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12:00 AM
I've so far just used Racket and C, so I know nothing of the sort. Sorry :/
 
@Daminark NaP (not a problem)! I am doing the course right now, was just curios about how to represent methods as functions)
 
@Balarka the part I just got through feels a bit like it's building up to sheaves
 
btw, is some mathematical class also a set?
 
@Daminark hehe yes
 
With all the $f^*(\psi)$ stuff
 
12:03 AM
some book has told me set $\ne$ class, but I didn't go deeper there
 
@Daminark ight
i'm just gonna kick off the r from ight everytime i wight ight
 
In a set theoretic context you've got this von Neumann/Godel set theory, you start with the idea of a class, and a set is a type of class that can be contained in another class. So some types of classes are sets, others are not, they're called proper classes
 
"you've got this von..." sounds very German, @Daminark ;)
 
Lmao, beautiful
@Balarka sounds horrendous
Also it's not wight, it's wite :P
 
i shall move away from keyboard for 5 minutes and rage a bit, excuse me
 
12:08 AM
kk
 
sounds pretty abstract
I think I will not go deeper in classes this time, too
:)
thank you, bye-bye, @Daminark
 
See you!
 
@Daminark Rage
 
Oh for fuck's sake...
 
feeling a strong 9 on that one
the best part is where he blends the smoothie
 
12:14 AM
I'm a bit nervous to watch the rest
 
looool
 
Anyway, so this defines poles the way my physics TA did
With the whole business about infinite limits
 
Right
The geometric picture of poles to me are just points where the function blows up straight to infinity
 
What about sin(1/Re(z)) as z -> 0
 
@Daminark Here's an interesting point to make. If $f : X \to \Bbb C$ is meromorphic with say a pole $p \in X$, locally near a deleted neighborhood of $p$, it looks like $1/z^k : \Bbb C \to \Bbb C$. So you can extend $f$ to an actual holomorphic map $\tilde{f} : X \to \Bbb P^1$ by sending $\tilde{f}(p) = \infty$
So meromorphic functions always extend to holomorphic maps to $\Bbb P^1$
 
12:25 AM
for integration by parts, can one set U to be something that will derive into zero?
nvm thats a stupid question
srry.
Wow, that was dumb. Please ignore me
 
@Balarka and that was the next theorem!
 
mhm
 
Let R be the rectangle joining the midpoints of the sides of quadrilateral Q having vertices
(±x,0) and (0,±1). Calculate: limx->0^+ = (PerimeterofR)/Perimeter of Q
 
Any good exercises here to look into?
Oh there aren't many, I'll try all
 
2 and 4+5 are useful
 
12:35 AM
this is confusing me
How would I go on to solve this? the perimeter of R would be 2l+2w thats all i can put in right now...
 
12:49 AM
Hey @Balarka I'm back
 
@MATHASKER You can just plug in x=0 directly, I think, neither of those perimeters go to 0
 
I thought I killed you!
 
You'd want Pythagoras
 
Not yet >:)
 
His is an evil laugh
 
Hiiii
 
WiFi is down due to general construction fuckery
 
Nice new word.
 
In certain circles it's fairly common, probably
"Construction"
 
I will tell you a story now.
 
12:53 AM
Not actually construction, maintenance I think. A sewer broke and they've torn up a street trying to fix it. They must've cut an underground wire related to the internet at some point
 
A long time ago, I bought a loaf of bread from a bakery downstairs.
 
@Balarka so as $z\to\infty$, $f(z)\to \frac{a}{c}$, so I think the way we extend to a holomorphic function on $\mathbb{P}^1$ is to let $f(\infty) = \frac{a}{c}$ (if $c=0$, call that $\infty$)
 
Slice by slice, I ate the bread over a couple of days.
 
@AkivaWeinberger plug in zero to what doe, they don't give me an equation
 
x
If you do want an equation, you can draw it and do Pythagoras for the outer thingy
 
12:54 AM
Suddenly, one day, I noticed a cockroach, dead and in perfect shape, fossilised between two of the slices. End of story.
 
Q
WHOA
EWW
Jesus
 
And for those points such that $cz + d = 0$, let $f$ be infinity
 
X like the point (+x,0)(-x,0)??
 
Yeah
Poor cockroach died in the most ironic of deaths
Spend a life trying to find food and die trapped inside it
 
but that would be zero zero, that would not create a quadrilateral doe would it?
 
12:56 AM
Probably doesn't really matter
 
@AkivaWeinberger It must have gotten into the dough. Bad bakery.
 
Why does the integral from 0 to inf of 3x/(4+x^2) diverge??
 
@Daminark Right
But $f$ also has a pole
Which you have to send to infinity
 
At a glance, seems like it's O(1/x)-ish
 
then how would i find the limit doe without a graph to look at or equation? @AkivaWeinberger
 
12:57 AM
Well they'll just be when $cz+d=0$, right? So at that point we're good
 
@10Replies Hm So 4+x^2 is, for x large enough, less than 2x^2
 
right
 
So 3x/(4+x^2)>3x/(2x^2) for x large enough
which means 3x/(4+x^2)>3/2 * 1/x
and we know the integral of 1/x from blah to infinity diverges
('cause the antiderivative is ln and ln(infinity)=infinity)
I suppose you could just actually find the antiderivative of 3x/(4+x^2) directly but that seems more annoying
 
Ah thanks. That makes sense
What about integral 15/(x-7)^2 from 0 to 7
I tried to solve it by integrating and then rotating the integral
 
And conveniently, the derivative is the determinant of the matrix over $(cz+d)^2$
Well that's away from those blow up points
 
1:06 AM
4 mins ago, by 10 Replies
What about integral 15/(x-7)^2 from 0 to 7
nvm I figured it out
 
@Daminark Right
 
I mean at the blow up points it should be fine anyway but the worry is that the derivative is also infinity...
 
You have to switch charts to infinity in P^1 and take derivatives then
 
1:32 AM
@Daminark But yeah, you're claiming as $f' \neq 0$ (because the determinant is nonzero, as our matrix was in $\text{GL}(2, \mathbf{C})$), $f$ is locally a biholomorphism everywhere (by inverse function theorem). Why is it globally a biholomorphism?
There's actually a neat observation here.
 
2:14 AM
Hey I'm back, briefly. And I imagine you can show that the composition of Mobius transformations is a Mobius transformation
 
yup
 
So then just take that corresponding to the inverse matrix?
 
right
 
Alright sick
In the case of 4, the idea should be that each one can be given as a Z-linear combination of the other, so you want an invertible integer matrix
But the thing that's toying with me is that we have to have derivative 1, instead of possible -1
 
2:30 AM
Here's a probability question I should know.
Let $x$ be random real number in (0,1). Then pick a random number $y$ in $(0,1-x)$. How is $y$ distributed?
 
@Dodsy @Abcd @LeakyNun I choose to leave the site and you people send me a bunch of email notifications excusing me of having duplicate accounts? If you have an issue contact a moderator. I'm not your babysitter and it's not me job to mediate when you people decide to get in an argument. I've asked you to leave me alone, and that wasn't a joke. Stop trying to contact me. Thank you.
(No, I am not sticking around.)
 
fyi, dodsy's on a ban right now.
 
I only logged in because I was annoyed by that.
@Semiclassical for?
 
you can look into the transcript for that. I'm not going to weigh in.
 
im too lazy. Can you just summarize or link me?
 
2:37 AM
i'll find when it all went down
 
kk
@Semiclassical it's an illegitimate ban
someone flag spammed him
too bad
for that accusation I should flag him. that was a bit offensive him accusing that guy so insistently. I'm too lazy and don't really care. I just want people to quit bugging me.
 
I dunno
I'm not going to weigh in further than that, though.
 
no it goes further
apparently Dodsy was spamming flags
nvmd
that's not the case
not evident anyways
he was never on after that point
@Semiclassical since I'm not going to be back on, please give @Dodsy a figurative slap in the back of the head for me would ya? He's acting all paranoid about me again claiming other users are me. Kind of annoying to everyone. Even if he thought it was a username change it was still kind of dumb of him to act like that.
 
way too much paranoia in general in this chat
i really dgi
 
me neither
i mean people assume so many things without reason
 
2:50 AM
Is there a way to disable the email notifications? Might be less of a headache than dealing with this situation
 
@Daminark Yeah I'm not entirely sure why we aren't allowed $\det A = \pm 1$ in this context either. If I were to restrict that I'd assume $\omega_1/\omega_2$ and $\omega_1'/\omega_2'$ are both in the upper half-plane.
 
sure there are flag spammers, but they could be brand new users that don't even use chat. Heck, it would be MO users playing a mean prank for instance.
just giving a random example to prove my point
 
eh, the latter strikes me as pretty paranoid
 
we have no evidence that any one of us is the flag spammer
 
(Just switched laptop; really uncomfortable to work with but so it is)
 
2:51 AM
@Semiclassical Just saying that it could be any user from anywhere on the site being a troll. Even a user from outside MSE.
 
@Balarka do you have Washington on this laptop?
 
i don't know what evidence the mods have access to re: flags so I can't say one way or another.
 
@nitsua60 see the above comments and the linked reply I originally made. I don't wish to report it using a flag as it's probably not flag worthy. However, I would appreciate it if you could find the time to make some kind of post warning people about random accusations. Dodsy was being quite rude with ABCD.
@Semiclassical only staff can see who made flags
 
all mods can do is see that they occurred
 
2:51 AM
@Daminark Unfortunately no. But let me use the browser version.
 
which is the annoyance
hence, we know an excessive amount is being done
more than normal
but we don't know if it is one person or a group or if it's a spambot or what
my money's on it being a spambot
if only because that would explain the erraticness
but anyways I've been here far too long already
im leaving
 
happy trails
 
See you
 
@Daminark But yeah the upshot of #5 is the really nice fact that homothetic lattices give biholomorphic torii
In fact the converse is also true.
So there are inequivalent complex structures on the torii coming from non-homothetic lattices
 
I think that kind of stuff is realized concretely in the realm of elliptic function theory?
 

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