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12:12 AM
@Fargle You get Hawaiian if you collapse $\{0, 1, 1/2, 1/3, \cdots\}$ in $[0, 1]$ though
 
@BalarkaSen I buy that.
 
What do you get if you collapse a Cantor set in $[0, 1]$? :3
 
A headache.
2
 
The answer is "Nobody gives a shit" but close enough
 
What does "collapsing" mean?
 
12:19 AM
Floopsquish
 
Ah, thought so.
 
"Floopsquish"?
Hey there @apnorton!
 
Hey!
 
Haven't seen you in years
 
Yeah... I kinda took a break for a while but I'm thinking about coming back more regularly now :)
How are you?
 
12:21 AM
You graduate by now?
I got stuck being room owner ... as you might have noticed. So much for retirement and no responsibilities.
 
Yeah! I finished undergrad in 2017 and am now working as a software dev
 
I figured it should have been a while. Still in greater VA?
 
Yep! Right outside of DC
 
I like DC (well, except for some obvious residents).
 
:P haha
Honestly I'm trying to decide whether I want to stay in private industry or if I want to go to grad school and eventually be a professor
The hard part is that I'm cautious and don't like making big changes, so I want to be really sure before I do something like that :P
 
12:25 AM
CS faculty make more money than math faculty, just cuz you guys can make so much in industry.
 
Mhm. At some point tho the industry has to change what it's paying people. The skill that I see being exhibited vs the salaries people are getting is... concerning sometimes. I felt like people cared more about doing excellent work simply because excellence is good (as opposed to "do what you can to make it appear like you did good work") when I was at school, but idk if that's all of academia or just the professors I was around.
 
The whole world is excelling at mediocrity and cheating now ... look at the leadership.
 
:'(
 
Hell flks
 
Do you know of any good books/resources that might help someone figure out whether they'd enjoy being a professor? The one concern I have right now is that I do kinda want to stay close-ish to the east coast but a lot of good CS schools are over on the other side of the continent... but there are a few places I'd consider for grad school over here, too.
I did a research project in undergrad that resulted in a conference paper, then I worked with a professor on a conference workshop after I graduated, too --- both of those experiences I really liked, which I think/been told is somewhat representative of what researchers do in their day-to-day
 
12:36 AM
More on the west coast, yeah. No, there are two basic issues to contemplate. One is research and pressure to publish. That's significant. The other is teaching and working with students. And, there's also general university responsibilities, committee work, etc.
 
I'm pretty confident I'd enjoy the teaching aspect, at least from what I saw when I was a TA and worked closely with a couple professors supporting their classes
 
I loved teaching, and still do. But the fact that I was getting more students I couldn't motivate to work hard and pass my courses really bothered me.
 
That bothers a lot of the professors in where I do my undergrad as well
I think that's just the woe of being a good teacher
 
I've noticed that sort of stuff, too.
 
But that changed with the changes in society, etc. The first 25+ years of my career, granted plenty of students avoided me, but I motivated most of my students to work their butts off and do better in my classes than they did with most teachers. That changed toward the end. I still had amazing students, but others I couldn't get to do anything.
 
12:43 AM
^^ I can also see the impact of this change in society where I work --- there are a lot of my peers who are super lazy, while it's less common to see in the people who've been around a while (of course, this snapshot has some massive sampling bias)
It's kinda scary when I think about how some of the people who I TA'ed managed to squeak by with a degree, and are now probably working on software running my bank accounts, flying planes, etc.
 
Yup ... and look at the role models in the world now.
 
Speaking as someone who hasn't always been the best student, I think the change has a lot to do with a sort of general malaise or exhaustion with the socioeconomic state of things. College is "what you're supposed to do" for many, and it's seen as a job factory. Society in general is also much more atomized and individualized than it used to be, as a result of the same factors.
 
Mhm
Very much that, too ^^
 
In my specific case, it was very different---the fact that I was seen as a precocious young brat meant that I was, until a certain point, literally allowed to be lazy. Broader changes in educational standards have had something like that effect for many people---people being pushed forward through school even though they may not have adequately learned what they needed to. Many people who get pushed forth in this way mentally check out even before high school.
And I don't blame them.
NCLB and the things it precipitated are kind of what I'm thinking of.
And then there's something to be said for the erosion of the "triangle of educational responsibility" between teacher, student, and guardian.
 
$K/F$, $\mu_n \subseteq F$, $1 \to \mu_n \to K^\times \xrightarrow{n} K^{\times n} \to 1$ induces $1 \to H^0(G,\mu_n) \to H^0(G,K^\times) \to H^0(G,K^{\times n}) \to H^1(G,\mu_n) \to 1$ by Hilbert 90, i.e. $1 \to \mu_n \to F^\times \to F^\times \cap K^{\times n} \to H^1(G,\mu_n) \to 1$
hey @loch
since $G$ acts trivially on $\mu_n$, $H^1(G,\mu_n) = \operatorname{Hom}(G,\mu_n)$
 
12:53 AM
Yikes, I got a phone call and I missed all this ...
 
hey what on earth is the map $F^\times \to F^\times \cap K^{\times n}$ if not still $x \mapsto x^n$? I guess we can't just change that term to $F^{\times n}$ because we would lose exactness
 
Oh, you paged loch. Good.
@Fargle @apnorton: What's worse is that so many high school teachers and college teachers just give up and have no standards, because it's just easier to avoid all the whining and parents complaining. So they teach easy courses and give high grades. Guess who didn't fit into that one?
 
college teachers?!
 
Yeah. I think the issue is deep and systemic at this point.
 
Yes, definitely, in the US.
And in Europe the system is so different, anyhow. Like professors just lecture and hardly get to know students.
 
12:59 AM
From what I saw in my experience, I feel like there's a significant issue with an "entitlement to good grades" especially in college.
(which leads to the whining and obnoxiousness)
 
YEs, @apnorton. When I was a kid, the teacher had authority and parents always backed up the teacher. Now parents attack teachers — even up to and including college teachers.
 
Thank goodness for ferpa to give even a slight tool for defending against parents
 
Yup.
 
One of the professors I TA'ed for failed a student for cheating and they sued the school. Idk how often that happens, but it was surprising to me :\
 
Most universities have very careful academic dishonesty policies. You can't do that sort of thing without going through a hearing by the academic dishonesty board. That protects against that sort of thing.
 
1:03 AM
well we know that $1 \to \mu_n \to F^\times \to F^{\times n} \to 1$ is exact so we can "divide" that exact sequence by this exact sequence to get $1 \to F^{\times n} \to F^\times \cap K^{\times n} \to \operatorname{Hom}(G,\mu_n) \to 1$
i.e. $\operatorname{Hom}(G,\mu_n) = (F^\times \cap K^{\times n}) / F^{\times n}$
 
hi @LeakyNun
 
Geometry: A Geometric Approach
11
 
aha if $K$ is algebraically closed then $F^\times \cap K^{\times n} = F^\times$
@loch is there no high-tech proof of hilbert 90?
 
i dunno
 
What does it say again
$H^1(\text{Gal}(L/K); L^\times) = 0$?
 
1:08 AM
@BalarkaSen $H^1(\operatorname{Gal}(K/F), K^\times) = 1$
 
Gotcha
 
:)
 
So in a split extensions $1 \to G \to X \to K^\times \to 1$, all the complements of $K^\times$ are conjugate?
 
according to wiki if $G=\operatorname{Gal}(K/F)$ and $H = GL_n(K)$ or $H = SL_n(K)$ then $H^1(G,K) = 1$ also [where this special case is $H = GL_1(K) = K^\times$]
 
That's neato
 
1:11 AM
what is a complement?
 
All the sections of that extension, if you like. They are all conjugate in the group
Interesting, I don't have a proper intuition for this
How do you prove this?
@BalarkaSen That was backwards, by the way. $1 \to K^\times \to X \to G \to 1$ is the correct thing.
 
I'm not sure either is correct...
$H^1(G,K^\times)$ isn't $\operatorname{Ext}(G,K^\times)$ right ($G$ isn't even abelian)
wiki says extensions are classified by $H^2$
 
Conjugacy classes of complements of split extensions $1 \to N \to A \to H \to 1$, where $N$ is a $H$-module, are in 1-1 correspondence with $H^1(H; N)$, no?
@LeakyNun Not the same thing
 
ok then
 
$H^1(G; K^\times)$ is still classical group cohomology, right? Abelian-ness of $G$ is not necessary
 
1:21 AM
yes it is classical group cohomology
 
Mmk, then what I have is correct I think. Just not useful maybe
 
in The h Bar, 41 secs ago, by Secret
Don't recall sets in set theory have this cutoff property
 
For $\Bbb C/\Bbb R$, this says in $\Bbb C^\times \rtimes \Bbb Z_2$, $\Bbb Z_2$ has a unique complement. Which looks correct.
 
@BalarkaSen let $\varphi : G \to L^\times$ be a cocycle. By Dedekind's independence, $\{\sigma \mid \sigma \in G\}$ are linearly independent, so $\sum_{\sigma \in G} \varphi(\sigma) \sigma$ is not the zero function, so there is $a$ such that $b := \sum_{\sigma \in G} \varphi(\sigma) \sigma(a) \ne 0$. Then for any $\tau \in G$ we have $\tau(b) = \varphi(\tau)^{-1} b$, i.e. $\varphi(\tau) = \tau(c)/c$ where $c := b^{-1}$, so $\varphi$ is a coboundary
 
Oh that's cool
 
1:33 AM
computation skipped: $\varphi(\sigma \tau) = \varphi(\sigma) \cdot \sigma(\varphi(\tau))$ so $\tau(b) = \tau(\sum_{\sigma \in G} \varphi(\sigma) \cdot \sigma(a)) = \sum_{\sigma \in G} \tau(\varphi(\sigma)) \cdot (\tau\sigma)(a)$ $= \sum_{\sigma \in G} \tau(\varphi(\tau^{-1} \sigma)) \cdot \sigma(a) = \sum_{\sigma \in G} \tau(\varphi(\tau^{-1})) \cdot \tau(\tau^{-1}(\varphi(\sigma))) \cdot \sigma(a) = \varphi(\tau)^{-1} b$
where $\varphi(\operatorname{id} \circ \operatorname{id}) = \varphi(\operatorname{id}) \cdot \operatorname{id}(\varphi(\operatorname{id}))$ implies $\varphi(\operatorname{id}) = 1$
and $\varphi(\sigma \sigma^{-1}) = \varphi(\sigma) \cdot \sigma(\varphi(\sigma^{-1}))$ implies $\sigma(\varphi(\sigma^{-1})) = \varphi(\sigma)^{-1}$
 
I like writing everything additively when doing these computations, but I realize writing $K^\times$ additively is a sin
 
then what is $\sum$ lmao
 
is a \sin ?
 
smacks Ted
and then generalize to infinite $G$ by the fact that it is profinite...
but I would have to learn infinite Galois theory
@BalarkaSen why does $H^1$ commute with inverse limit?
eh... wiki says $G \to H$ induces $H^n(H,M) \to H^n(G,M)$
 
Isn't that just cuz cohomology is a contravariant functor?
 
1:43 AM
@LeakyNun Hm, I don't know.
 
but we're changing base now? @TedShifrin
 
I don't know what that means. If you precompose an $n$-cochain in $H$ with values in $M$ with the map $\phi\colon G\to H$, you don't get an $n$-cochain in $G$? ... I don't think about group cohomology.
 
You do. I don't follow what Leaky is trying to say
 
Thanks, a @Balarka.
 
I see
45 mins ago, by Leaky Nun
i.e. $\operatorname{Hom}(G,\mu_n) = (F^\times \cap K^{\times n}) / F^{\times n}$
I wonder what happens if $n = \operatorname{char}(F)$
 
1:50 AM
fields of finite characteristics are a myth
 
LOL
 
in that case $\mu_n = 1$ so... $(F^\times \cap K^{\times n})/F^{\times n} = 1$?
that doesn't look right does it
so take $F=\Bbb F_2(t^2)$ and $K = \Bbb F_2(t)$...
what even is $F^{\times2}$
so $t^2 \in F^\times \cap K^{\times2}$
so $t^2 \in F^{\times2}$? so $t \in F$? that's a contradiction
what went wrong
ah I'm an idiot, $K/F$ is not separable :P
 
thats what happens when you believe in myths
 
LOL
The power rule is difficult for Leaky.
 
so we should look at Artin-Schreier extensions
e.g. $F = \Bbb F_2(t^2+t)$ and $K = \Bbb F_2(t)$
and then god knows what $K^{\times2}$ and $F^{\times2}$ are
 
2:14 AM
What is the derivative of $\abs{z}^ae^{i a\arg z}(z+1)$?
 
any idea how I would go about proving f(b) = f(a) + f'(a)(b-a) + f''(c)/2(b-a)^2 as an extended or modified version of MVT? looks a lot like a taylor expansion but it's not quite
 
2:51 AM
Looks like the Taylor Remainder Theorem (maybe that's just something my book called it? idk) to me
 
3:10 AM
any ideas on how to get to it using rolle's/MVT? working from Spivak
mainly chapter 10/11
 
3:37 AM
@LeakyNun, how to show in Young's inequality, if equality occurs, then $a^p=b^q$?
 
@kylecampbell I feel like you'd just apply the MVT twice?
E.g. given a, b, f that blah blah blah, there exists c st $f(b) = f(a) + f'(c)(b-a)$
 
@apnorton yeah, and then apply mvt to find point between a and c, i think.
@kylecampbell, perhaps, looking at Th 5.15 of baby Rudin may help.
 
So there then exists a $a < d < c$ such that $f'(c) = \text{something with } f''(d)$
One of these days I should get Rudin and read through it
 
@apnorton, do you have any idea, why in proof here, equality holds if and only if $a^p=b^q$?
Well, I have done plugging and chigging, so I need only one direction proof: equality implies $a^p=b^q$
 
Hmm let me think
 
3:53 AM
Thank you
 
I don't know :(
 
ok, np
 
I'm probably going to keep playing around with it, but analysis/real inequality stuff was never my strong suit
 
hmm
 
4:09 AM
This chat does seem to skew hard away from analysis.
2
 
Please someone provide a link that shows that $f(x)=x^p$ continuous where $p>0$ is fixed.
(or refer me to a book or someything)
 
Is anyone still here?
 
4:25 AM
Also I'm still here, haha
 
can someone check this for me?
 
@apnorton thanks
@CaptainAmerica16 that is correct
 
@CaptainAmerica16 Seems good to me, but I think that sort of proof really depends on theorems/lemmas you're allowed to use/have already developed.
 
Thanks, I wasn't sure if I made a leap in logic.
 
yeah, depends on what properties of real numbers you have to work with
 
4:28 AM
@apnorton It's based on Spivak ch. 1
 
I don't have spivak :(
 
I have a PDF if you want
 
then yes it certainly should apply from his P1-P12
 
@kylecampbell Cool :D
 
@CaptainAmerica16 This looks right, based on what I remember from his book.
You might do well to indicate which fact tells you that if $c > 0$ and $b - a > 0$, then $c(b-a) > 0$.
 
4:30 AM
Got it.
 
This is an element of care in proof-writing that will help a lot going forward, so you don't just have a jumble of facts and stuff in your head.
(And will also make your instructors' lives easier. :P)
 
Just state the fact outright? - assuming the reader doesn't know the properties in Spivak
 
I would assume the reader has access to whatever textbook you're working out of, so you can just refer to the fact by its "code".
 
That's just me personally.
 
4:32 AM
Thanks for the advice
 
Chapter 1 of Spivak makes life entirely easier for any later chapter too, a solid understanding
 
Oh yeah, for sure.
 
@Silent you're just asking for a proof that x^n is continuous for any real n on R?
 
I've actually forgotten how to show that, so now I'm interested, too ^^
 
yes, any real power, not just integral power.
 
4:37 AM
Wait how do we even define $x^p$ for $p \in \mathbb{R}^+$?
 
@kylecampbell does spivak has some proof? u know, i don't have hard copy with me, hence searching is difficult for that proof in book.
@apnorton Very nice question.
I had to struggle for this a lot.
 
not that I know of in Spivak
 
@apnorton $\mathrm{sup}\;\{ x^{r}\}$ for $r$ rational less than or equal to $p$
 
@apnorton But, then I saw Tao's Analysis I. In Ch 1 he does a good job, although you will have to fill so many gaps.
 
I'm looking at Rudin's thm. 5-15 for that Taylor Remainder thm question, but the problem is, is that all I have to work with is f and f' are continuous on [a,b], and f'' exists on (a,b). So conclude from those cond. that there exists a c in (a,b) s.t. f(b) = f(a) + f'(a)*(b-a) + f''(c)/2*(b-a)^2
 
4:49 AM
Hm
I can get close but not quite by just using the mvt twice
f(b) = f(a) + f'(a) (b-a) + f''(c_2) (b-a)(c_1 -a)
 
yes! close but not quite
 
5:31 AM
Hi, can someone help how one can go from eq. (7) to (8) in this paper journals.aps.org/pre/pdf/10.1103/PhysRevE.64.021604
It says "solving by quadrature"
 
what is the paper's title?
 
Is there an arxiv link?
 
 
1 hour later…
6:57 AM
@kyle: It is precisely Taylor's Theorem with remainder. So, yes, you can prove it from scratch.
Use the Cauchy Mean Value Theorem, considering the two functions $R(x)=f(x)-f(a)-f'(a)(x-a)$ and $g(x)=(x-a)^2$.
 
Ted, I can't sleep until I figure out this problem.
Actually, two problems.
 
how to show in Young's inequality, if equality occurs, then $a^p=b^q$? @TedShifrin
 
@Silent How did you get that hat on your avatar?
 
@CaptainAmerica16 its hat, winter bash. I don't know what are qualifications to get one. There was a message to me from stackexchange, and I wore that hat.
 
@Silent I knew it had to be something special. I kept seeing too many people with one.
 
7:10 AM
yeah, it you get this privilege every year around christmas-new year. I does not last after new year, i think
 
@Silent Hm...I'll have to do some research on this.
 
:)
 
7:40 AM
@MatheinBoulomenos,
35 mins ago, by Silent
how to show in Young's inequality, if equality occurs, then $a^p=b^q$? @TedShifrin
 
Hey @TedShifrin thanks. I'm not quite seeing how the third term goes to f''(c)/2*(b-a)^2.
 
 
3 hours later…
10:21 AM
@Fargle This chat does seem to skew away from hard analysis.
 
10:58 AM
Hello.
 
 
2 hours later…
12:45 PM
I have a rather simple polynomial with all odd orders present, and the coefficient of all orders is identical
So it's basically x^(2k+1) from k to infinity, and some constant term
Can anything be said about whether the polynomial is solvable? I'm thinking about proofs in Galois theory about polynomials of order 5 and higher
 
 
3 hours later…
4:02 PM
Hello!!

Let $m\geq 4$ be an even integer and $a\in \mathbb{Z}$, $a\geq 2$. I want to show that $\frac{m^a}{2}+\frac{m}{2}-1$ is composite.

We have that $m=4+2k=2(k+2)$ for some $k\in \mathbb{N}_0$.

Then $$\frac{m^a}{2}+\frac{m}{2}-1=\frac{[2(k+2)]^a}{2}+\frac{2(k+2)}{2}-1=2^{a-1}(k+2)^a+(k+2)-1=2^{a-1}(k+2)^a+k+1$$

How can we show that this is composite?
 
@CaptainAmerica16 Hat info: winterbash2018.stackexchange.com
 
@apnorton Thanks, I realized I had two hats today. I want more though.
 
@user55789 if what you mean is $\sum_{k = 0}^\infty x^{2k + 1}$, that's not a polynomial
 
Hey @TedShifrin !! Do you maybe have an idea for my question above?
 
 
2 hours later…
6:07 PM
If we take linear transformation derivative at a boundary point then it doesn't have to be unique? Is there some example of a function to check this?
 
6:48 PM
@BalarkaSen If you collapse the Cantor set in $[0,1]$ you should get another Hawaiian earring, no?
 
Oh, I seriously doubt that, DogAteMy.
The quotient is surely very un-Hausdorff.
Hmm ...
 
What two points can't be separated?
 
Well, I'm troubled by collapsing an uncountable set to a point.
I dunno.
 
Enumerate the components of $[0,1]/C$, send each one to a ring of the Hawaiian earring
The Cantor set turns into the basepoint, and any neighborhood of it will swallow all but finitely many of the rings
 
I don't think that's going to be continuous.
Something's wrong.
 
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