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04:00
I get the feeling it isn't.
Maybe it is.
Hmph.
if $k$ is a field, $K/k$ is algebraic, and $\overline{k}$ is an algebraic closure, then $K/k$ is separable $\Leftrightarrow$ $K\otimes_k\overline{k}$ is reduced.
lol
@AlexanderGruber what are the rules about comments
are there any?
@Mike there are some
in general abusive behavior isn't used
you're not supposed to leave off topic comments (though some wiggle room exists on this rule)
hmm
you're supposed to avoid useless comments like "+1" and take long conversations into chat instead of making deep comment threads
04:05
"Or, as a third option, you could do your own homework." is this abusive
borderline
depends on the user's history
he has posted three PSQS in the last hour
and nothing else
also depends on the commenter's history
abusive behavior is a judgement made over time
Oh god @Mike this is 100 times worse than my confusion before
the only thing that'll get you suspended in one comment is posting racist stuff, links to porn/death, super nasty comments, etc.
04:07
someone else posted a pleasant response that hopefully would have the same effect
@Anthony do you know anything about path connectedness?
We went over it.
That's what I'd look at here.
Woooo.
anyhow, if you think the commenter is over the line, flag it and we'll look into it
there's only three questions in his user history. i know that's not actionable, so i won't bother
04:09
@Mike it may be worth sending him a non-suspending message
Are the mods busy with April fools pranks @Alex?
we do that a lot for new users who might otherwise not know why their quesitons are being losed
@Mike So I think my misunderstanding is coming to haunt me again.
@Anthony Do you know the theorem that path connectedness implies connectedness?
Yeah.
But so like, it seems like it is disconnected, it's so hole-y!
04:11
Then prove that $\Bbb R^2 \setminus \Bbb Q^2$ is path-connected.
Darn it guys stop pinging me.
(joking)
Disconnected doesn't mean "has a lot of holes", though. I think we would agree that $\Bbb R^2 \setminus \Bbb Z$ is connected.
@AlexBecker never.
@skullpatrol not so far!
I get every @Alex ping.
04:12
@AlexBecker me too
But so is the path I'm making in R2? Or in S?
so does alex youcis
you know there's just a user called "Alex"?
@Anthony what's S, even?
Eer R^2/Q^2
04:12
path-connected means there is a path connecting any two points in your space
@AlexanderGruber There are several actually. But IIRC the system only cares about the first 4 letters.
a path is a map $[0,1] \rightarrow X$, where $X$ is your space
Sorry Alex's
@AlexBecker did you see any april fools pranks on here today?
you have to remember that a topological space is it's own thing - in general, there's no "ambient space" it lives in. this space doesn't live in $\Bbb R^2$, it's its own thing.
man i'm tired.
04:13
Then what is connectedness even good for?
@AlexanderGruber I saw one which a few people flagged as spam. The user deleted it, so I decided to leave it be.
Even though I disapprove of using this site as a platform for pranks.
@AlexBecker MSE pranks don't tend to work very well anyway.
Aghhhh Thanks for your help @Mike
@Anthony what do you mean what is connectedness even good for?
I don't know
04:15
@AlexanderGruber Yeah, not sure how I feel about the system-wide "unicoins" either.
I get the feeling that connectedness conveys- but something like [1,2] [3,4] is disconnected because it's the union of two disjoint open subsets of [1,2] [3,4]?
informally, disconnectedness is exactly what you think it is
But then like R^2/Q^2 has so many holes, and path connected means you can move from one part to the next? But not in R. But move in what sense?
okay now you're getting hurt by thinking informally
path connected means that there's a path in R^2/Q^2 between any two points
I guess I just don't know what a path is?
04:18
Then look at the definition in your notes.
Isn't it a continuous map from [0,1] ?
Yes.
But continuity is mandated by the metric on R^2 isn't it?
i mean, yes? R^2/Q^2 has the subspace topology coming from being a subspace of R^2
So how can there be a continuous map to it?
That seems so unlikely to me
04:21
that's...
first off, any constant map is continuous
here's an example of one that's not constant
$\phi(t) = (\pi, t)$
Oh shit
The space doesn't need to continuous for the map to be, right?
continuity is a property of maps, not spaces
there's no such thing as a continuous space
Ugh
Why can you have continuous maps to crappy spaces :/
man i'm so tired.
maybe i should go to starbucks.
This late?
04:31
it's 9pm man
Lol
Is that not late?
:(
no
I just realized R^2/Q^2 is just like a grid, ain't it.
but i'm also tired all the time
Noooooo. :(
04:32
yeah - a really, really gross grid
But a grid. I thought it was like the discrete metric.
Er.
Not a grid.
I was unhappy.
well Q^2 is like a grid
so this is like what happens when you delete a grid
Oh. So it's not a grid. There are gaps....
Here's the way I would think of it
First, think of what R^2\Z^2 looks like
It's just a lot got a lot of discrete holes everywhere, but it's still sensible
Yeah.
04:35
Well imagine if you scaled that copy of Z^2 down
Say, (Z/3)^2
It's the same thing, but now there's more holes, and they're more densely packed.
Mmhmm.
You can (informally!!) think of R^2\Q^2 as the "limiting value" of that process.
But so it's not a grid, it's dense holes.
Yeah.
But those "holes" are laid out nice and uniformly
You can dodge them pretty easily.
If you also take out the irrational pairs it's not path connected?
04:38
Hell no it's not.
Woooo.
Also, what's a good intuition for why the topologist's sine curve isn't path connected?
A path can't run down the whole length of the curve.
Which it would have to to reach the point at the "end"
A path can only run down part of it (think compactness: a map $[0,1] \rightarrow \Bbb R$ has bounded image)
But so the only trick going on is that infinite length has been scrunched up into finite space?
Pretty much, yep, and then a point added "at the end of it"
05:27
April fools' day is not the same without robjohn around :(
it's a shame he died in that car crash
05:54
@anon back in your original form, i see
06:36
Anon did the best April fool's prank when he created a duplicate robjohn :D
good times
we had almost the same rep then
Yep
Now they are trying to make us all into unicorn believers >8(
do unicoins stay?
06:59
Please help with this question:
Find a general term (as a function of the variable n) for the sequence {a1,a2,a3,a4,......}={8/3,64/9,512/27,4096/81,.......}.
My answer:
an = n^2
Yet apparently this is wrong.
@AÌ·nÌ·dÌ·yÌ· The numerators are powers of 8 and the denominators are powers of 3.
@AÌ·nÌ·dÌ·yÌ· consider the ratio $a_{n+1}/a_n$
Also, such questions are generally stupid, because the pattern can be anything, lol.
They teach this kind of stupid thing in schools, sadly.
Well, I can't think of a formula for an.
@JasperLoy I'd replace stupid with ambiguous
07:02
I do not know why.
@AÌ·nÌ·dÌ·yÌ· Well, I have already given you a hint, think for the next few hours.
Cool, thanks for the hint @JasperLoy
@ccorn That's too political.
@JasperLoy Yes, the hints should be enough for now
Off topic: $k$ is an integer. If $f(\pi kx )$ has domain $[0,1]$ on the real line and $f$ (when defined on the real line) has zeros at all integer multiples of $\pi$, how many roots does f(\pi kx) have on $[0,1]$? My response: $\pi kx$ is an integer multiple of $\pi when $kx=j$ is an integer. Wouldn't the answer be infinity since there are infinitely many rationals between 0 and 1?
"rationals" are $x=j/k$
07:07
From the category of answers deserving more upvotes, consider this hint by Gerry Myerson with its awesome density of useful hints in just a few sentences.
@TheSubstitute Remember that $k$ is given, thus fixed.
@TheSubstitute The second caveat: Both endpoints $0$ and $1$ are included, which gives one extra repetition that needs to be taken into account.
Thanks, I was allowing $k$ to vary.
:14613507 See it's good to leave the fun to you.
07:34
@TheSubstitute Third caveat: You may need to also consider $k=0$ and $k<0$. The case $k=0$ may seem strangely out of place, but it is instructive insofar as it hints to the rationale of characteristic 0 in algebra.
07:52
Are there any issues with considering the 'class of all cateogries' or the 'class of all functors'? (Note that I don -note- speak of the category of all categories or the category of all functors)
dunno
not sure why you'd want to
08:07
@Mike: are you speaking with me? And why does it matter why I want to? The motivation is to have a mathematical object $Cl$ and and make "let $F$ be a functor" be written as let $F\in Cl$. But the question is a mathematical one.
Off topic: Oxford dict about class‌​: A set or category of things... obviously useless for math
@ccorn: big surprise
That's a silly motivation, IMO :P
But math is silly.
@Mike: I really don't understand why you debate the motivation for it, it's a mathematical question.
"Are there size issues with 'the class of all functors'."
"class" is not a well defined thing. It's just informal language (at least, in ZFC). I can't say anything about that unless I know what a class is.
And I'm not debating the motivation for it (and definitely not in a hostile way), I'm just chatting.
08:18
@NiftyKitty95 Besides size, classes may have inclusion issues. I am definitely not an expert in these things, but I suppose that things are fine as long as everything can be reformulated in terms of the defining properties. That is, if the notion of class can be eliminated completely. But then, it's a non-issue. That might explain Mike's initial response.
Away from keyboard
Well, if you formulate a cateogry, e.g. the category of sets Set, then Ob_Set is not a set. Set isn't an object in ZFC of course. But if it's possible to formalize Ob_Set, then it must also be possible to formalize classes.
Sure, there are set theories in which one talks about classes. I know one is abbreviated as NGB but I don't know its name. You can also do it with Grothendieck Universes without leaving ZFC, IIRC.
vonNeumann-Gödel-Bernays
In the former case you probably do run into issues with talking about a "class of all classes" just like you do with sets in ZFC. Russel's paradox is hard to do away with. In the latter you just don't mention classes.
no class is member of a class
08:27
(These are things I was told once. I'm not an expert, I'm someone with a vague knowledge of something passed down by people with less vague knowledge)
(I'm sure this is clear to any observer :) )
no proper class is member of a class, is what I should have said. i.e. the "class of "
in NGB?
also
All sets are classes. And from all the classes: A class is a set if it's in a another class
Implying there can be classes which just happen not to be member of another class and then these aren't sets
I want to put functors in a class - mostly because I don't know if they fit in a set, and because I don't quite know what a functor is, if I don't model it via set theory
@kitty which textbook/prof are you following?
@niftykitty
08:48
@skullpatrol: None in particular
I look for good definitions in the nLab, Wikipedia, I know Joy of Cats, Goldblatt, Simmons (informal)
it's for my own notes, I don't quite know under which "type" to introduce functors etc.
I don't mind forming the category of all functors, even if 'all functors' is larger than any the class of all my sets, but I just don't want to be formally inconsistent
or make a defintion so that what I naively set out to be 'the class of all…' or 'the category of all..' is in fact not a class resp. category.
09:17
I've proved that every (non-elemantry) tree has at least two maximal independent sets, with equality only for stars. I'm trying to extend that proof now for general graphs, with the assumption that diamater>=2 holds the same 'status'. Any suggestions/hints on how to approach this?
I've considered heading to the u,v vertexes and the path between them that forms the diameter, but not sure what to do with it actually..
Actually, I don't think it's necesserily true..
If I have equation $y'(x)=y(x)+c e^{y(x)}$ and $y(0)=y'(0)=2$, can I just plug in that to get $2=2+c e^2 \implies c=0$, and then that my particular solution has to satisfied equation $y'=y \implies y=ae^x$ and from $y(0)=2$ I get $y=2e^x$. I think that this way isn't right?
@Cortizol Yes, you can plug that in and determine $c$, assuming $c$ is a constant.
@Cortizol: yes
09:36
I think this is some cheating, but that is my problem in my head. Can we find some function that $y'(0)=y(0)=2$, but $y\not\equiv 2e^x$?
Nice, the simplicity
As simple as possible, but not simpler :-)
09:53
I think I may try to prove this via construction. Every connected graph G with $diam(G)\ge 2$ has a maximum path (the diam path) which is also a maximal path...
Which is also bipartite
And I can construct the two independent sets on the path: All the even vertices on one hand, and all the odd vertices on the other hand.
Too clumsy to try adding to these sets the vertices outside the path. Gah, I'm lost..
10:11
Greetings
A very nice inequality $$2^{n}+1<2^{n+1/2^{n-1}}$$
Greetings
@skullpatrol Hi :-)
r9m
r9m
@Chris'ssis $2^{n}+1< 2^n(1+\frac{1}{2^{n-1}}) < 2^n(1+1)^{\frac{1}{2^{n-1}}} = 2^{n+1/2^{n-1}}$ :D
Hi @chris'ssis :-)
10:28
@r9m :-)
@r9m Did you figure out what that smile meant?
r9m
r9m
@Chris'ssis now I'm confused .. :O
what did the smile mean ? .. did I do it wrong ?
In real analysis, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x. The inequality states that :(1 + x)^r \geq 1 + rx\! for every integer r â‰¥ 0 and every real number x â‰¥ âˆ’1. If the exponent r is even, then the inequality is valid for all real numbers x. The strict version of the inequality reads :(1 + x)^r > 1 + rx\! for every integer r â‰¥ 2 and every real number x â‰¥ âˆ’1 with x â‰  0. Bernoulli's inequality is often used as the crucial step in the proof o...
@r9m so the middle step is from Bernaulli's inequality .
@Complexanalysis No, it's not. Check the second inequality for $n=2$.
It's wrongly applied.
:14617233 Exactly! This is the usual mistake of the beginners. :-)
10:43
@Chris'ssis i guess you wanted to say for $n=1$ .
@Complexanalysis Yeah.
@r9m I think you did things in a hurry :D
r9m
r9m
@Chris'ssis ya .. I understand now .. :)
@Chris'ssis for $n=1,2$ the inequality clearly holds , then for $n>2$ @r9m 's argument goes well .
@Complexanalysis for $n=2$, the difference is $-0.343146$. I don't know what you're referring at.
r9m
r9m
@Chris'ssis for $n=2$, $4\sqrt 2 > 5 $
10:57
@Chris'ssis i said for $n=1$ and $n=2$ we can check manually by substituting . for the rest we can rely on bernoulli's strict inequality case .
@r9m is $5$ or $6$ up there?
r9m
r9m
@Chris'ssis you said $2^n+1$ .. so $2^2 + 1 = 5$
@r9m i was wondering what $ (\mathbb Z \times \mathbb Z ) / \langle (1,3) \rangle $ ?
@r9m let me check the whole thing again
@Complexanalysis @r9m This inequality doesn't hold! Am I crazy here? :-) $$2^n(1+\frac{1}{2^{n-1}}) < 2^n(1+1)^{\frac{1}{2^{n-1}}}$$
@Chris'ssis i think you are right . this doesn't seem to hold .
isn't $(1+x)^r < 1 + rx $ holds for $r > 2$ .
@Chris'ssis your assessment is very right i would say .
11:11
@Complexanalysis OK :-)
for $n > 1$ the inequality is in-fact the other way around , isn't it ? , ie $(1+ \frac{1}{2^{n-1}} \ge 2^n(1+1)^{\frac{1}{2^{n-1}}})$
@Complexanalysis Not really. You need to remove $2^n$ in the right side or to multiply $2^n$ in the left side.
@Chris'ssis yes yes , forgot .
11:35
hello
what it means a formal series ?
i guess a series with infinite terms
Gah, this drove me crazy. Not sure what to do with it anymore.
0
Q: Diameter and maximal independent sets

StudentmathI've proved that every nontrivial tree has at least two maximal indepndent sets, with equality only for stars via the bipartition of the trees. I am trying to extend that proof to general graphs, and I think the following is correct: Every connected graph G with $diam(G)\ge 2$ has at least t...

It seems like every statement using the diameter is simply wrong.
 
1 hour later…
12:43
@Mike can you help me in formal series ?
13:15
I didn't see @robjohn today ...
@Chris'ssis He's in California, it must be about 6 a.m. for him. He'll come later, probably.
@Chris'ssis or @DanielFischer can you help me ?
please
@DanielFischer I see, thanks! In general, I've almost never figured out when he isn't on-line. My impression has been that he's around all the time. :-)
@Chris'ssis He's a mod, it's his job to generate that impression.
@DanielFischer Yeah, true.
13:21
@DanielFischer do you know the recent fever about this game ? gabrielecirulli.github.io/2048
@Complexanalysis I've noticed it kind of caught on.
@DanielFischer its quite addictive .
@Complexanalysis Aha. Good thing I quit soon enough.
13:34
Whats up?
someone help me please
@Vrouvrou Most people don't know about that. Did you try Overflow...?
no,
>_<
hi @PedroTamaroff
@Complexanalysis Hello.
13:43
what is the adress of overflow ?
@PedroTamaroff wat's up ?
But try to be extra clear and say you have cross posted in MSE.
@Complexanalysis Studying.
@PedroTamaroff cool >D
13:56
@PedroTamaroff Do you know the following lemma which says that If $ f\in C^2 , Df(x_0) =0, D^2f(x_0 \neq 0)$ then there exists a local diffeomorphism $\phi$ such that $f \circ \phi^{-1} (y) = \sum_{i=1}^n \pm y_i^2$ .
@PedroTamaroff I have something you might like. Prove that $$2^{n}+1<2^{n+1/2^{n-1}} \text{(no calculus)}$$
we are taking $f $ to be from $\mathbb R^n \to \mathbb R $
@Complexanalysis No, not really.
@anon
14:27
@PedroTamaroff $$\sum_{k = 0}^n\binom{n}{k}k^2x^ky^{n - k} = nx(x+y)^{n-1} + n(n - 1)x^2(x+y)^{n-2}$$
Odd... It looked like I was still attached to the room, but when I posted a comment, it stayed green...
Then, when I refreshed, I saw a lot more and my avatar floated back in
I guess I was disconnected
I wonder when that started. I was wondering why I never got any pings.
@robjohn Good morning.
Got that ping?
@DanielFischer good morning. I hope your day is going well.
@DanielFischer yes
@robjohn Not bad, not bad, thanks. I hope the same for your day.
@DanielFischer I hope so, too. Just got back from walking the dog. Looking for a few good questions.
@robjohn Good luck.
14:38
The Mathines... we're looking for a few good questions.
Perhaps I am dating myself
@robjohn I don't think that's true Rob.
@PedroTamaroff uh, why?
Oh.
You were doing it with $y$.
Sorry.
=)
The argument is obviously symmetric.
@PedroTamaroff the equation I referenced had a quadratic in $x$ on the right... with no $y$ mentioned.
14:53
@robjohn Hey :-) I wanna show you something marvellous.
@robjohn Oh, we have $y=1-x$ there. =P
@Chris'ssis howdy.
@PedroTamaroff Oh... I didn't see that.
oops , i didn't intend to post it here .
$\binom{n}{k}k^2=n(n-1)\binom{n-2}{k-2}+n\binom{n-1}{k-1}$
$$nx + n\left( {n - 1} \right){x^2} = \sum\limits_{k = 0}^n \binom nk {{k^2}{x^k}{{\left( {1 - x} \right)}^{n - k}}} $$
That's certainly true.
15:03
@PedroTamaroff Yes. Your formula is correct
:14623271
I missed an "a" there ... :-)
@Chris'ssis Did you notice that each of the integrals, not only the limit, is equal to $0$?
@robjohn yeah
@robjohn Do you refer to the initial integral expression?
@Chris'ssis actually, maybe not. I think for $n=3$ we get $\frac{4\pi}8$
No, they are not all 0! $\pi/2$ for $n=3$.
(simplified form)
@robjohn This question is too nice to be real. (imho)
I asked that in the past here.
15:14
@Chris'ssis it is $0$ for $n\equiv1,2\pmod{4}$
@robjohn It seems so.
It is $2^{-n}$ times the coefficient of $x^{n(n+1)/4}$ in $(1+x)(1+x^2)(1+x^3)\cdots(1+x^n)$
and $n(n+1)/4$ is an integer only when $n\equiv0,3\pmod{4}$
@DanielFischer you got some time ?
@Complexanalysis I think so. What for?
wat4?
15:22
Hehe, the vote counts of the two undeleted answers here are not very close ;)
Define $y=F(x)$ as $y = (f(x) - f(x_0) , x_2- x_{20} , ..... x_n - x_{n0}$ where each entry is the diagonal entry of $n \times n$ , $x_0 = x_{10} , ...... , x_{n0}$ is a particular vector in $\mathbb R^n$ . my target is to show that if $det DF(x_0) \neq 0$ then $f\circ F^{-1} = f(x_0) + y_1$ . @DanielFischer I couldn't verify the last relation with the inverse i have computed .
I think I might also nicely bound it from above $$\underbrace{\sin\sin ...\sin (x)}_{n - \text{times}}\ge \frac{x}{nx+1}, \space x\in (0,\pi/2)$$
@Complexanalysis Sorry, I don't get what is what. Start with $f$, that is a (continuously?) differentiable function on some open subset $U$ of $\mathbb{R}^n$, right?
@DanielFischer yes lets assume $f \in C^k(\mathbb R^n , \mathbb R)$ .
15:38
@Complexanalysis Okay, and $x_0$ is an arbitrary point?
@DanielFischer yes , but $Df(x_0) \neq 0$ .
@Complexanalysis Okay. Without loss of generality, $x_0 = 0$ and $f(0) = 0$. Then $F(x) = (f(x),x')$, where $x = (x_1,x')$. That right?
@DanielFischer exactly . I assumed that $F$ is the diffeomorphism , since $DF(x_0) \neq 0$ as well.
hi @robjohn can you see this answer please mathoverflow.net/questions/162207/…
@Chris'ssis $|\underbrace{\sin\circ\sin\circ\cdots\circ\sin}_{n\text{ times}}(x)|\le\underbrace{\sin\circ\sin\circ\cdots\circ\sin}_{n-1\text{ times}}(1)\to0$
15:42
@Complexanalysis Well, if $\partial f/\partial x_1(0) \neq 0$, then $F$ is a diffeomorphism in a neighbourhood of $0$.
@robjohn nice
@DanielFischer yup . Now what would be $F^{-1}$ , i got it wrong thats why couldn't varify $f\circ F^{-1} (y)= f(x_0) + y =0+ y_1$.
@Chris'ssis Did would say that is a paraphrasing of his argument. I think it is a clarification.
PLEASE what is $(1+t)^{-1}$ using taylor
@robjohn You got 11 stars thanks to me.
15:49
@Complexanalysis Since $F$ doesn't change the last coordinates, $F^{-1}(y) = (g(y),y')$ for some $g$. So $f(F^{-1}(y)) = f(g(y),y')$. On the other hand, $(f(F^{-1}(y)),y') = F(F^{-1}(y)) = y$, so $f(F^{-1}(y)) = y_1$.
@Vrouvrou $1-t+t^2-t^3+t^4 - + \dotsc$
@Sawarnik you mean because of my reply to your comment?
@robjohn Ya. Your 11 stars wouldn't have come if I didnt comment.
That attracted at least one mod from another site and one community mod
@DanielFischer yup , thanks .
@robjohn i dont understand the answer can you help me
please
15:55
@Vrouvrou look, all I know about the $M_k$ and the $\beta_k$ are what you give in the question. I don't know what the Morse numbers are apart from that. From what you give in the question, I cannot answer the question.
When using the Cauchy Riemann equations for determining whether a function is differentiable, most of the books I read tell you to split the function into that $f(z) = u(x,y) + iv(x,y)$ thing, but is there a good reason for this? Can't I just take the partial derivatives of $f(z)$ along the two axes and get two complex numbers?
no it's juste that i dont understand in the answer given what it means by " To compute these Taylor coefficients use the know expansion $(1+t)^{-1}=\sum_{q=0}^\infty(-1)^q t^q.$"
@robjohn
@GregRos you can do that but it is simpler when moving from functions on $\mathbb{R}^2$ to do the other

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