hwat, how do people survive without coffee

my special gimmick is that i don't drink coffee on weekends

i can barely move on weekends, but mondays... man i love mondays

You addict, you!

munchkin complained that her day at school was "exhausting," and that she had "a hundred y five" emails, "a hundred y five" meetings, and also that she had to talk to the cat

Emails y meetings — poor dear.

and with all of that, the cat wouldn't leave her alone!

Olivia knows best!

@LukasHeger This point is adopted when you consider groups schemes G and passing to the stack BG, and also more natural in higher categories.

one, given f having f' with a zero of order k at z_0, apply B to g(z) = f(z) - f(z_0), which has zero of order k+1 at z_0, to learn that g is locally (k+1)-to-one, and wouldn't f and g have the same local mapping properties? anything go wrong with that?

@leslietownes Oh, I see. Thank you

I was thinking about local primitive but $g(z) =f(z) -f(z_0)$ is much easier approach.

hello from somewhere ovr the north atlantic

ay

ahh, we've finally managed to deport him.

i'll quote arnold

i suspect a sardine would feel at home here. in case there was too much room, there is a box of some sort right where my right foot would go.

covector? what's wrong with a plain ole vector?

The example 11.36 is ${x dy- y dx\over x^2+y^2}$ on $\Bbb R^2\setminus\{0\}$. $\omega = d(\tan^{-1}y/x)$ for all $x\neq 0$ on $\Bbb R^2$ and the line $x = 0$ divides $\Bbb R^2$ into two half planes.

$\tan^{-1}(y/x)$ is undefined on the line $x = 0$. That's why $\tan^{-1}y/x$ cannot be a potential function of given $1$-form on $\Bbb R^2\setminus\{0\}$ right?

yes. and it's just one example, with other ways of splitting up the plane you could find other functions that come pretty close, but they'll all have some problem along some path that includes the origin.

the books description of it as an attempt at a smooth 'angle function' is a pretty good one. you can define angle as a function locally anywhere away from 0, and continue it in various ways by making arbitrary choices, but when you try to close up around the origin something will go wrong every time.

Any closed form is locally exact. Thank you @leslietownes

lotsa great stuff in that one example. it's what you run into in defining the logarithm in complex analysis. some would call it the principal (no pun intended) motivating example for a lot of phenomena in complex analysis.

and who knows what else.

always leaves a residue in my mouth

That's one thing in common with complex analysis. Leaving a residue at a pole

some many avenues here...

well, good night/morning folks, i'm going to try and see if i can sleep in the sliver of allocated space. i feel at one with the aircraft. airbus should be shot for producing such a cramped vehicle.

Suppose, for a given $\epsilon>0$, I know $|x-\alpha|<\delta\implies |f(x)-f(\alpha)|<\epsilon$. Does the implication also hold if $|x-\alpha|\le\delta$?

Depends what $\delta$ is

@schn It doesn't have to, but it can always be chosen to if one such $\delta > 0$ already exists

say, take $\delta_0 = \delta/2$

I've had two answers to my question on Spivak's proof of theorem 1 in the appendix of chapter 8 of Calculus, yet I can not understand why $f$ is $\epsilon$-good on $[\alpha-\delta_0,\alpha+\delta_0]$. For all $y,z\in[\alpha-\delta_0,\alpha+\delta_0]$, we have $|y-\alpha|\le\delta_0$ and $|z-\alpha|\le\delta_0$. What do these two inequalities imply?

By continuity at $\alpha$, we have $|y-\alpha|<\delta_0 \implies |f(y)-f(\alpha)|<\epsilon/2$ and similarly for $z$.

According to physics trump is second black president because brown doesn't exist it's just orange but darker.

If anyone here excels at (or enjoys) puzzling indefinite integral challenges, here's one for you!

ted: just use u = what mathematica came up with

funny how i clicked on that despite neither excelling at nor enjoying indefinite integral challenges

a morbid impulse that underlies so much of the digital economy

I'm sure you'll blame Munchkin by the end of the day.

You could go read the Post article about the latest brou-haha with sports figures and anti-Semitism and racism.

So, one of the so-called improvements to the cite is that our list of notifications now has a "Mark all read" thing to click on. It doesn't seem to remember when I've read things and so I have to keep clicking on that. Very annoying.

@leslie Here's a good one for you.

mm, yes. there's some confusion in there. idea of introducing new variables is a good one, but linear algebra maybe not the best tool for then untangling the nonlinear part.

not like that, anyway.

something sorta related is looking for eigenvalues by trying to put A - lambda I in row echelon form. symbolic packages can sometimes do it, but the cases they come up with in dividing by matrix entries are not particularly relevant to what the eigenvalues are, so you have to then case-analyze these nasty expressions to get eigenvalues out.

so, perhaps not surprisingly, it isn't a great way of computing eigenvalues.

cool idea though, just needs a little more stress testing before it leaves the garage.

Is it required in this answer (math.stackexchange.com/a/990626) that we say “Now, $\pi_1(U)$ is a union of opens”? Because the chosen x in $\pi_1(U)$ was arbitrary, thus by definition each x in $\pi_1(U)$ is contained in an open set of X, so that $\pi_1(U)$ is open in X.

another thing i've seen a few times is people trying to replace integer programming with solving large linear systems with linear algebra, and forgetting that "oh, and also it should be an integer" is not a trivial constraint and often the whole point of the problem.

@NotTfue huh?

@leslie Actually, in my haste, I screwed up the inverse. My solution does in fact turn into $x=2/x$, tautologically.

@Jakobian He messed up. The entity to which he refers is orange, not brown.

@NotTfue but beige is light orange and therefore light brown so by that logic we've had 47 black presidents

I await the first vantablack president

ted: someone posted a solution to the indefinite integral. haven't worked through the details but looks plausible.

I posted a solution. Or is there another?

Looks like another huge-rep user has been suspended, presumably for PSQs.

there's another that seems to lead to your result via integration by parts. i don't know how you'd guess those parts without seeing your result.

Oh, it's the same guy who posted the first answer with Green's.

I couldn't even take the derivative and combine terms to get it back to the original.

Oh well. I stand by my original comment that one should not expect to find an antiderivative :)

I have never found a complicated indefinite integral I think

not more complicated than $\int \frac{1}{\operatorname{tan}^4(x)} \mathrm{d}x$

no wait I did $\int \sqrt{\tan(x)} \mathrm{d}x$ at some point

you have good taste, lukas.

I think I couldn't get a decent GRE score because I'm not good at calculus

our "calculus/analysis" class was weird

when I was in high school I was one of those kids that were excited about integration

I've used to calculate a bunch of integrals, but seeing what some people integrate, it was nothing compared to that

We know that if a $Y = g(X)$, where $g$ is a strictly increasing function of a random variable $X$, we can compute the cumulative distribution of $Y$. What if $Y_t = g(X_t)$ where $X_t$ is a process? Can we say something meaningful about the process $Y_t$ as well?

@Jakobian Man, I completely misread that the first time. I read it as "When I was in high school, I was excited about integration."

I have no idea how it is that my brain is that miswired.

@XanderHenderson that's kind of what I said