$\Bbb C - 0 \to \Bbb C - 0$, $z \mapsto z^2$

That gives you a covering map

he said that yes for the root test but no to the ratio test, and that a good exercise would be to prove that

so maybe ill try that if i finish my hw early

I guess the point should be that each portion of a riemann surface is locally just a copy of C

so one should get the same classification of singularities on the Riemann surface as on C.

Of course.

(things get weird if you start considering degenerations of a riemann surface, of course. but that's different)

Hm, I have never really thought about how these multifunctions define global functions on a branched cover.

Locally, the story is $\sqrt{z}$ is a well-defined function on a small open ball $U \subset \Bbb C$ away from $0$.

yeah.

Because that's just the lift of the identity $f(z) = z$ function $f : U \to U$ to the cover $p : \Bbb C \to \Bbb C$, $p(z) = z^2$, right?

tbh i'm weak on formalism for this point.

By lift I mean a map $g : U \to \Bbb C$ such that $f(z) = p(g(z))$.

ah

Which is the squareroot, as you can see

mhm

The point being locally you can lift map by "the holomorphic map lifting lemma", because $p$ is a covering map (read: biholomorphic) away from $0$.

That's how I think about things at least, coming from a topology background

the usual viewpoint i have is that you start with a patch U of C on which the function is holomorphic

Right, and I'm trying to analytically continue aren't I?

and build up a chain of overlapping patches

right

Ok, excellent, we're on the same page slash picture

right. intuitive vs. formal descriptions

ugh. sometimes student work annoys me

there were two ways to interpret a certain problem. one made the calculation easy (and tbh I think that was the intended route)

the other gave a result which couldn't be simplified too much.

they went down the harder way, of course? :P

actually, most went down the easy route. so for most this doesn't matter

ah

now, if they did the harder version and left it more-or-less as is, I'm fine with taht

problem is, some people tried to have it both ways

to wit: You've got $$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{i (kx-\hbar k^2 t/2m)}\phi(k)\,dk$$

that's the general case, at least.

the special case is when, instead of taking a sum over infinitely many wavenumbers $k$, you just suppose that there's a single $k$ and so write $\Psi(x,t)=Ae^{i(kx-\hbar k^2 t/2m)}$

in that case, it's simple to show that the probability current, defined as $$J(x,t)=\frac{i\hbar}{2m}\left(\Psi^\star \frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^\star}{\partial x}\right)$$

(might be off by a minus sign, blarg)

i see

in the single-k case, that reduces to just $J(x,t)=\hbar k|A|^2/m$

which is nice enough, and has an easy interpretation: the probability current of a plane wave doesn't vary in time or space.

problem is, in a 'real' free particle, you'd instead have some distribution of wavevectors $\phi(k)$.

e.g. $\phi(k)=e^{-a k^2}$ (times a normalization constant)

in that case, the above formula gives you a product of double integrals.

can't do much with that.

in particular, you had better not say that $\frac{\partial}{\partial x}\Psi(x,t)=i k\Psi(x,t)$ when $\Psi(x,t)$ is that integral above.

ah ok

The whole point is that you don't have a well-defined $k$ in that case.

right

One can justify some approximations in that direction; for instance, if $\phi(k)$ is Gaussian then so to is $\Psi(x,t)$

gotcha

and if you say that Gaussian is narrow, one can approximate that by a delta function $\phi(k)=\delta(k)$. That recovers the one-mode case I gave earlier.

So there's stuff one could say, but it's definitely not as simple as a few people made it out to be

some people also tried to multiply the two integrals they got together without distinguishing the integration variables. $$\left(\int_{-\infty}^\infty f(x)\,dx\right)\left(\int_{-\infty}^\infty g(x)\,dx\right)\neq \int_{-\infty}^\infty f(x)g(x)\,dx,$$ people

lol

Might anybody here know of a text which proves the following inequality: math.stackexchange.com/questions/2447778/…

there's also one or two people who are doing $\partial/\partial t$ instead of $\partial /\partial x$

which is just kinda careless

when I apply the Euler-Lagrange equations to $\int \ln f(x) dx$ so to find the function that minimizes the functional I obtain that $\frac{1}{f(x)} = 0$ but this is not what I am expecting at all since $\ln f(x)$ explodes as $f(x)$ gets larger, am I not applying the Euler-Lagrange theorem correctly?

@aidan amazon.com/Inequalities-Cambridge-Mathematical-Library-Hardy/dp/…? lol

a guess

@GFauxPas I'll try to get a copy of that book.

I never read it

Ive just heard of it

this will sound totally insane, but it came up in discussion on PPCG's chat. Is 99.9999999... = 100, and can it be proven mathematically? If the answer is no, then is there evidence?

the answer is yes, but to see that you need to chase definitions: start by defining the left side

@MikeMiller i'm not caffeinated today - can you clarify what you mean by that need for a definition? In the strictest sense, 99.999(repeating) would not equal 100, in my limited mathematical understanding (I'm not a math expert xD)

@ThomasWard the easiest way to show this is to start with 99.9... and divide it by 10

then you'd have 9.9...

if you take the difference of those, you get exactly 90.

so 99.9... is a number which, when divided by 10, is 90 less than what you started with.

but 100 / 10 = 10 is 90 less than 100.

so 99.999... and 100 behave the same in that regard.

oh geez it looks like the PPCG people have followed me >.<

saw that coming

algebraically: the above relation amounts to x/10 - x = 90, and you can do simple algebra to show that x=10 is the unique result.

@ThomasWard take us to the mathematicians :P

The point is the recurring decimals need a definition.

ahhh, I see both points now.

If 0.999… and 1 were different, there'd be something in between them.

Usually, you can say $0.9999\ldots$ is $\sum_{i = 1}^\infty 9/10^n$.

Which also needs a bit more rigor to justify; it's a limit of the partial sums of that "infinite sum", but let's not worry about that

yeah.

But once you have that, that's a geometric series, so you can sum it. Once you do that, you'll get $1$.

i'm cheating a bit, in that I'm relying on the fact that multiplying/dividing by 10 just shifts all the decimals over by one

strictly speaking one needs more justification than this.

@cairdcoinheringaahing Alright, changing approaches real quick. Do you agree that 1/2 + 1/4 + 1/8 + 1/16... == 1?

Hello, i need someone to explaine me what means this set $\{u\in X, u^+\neq0, |supp(u)\cap \Omega|>0\}$ where $\Omega\subset \mathbb{R}^N$ $supp(u)=\overline{\{x\in\mathbb{R}^N, u(x)\neq0\}}$

@Vrouvrou Which part of it do you need clarification?

$|supp(u)\cap \Omega|>0$

@Vrouvrou Is the absolute value notation defined in the text? Look for that.

i think that it is the measure ? there is nothing about that on the paper

So is your question answered now? Looks like it is.

it means that the intersection is non empty ?

It means what it means, whatever that absolute value means.

???

@DJMcMayhem No, it approaches one with the more precision you calculate it to

@ShaVuklia Glad for you. Well, I'm preparing some magical results for solving some unsolved problems (from the mathematical area I'm interested in).

After a while it is hard to go on without some magic in place.

Some time ago Hardy said: “None of the proofs of the Rogers-Ramanujan identities can be called “simple” and “straightforward”; and no doubt it would be unreasonable to expect a really easy proof”

ugh I hate hardy.

It's under work the thing that will change this perception.

I just think he was always going around crying about how hard math is and how he never did anything of importance and that he was old.

Hello @Dodsy and @Waiting.

hey jasper.

@Jasper Hello amazing JASPER!

@Ted, @Eric Hi

yo

hi everyone.

Do any of you know how to construct a vector field with no zeroes on a noncompact manifold? I can convince myself that I can do this if I have a vector field on the manifold with finitely many zeroes

By pushing the zeroes along an embedded ray $[0, \infty) \to M$ starting at the zero

@Waiting Do you like anything other than limits, series, and integrals in math?

@BalarkaSen what's $M$?

the manifold

oh, right.

M is for McDonalds, the fast food restaurant.

:o

But I don't really know how to come up with a vector field with finitely many zeroes on a noncompact manifold

Sounds hard dood

Is dood the cool spelling for dude?

yeah :)

is Jasper the loser spelling for Dodsy?

LOL

@Jasper imgur.com/a/HKawt

@cairdcoinheringaahing That means you're not thinking about it in terms of limits. If you're thinking about purely in terms of arithmetic, then you're right than 0.999... != 1, it's simply undefined because you can't sum something to infinity.

@Balarka i recall from Spivak that you can construct one if $M$ is open using triangulations or smth

no

o.

So it's either defined it terms of limits (where summing something to infinity is perfectly reasonable, and 1/2 + 1/4 + 1/8... == 1 and 0.999... = 1) or there's literally no answer because infinity is a meaningless concept

@EricSilva Huh, really?

@Jasper Yes

yeah i seem to recall there was such an exercise that explicitly constructed one

@Waiting I don't really have any favourite area now. I try to like all kinds of math. =D

but it's easier to show one just exists in that case then to work out what one looks like

well iunno if it's obvious why one exists either

I thought the triangulation thing was usually used to come up with vector fields with total index $\chi(M)$

but that's for compact dudes

@Dodsy Spivak wrote like 8 books

:40239623 I think they are talking about another book.

Wow that was silly of me

which is just a v - e + f counting thing

yes, I caught that :)

@Jasper did you read the problem set?

@BalarkaSen Yes that's what I said earlier. But that's not smooth and requires triangulations.

@Dodsy I clicked on it and wondered why you showed me.

@Jasper eh, see what you thought.

@0ßelö7 How do you do it with triangulations? I dunno

Dami likes problems like this but he's not online.

(The original question is 0celo's by the way)

The idea is in GP.

@Dodsy I think the problems are interesting and doable at first glance.

it's from putnam meeting, they give us problem sets. @Jasper

But I think you have a hard time with the "finite" part on a noncompact guy.

Huh? IIRC GP only comes up with triangulation in the context of showing there's a vector field with index $\chi(M)$

@Dodsy You want to do Putnam?

@Jasper Any mathematical area requires a huge amount of investment. Walking through many mathematical areas is fine, but excelling in any of them means sacrifice, huge investment, that simple.

@BalarkaSen yeah and the vector field has isolated zeros

I don't at all recall noncompact business in GP

I don't think I'll write the test this year, but I'm gonna go to the meetings.

@0ßelö7 Oh

For sure

I'm saying that's a program for getting a VF with isolated zeros.

@Dodsy I am too stupid for Putnam. I never took part in any competitions.

@0ßelö7 Right

I am dumb today

gotta go do a quiz.

@Jasper I am also too stupid for Putnam. :)

@Dodsy Yeah, we just go there and Put our Name.

haha yes.

@Balarka i seem to recall the triangulations were just for isolating zeroes and then you have to connect them all and cleverly do some isotopy business

@Dodsy Hey guess what, I am getting better at jokes.

this is off the top of my head so idr, but maybe check vol 1 of spivak, it's probably there that i saw this

@Eric i think i see this. i am only convinced of doing the isotopy business for finitely many zeroes right now

or at least, that's the case i can actually write down this

right

i should be doing my rep theory hw :D

just do eet

Vol 1 of Spivak has some very sketchy things in it

Very mysterious things

lol

@0ßelö7 this was a problem, so whether or not it ends up sketchy is in part up to the reader

Jasper, an amazing fact about going deeply in a certain mathematical area, is that after a while you can coonect in a single picture you have learned, discovered, intuition is very keen, entirely special, the results simply come to you without any effort (or at least this is the perception). Those kinds of wow results.

let's just say you take a vector field and bimbadaboom the zeroes to infinity

and leave it at that

> bimbadaboom the zeroes to infinity

i endorse the use of bimbadaboom as a technical term

@Waiting Yes, starred.

@Jasper the results come to you even when doing other activities. Now I understand why Ramanujan had notebooks. You need notebooks always with you to write those results in there. It's not that you stay in a room and discover results. The results come to you even in dreams.

@EricSilva You can put ridiculous exercises that require some ingenious construction and give no hint. That's on the author, not the reader. (I have a specific exercise in Spivak 1 in mind.)

sure, but this exercise had an outline the reader was meant to fill in, in particular

@Waiting I star many of the things you say in this chat, lol.

@Jasper hehe, I see. :-)