10:01 PM
Awesome
Than I think I might just have my proof..

Hello people :)
:D
Long time @studentmath @TedShifrin @Sawarnik :D

@Sab!

I'm getting a B in Calculus I feel my career is over now

@DanielFischer Since you mentioned the problem of finding good questions, I experimented with using SEDE a bit more: Unanswered questions with a tag favorited by users with a badge in that tag

Nah, doesn't work..

10:05 PM
I like the new name.

It's not as skewed toward insanely difficult stuff as Sam Saffron's query

@WeaponofChoice Gracias.

Ideally, something like this would be built into the system: when a tag-badge holder upvotes or favorites a post with that tag, that ought to mean more than a random upvote. And be used to promote these questions somehow.
Oh well. We'll see what the SE Data Science geniuses come up with, when they ship the new content-quality algorithms.
So far, I'm not optimistic...
We’re currently analyzing the data at #SEMeetup2014 http://t.co/HqMWDLj1EG

@Sabಠ_ಠ Hey .. but just as I was about to sleep..

i can't even sleep im devastated

10:09 PM
@Sabಠ_ಠ :O :(

my class tests mark (5 tests) 50%
my final paper 1 74%
my final paper 2 100%
that's how lazy and stupid I was. Now I get a B in the class

well really sorry sab :( .. but i got to go now, take care :)

c ya ;)

bye:)

Hi, I have a differential equation, $dy/dx = -1+2xy$, I need to find two values, $a$ and $b$ that differ only by $0.01$ but, when they are set to the initial condition of $y(0)$, they make different graphs. How would I proceed..?

10:27 PM
@DanielFischer

@PedroTamaroff ?

@DanielFischer I have a question.
In the Riemann sphere, $\infty$ is not a distinguished point.

@PedroTamaroff Distinguished in what way?

The sphere is homogeneous.

Yes.

10:29 PM
Can we define differentiability at $\infty$?
Can we make it so that a function from the Riemann sphere to itself is "holomorphic" on the sphere?

Sure.

How so? Aren't functions with poles at a point rather good in the sphere?

@PedroTamaroff You know manifolds, don't you?

@DanielFischer Try me.

f(z) is differentiable at infinity means f(1/z) is differentiable at 0, no?

10:30 PM
$\widehat{\mathbb{C}}$ is a manifold. A complex manifold.

@DanielFischer OK.
@anon I think we can use $z\to 1/z$ as a coordinate change or something, @DanielFischer?
Much like the residue of a function at infinity is the residue of $f(1/z) d(1/z)$.

So a map between complex manifolds is holomorphic at $p$ if it is holomorphic in local charts.

@DanielFischer So what @anon said?

Right, $1/z$ is the usual coordinate around $\infty$.

10:33 PM
Of course, a function is differentiable in a pole when viewed as a holomorphic map into the sphere.
Since $\infty$ is not distinguished.

@DanielFischer Right. Which is nice!
So the only really bad functions are those with essential singularities?

@PedroTamaroff @DanielFischer Excuse my interruption, but either of you have an idea on what I can do? I really don't want to have to bruteforce this.

@PedroTamaroff Do you call all singularities that are neither removable nor poles essential, or only isolated singularities?

@DanielFischer We call singularities that are essential those which are neither poles nor removable.
Why "only"?

@PedroTamaroff Branch points, accumulation points of poles or zeros on the boundary ...

10:38 PM

I'll be away for a moment, gotta check my cake.

@DanielFischer Oh, OK.

@PedroTamaroff Basically, I'm given a differential equation, and then the question is asking for $a$ and $b$ such that $0<a<b<2$ where $a$ and $b$ differ by $.01$ I need to find them such that when I set $y(0)=a$ and $y(0)=b$ They make totally different graphs when the diff equation is solved. I have no idea what to do other than brute force it.

@Link I'm not really up for that. Sorry.

@PedroTamaroff It's fine, I was mostly curious what would influence the graphs to look totally different.

10:46 PM
sometimes you have to learn through "brute force" :(

@IceBoy Yea, luckily I have maple, so only 200 or so options to go through.
Haha. This is gonna be fun. Thank god I started it early.

@Link $y' = 2xy - 1$? That's a linear ODE, pretty well-behaved.
$$y(x) = e^{x^2}\left(y(0) - \int_0^x e^{-t^2}\,dt\right)$$

@DanielFischer Yup, it is. Could you explain why/how you got the term in parenthisis?

you've never solved y'+a(x)y=b(x) before?

10:53 PM
@JasperLoy Shhh... :-)

@anon I did it in high school. Fortunately, it is covered in my calculus books, lol.

@JasperLoy >_> <_< Sorry.

@IceBoy You look like Bart.

@JasperLoy that's not constructive criticism

use an integrating factor @Link (look it up if this is the first time you've been exposed to the term. although it's strange you'd come to this problem without first being equipped with the tools to do it...)

10:54 PM
@IceBoy It is called constructive correction of spelling.

@JasperLoy are we talking about "spelling"?
are we constructing a spelling bee here?
sometimes the details get in the way of learning :-)

@anon I can do it, just didn't see that form. I'll solve it now myself.
After solving it, how do I check for where the graph changes... is it something akin to critical points or something? (Not sure what influences the graph)

@Link The graph doesn't change wildly. The solutions to different but close initial conditions don't stay close together for long since $e^{x^2}$ grows pretty fast, but they're all of the same kind. Linear ODEs are pretty well-behaved.

@DanielFischer Got it. Thanks.