow :D :)
I thought my name was the worst ever. No one can pronounce it correctly and often come up with something absurd :(
@Calculus Correct :) :) :)
What was it actually haha?
10:32 AM
Correct actually. Just accent differences,
Oh correct?? Awesome!
What? But there's an 'r' there!
Or do you guys just have British accents?
@Calculus What is your name then?
Try to type out the pronunciation with the r columb
10:32 AM
Yes columb
(Where "war" is pronounced like the word. Which is not pronounced "wuh", at least not in my accent.)
That reads the same to me unless I americanise it
Where are you from?
10:34 AM
OK. In New York, 'r' is a letter that is always pronounced.
I will try to find a word counterexample
Oh apparently that is actually fact nevermind :)
lol, I had some typos ... (now all seems fine)
who needs rhoticity
Wait, Anthony, I thought you were this really quiet guy?
Sorry .. got disconnected :(
@columbus8myhw Here too :)
@Calculus What is your real name then? :D
10:45 AM
If you define $a_n=\underbrace{\sqrt{6+\sqrt{6+\sqrt{6+\dotsb}}}}_{n\text{ square roots}}$, then:
Hold on, still typing.
$$\lim_{n\to\infty}6^n(3-a_n)$$exists, and is approx. equal to $3.3656575397$.
@sawarnik Do you know who the upvoter is?
@JasperLoy You got it too?
10:47 AM
:18328261 It seems that there are two upvoters for me, lol.
Well, ok, got to go :D
That is, $$\lim_{n\to\infty}6^n(3-\underbrace{\sqrt{6+\sqrt{6+\sqrt{6+\dotsb}}}}_{n\text{ square roots}})\approx3.3656575397$$ exists. I don't know how to find a closed form for this—can you guys help me?
@Sawarnik I have further investigated the matter. It appears that this upvoter has two different accounts, lol.
Note: This is similar to $\lim_{n\to\infty}4^n(2-\underbrace{\sqrt{2+\sqrt{2+\dotsb}}}_n)=\dfrac{\pi^2}4$‌​.
Hi @Chris'ssis are you working on your book now?
10:54 AM
@JasperLoy No, I just worked on this one ...
A: Integral Contest

Chris's sisFirst, let $-\log(\sin(\theta))=y$ that yields $$\underbrace{\int_0^{\infty} \log(y) \log(1-e^{-2 y}) \ dy}_{\displaystyle \pi^2 \log(A)-\frac{1}{12}\pi^2 \log(\pi)}-\frac{1}{2}\underbrace{\int_0^{\infty} \log(\pi^2+y^2) \log(1-e^{-2 y}) \ dy}_{\displaystyle\sum_{k=1}^{\infty} \frac{\operatornam...

Any of you have any ideas on how to approach my thing? I've posted this on M.SE, but all I've got are proofs of convergence.
Of course, my work, approach there is perfect ... (maybe I didn't folow the time rule - by mistake)
@JasperLoy Because you got two upvotes on one post at the same time?
@Moronplusplus You have added another plus?
@Committingtoachallenge Yes, and around the same time. Do you have anything to do with it?
@JasperLoy Of course not, I only have one account
11:03 AM
@Committingtoachallenge But were you one of the upvoters?
@JasperLoy Nope, I don't serial upvote
OK, I don't mind getting serial upvotes myself, even though I raised a meta post about it. That post was because I could not tolerate the silly comments that followed.
Oh I haven't seen it yet
@Committingtoachallenge yes, a more advanced version.
OH nvm I have
@Moronplusplus When will it end?
11:06 AM
I'll get enlightened one day.
Are you guys ignoring my problem, or is it just that you have no idea how to approach it?
@columbus8myhw no idea.
@columbus8myhw I don't immediately know how to do it sorry Columbus, and I am only chatting while doing some work so I can't spend to much thought on it
Ah, OK.
This is odd. Say I have $m$ segments of length $n$, $m,n\in \Bbb N$, $m,n \to \infty$. In each of them there is certainly one positive answer somwehere in the segment. So the probability of two consequent segments having the two answers at distance greater than $n$ between them is $1/2$.
At most. And the proability of two consequent segments having two answers at distance smaller than $n$ is at least $1/2$.
11:09 AM
The relevant post is here, by the way. In this chat, I gave $L^2/6$ (where $L$ is the constant discussed in the problem).
@Studentmath "Answers"?
But then, the probability of some $c<m/2$ coupled segments having the two answers at distance greater than $n$ is $1/2^c$, which goes to $o(1)$ when $c$ is large enough. And the same for lesser tahn $n$, which means... what the heck.
The segment of length $n$ is made of positive and negative answers, totalling to $n$
@Studentmath I feel like I'm missing context.
@Studentmath Is this related to some post?
Nah, some problem I am working on
Each segment's answers are distribued according to Binomial distribution
11:13 AM
@usukidoll What's your question??
The sum of a rational number and an irrational one is always irrational. How do we prove that?
I am trying to follow an example from the book
@UserX By contradiction.
@UserX First, prove that $\text{rational}\pm\text{rational}=\text{rational}$.
by using the definitions of the Mobius function... each number like $\mu(2) $ is something based off of the definition, in that case it's a (-1). I figured that maybe since we have a prime number, the end result will be (-1)
11:15 AM
@usukidoll $\mu(\text{prime})=-1$, I believe.
ah there... I just needed a bit of a clarification because the book just jumped into the example... the book mentions the definition and then goes straight to the example... so now I know that all primes yield -1, but what about $\mu(6)$?
"The relevant post is here, by the way. In this chat, I gave L2/6 (where L is the constant discussed in the problem)."
Can someone clarify if each of the $y_1'$ and $y_2'$ equal $f_1(y)$ and $f_2(y)$ here math.stackexchange.com/questions/991496/…
Q: $\lim_{n\to\infty}\sqrt{6}^{\ n}\underbrace{\sqrt{3-\sqrt{6+\sqrt{6+\dotsb+\sqrt{6}}}}}_{n\text{ square root signs}}$

columbus8myhwWe have the following representation of pi: $$\pi=\lim_{n\to\infty}2^n \underbrace{\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dotsb+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}}}}_{n\text{ square root signs}}$$ which can be proven using the identity $\sin\left(\dfrac\pi{2^{n+1}}\right)=\dfrac12\underbrac...

@usukidoll We know that $\mu(n)=(-1)^r$ if $n=p_1p_2\dotsb p_r$. Since a prime has $r=1$, we know that $\mu(\text{prime})=-1$. Since $6=2\cdot3$, we know that $\mu(6)=(-1)^2=1$.
A prime has $r=1$? so that's just $(-1)^1 \rightarrow (-1)$ so all primes will have the result of $(-1)$ but what about when the result is $0$, like what happens if $\mu(9)$?
11:21 AM
$\mu(p_1p_2\dots p_r)=(-1)^r$ only when all of the $p_i$ are distinct. As your post says, "$\mu(n) = 0 $ if $p^2$ divides $n$ for some prime $p$" (that is, if the $p_i$ are not all distinct).
Hey Carry!!!
hmm $ \mu(9)$ is 0 because we have $\mu(3^2)$ where n = 9 and p =3 $3^2$ divides $9$
@usukidoll Yeah. Or, another way of thinking about it: $9$'s prime factorization is $3\cdot3$. Since those aren't all different, we have $\mu(9)=0$.
a soo desu ka. Wakarimasu ^_^
11:29 AM
@JasperLoy Simona Halep is playing the final game at Singapore ... :-)
@usukidoll Notice, by the way, that if $a$ and $b$ have no factors in common, we have $\mu(a)\mu(b)=\mu(ab)$. (If they do have factors in common, $0=\mu(ab)$, because—let's call one of the primes they have in common "$q$"—$ab$ has two $q$'s in its prime factorization.)
@Chris'ssis What game?
@JasperLoy Tennis
like 2 and 3 had unique factors of their own... but if they have a common factor like let's say a = 5 and b = 25 then it's 0 because a = 5 and b = 5 x 5
@usukidoll I mean, if we try to find $\mu(6\cdot15)$, it's zero, because $6\cdot15=90=2\!\cdot\!3\!\cdot\!3\!\cdot\!5$. $6$ and $15$ have a factor of $3$ in common—do you see now how $\mu(ab)=0$ when they have a common factor?
11:34 AM
probably not...
I'm a bit confused unless... 2 x 3^2 x 5 and we have one 3^2 which is 0 under the mobius defintion sorry it's getting late hence the typos
@usukidoll Yes, it's a multiple of $3^2$, which is another way of thinking about it. Hm, let me back up a bit:
@usukidoll "$n$ is a multiple of $q^2$ for some $q$" and "$n$'s prime factorization has primes which are not all distinct" are equivalent. Do you see why?

(Cont'd) Taking $90$, for example: It satisfies the first one, because it's a multiple of $3^2$. It satisfies the second one, because its prime factorization is $2\!\cdot\!3\!\cdot\!3\!\cdot\!5$, and those primes aren't all different from each other.
hey, do anyone knows about permutations and combination?
@carry A bit?
combinations are fun
no... is hard o_o
11:40 AM
pdes are hell ... pardon the language but it's the truth
no I have a stupid pde book and I only done half the problems (probably because lectures of 4.3-4.4) aren't done but gawd I hate green's theorem version of Fourier
in how many ways can six men and two boys be arranged in a row if
there are at least three men separating the boys?
so what if I have to take integration by parts 4 times with old fashion fourier. I get somewhere rather than nowhere and crying for an hour
ewwww that problem is nastayyy
can anyone help me with that que? >_< LOL!
idk bro. looks like pedoism to me..
is under the topic of permutations and combination :D
11:44 AM
What a night
> in how many ways can six men and two boys be arranged in a row if
there are at least three men separating the boys?

Um, $10$? I think? I suppose you can pretend that the three men who are between the boys are one big "superman" or something? EDIT: No, that doesn't work.
gross new topic x.x
@columbus that sounds TOO weird
ans is 14400
Wait, are they all distinguishable?
I was operating under the assumption that they are indistinguishable.
11:45 AM
distinfuishable? what do u mean by that?
As in, say two of the men are Rob and John. If they're distinguishable, then we have "Rob, John" being a different combination than "John, Rob". Otherwise, they're the same.
i think they are indistinguishable, which i call it (combination)
So, if (for example) we have two boys and three men, and if they're indistinguishable, we just have 10 combinations: "bbMMM", "bMbMM", "bMMbM", "bMMMb", "MbbMM", "MbMbM", MbMMb", "MMbbM", "MMbMb", "MMMbb" (where M=Man and b=boy).
Or, easier—show that it's $\binom52$.
Now, for YOUR problem (still indistinguishable, so not quite your problem):
What does $x \not \in x$ mean?
We can take each of those combos, and put an extra "MMM" in between the boys. That gives us 10 combinations, where there are three men in between each of the two boys.
11:50 AM
What's an example of a set $B=\{x\in A : x \not \in x\}$
$x\not\in x$ means that $x$ is not an element of itself. That is, if $x$ is a set (that possibly contains other sets), this means that $x$ does not contain itself.
Oh, I thought non-capital letters indicate variables
@UserX IDK.
@Jasper I got two votes on the same question too, so yes :D
Continuing: So now we have $10$, where they're all indistinguishable. How to move on to the distinguishable case? Well, there are $2!$ orders for the boys to be in, so I'll multiply by $2!$. There are $6!$ orders for the men to be in, so I'll multiply by $6!$. So we end up with:
11:53 AM
This is on the axiom of specification chapter, where it uses $S(x)$ as a condition. Can we use conditions on sets?
@columbus8myhw Did you put it in ISC?
$10(2!)(6!)=14400$, which agrees with what you guys got.
@Sawarnik What's "it"—my square root thing? What's ISC?
@Sawarnik Someone is having fun doing these things.
@sawarnik Any idea who it is?

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