Then, slightly salty, I replied basically "fine, but people from the TMP (theor. math. phys.) degree won't be able to take it now"

That sounds somewhat European/German to me.

He replied "Yeah, but algebraic geometry is not important for you anyways"

lel

@Danu ha

Because he knows exactly what I'm interested in :P

@Danu XD

I bet he doesn't even realize that there are pure math students in the degree. Oh well

I'm being called away to the kitchen...

So I guess I'm not taking algebraic geometry :\

@Danu :\

You could just... not go to lectures.

If the prof is not a good lecturer you wouldn't be missing much.

People don't follow textbooks here

brb

So how would I learn the material

I think this stuff will diverge due to having harmonic terms left behind every odd n

Stuff in physics diverging. What else is new?

Renormalize it.

@DanielSank : which right hand rule?

@IceLord your intro QFT class will not make you do Borel summation.

what are you even trying to do?

post the problem statement

I want to see what it is

I want to try it

@Danu :(

Yup, that thing is a fourier sine series that diverges

lol

@IceLord no the full problem

What exactly is this quantum system you are integrating. The presence of $\psi(t)$ and that $\sqrt{\frac{2}{a}}$ does remind somewhat of particle in a box?

So the $\frac{\partial \phi}{\partial x}$ gives you that divergent fourier sine series?

That might be the greatest video of all time.

The problem is that that series diverges, thus you cannot really interchange differentiation with infinite sums here

@IceLord what did you say

Because it diverges I am not even sure how to differentiate it since it will need some kind of closed form in terms of x for that series (closed forms for divergent series do exists, however I am not sure how to test that for a given divergent series). How do you get $/phi=that series$ is it a given thing in the question or calculated form previous steps?

That's not convergent.

Your $\phi$ doesn't converge, I don't know what you want from us.

Unless $\psi (t)$ somehow make the series converge. Is $\psi (t)$ arbitrary?

@Secret No.

It's a function of $t$, it cannot make the series converge unless it's $\equiv 0$.

ok nvm

@IceLord Problem is impossible, yeah.

You're probably reading it wrong.

Post a picture.

How do you get the $\phi$ like that, is it given or a previosuly calculated result?

Yes.

How could it possibly work out?

Did they give you the answer?

show us

But you cannot interchange sum and integrals if the summand or integrand is divergent? Ok I am quite confused now

if a series does not converge, it means the function literally is not a function

it's undefined

so what the hell would its integral even mean

(Assuming T is constant), $n^2$ will diverge like hell?!

lol

that's horribly divergent

also

the first term diverges too

You say that magnetic and electric fields come hand-in-hand (if magnetic heads rightward, electric points upward from the same point). Does it mean that I cannot have uniform electric/magnetic field without the orthogonal counterpart in the same volume of space?

$\sum \dot\psi=\dot\psi\sum=\psi\cdot\infty$.

In other words you have something that resembles $\infty - \infty$ for your $\mathcal{L}$

@Secret Yes but the second term is worse.

So it's $\infty-\infty^2\sim-\infty$, roughy.

I don't think that can be made converge without soem serious renormalisation/manipulation of sorts

Probably.

Also it's a function of time so the divergence is time-variable

If $\dot\psi(t)<0$, then you've got $-\infty-\infty^2$ and there's no hope.

There's more hope but more than $-\infty$ isn't great.

@0celo7 I forgot to ask, did you fix your thing about complement of codimension 2 subspaces?

So no. It diverges.

@BalarkaSen No, not yet. Been getting distracted.

Fair enough.

@BalarkaSen Been learning about rays on Riem. mflds.

Besse Einstein Manifolds has ~120 pages of "review" that I'm going through.

@Danu Books and homework?

@IceLord What's "this"?

@JohnDuffield I am at a loss as to how my answer is unclear.

@IceLord I'm getting in a car to go somewhere, sorry.

DIVERGENT

@BalarkaSen Do you agree?

@IceLord Oh that looks fun.

@DanielSank Well, I might eventually.

I still MUCH prefer a course over self-study

@Danu Indeed, but if there are other students in the course it's not entirely "self".

@0celo7 I have not been paying attention.

I found lectures completely useless in several of my courses.

@BalarkaSen The thing right above my @ you.

That's a divergent sum.

The lectures were more like introduction to the vocabulary, not much more.

@0celo7 We know that ;)

I agree, @0celo7

That moment when... you're a physicist and nobody cares :P

@IceLord Oh for heaven's sake, sure you can.

@DanielSank Don't be ridiculous.

What does the integral of a divergent series even mean?

It's not a function!

@IceLord You might want to know that @0celo7 is a very mathematical rigor oriented fellow.

Mathematically that integral is nonsense. But then physicists have their own logic...

@0celo7 It means you introduced a regulator, did the interchange, and then got rid of the regulator, but didn't write that all out explicitly.

@BalarkaSen It's not "physicist logic" it's "using advanced math implicitly without writing everything out explicitly.

@DanielSank Are you sure that regularization works here?

Howdy

Has anyone ever gone through and given an actual proof that regularization works?

@DanielSank Besides, the answer is divergent like $\sum (1-n^2)$.

You can't tell me that converges.

@DanielSank I can believe that can mathematically be made sense out of. That doesn't contradict that it is - as written - mathematically nonsense.

Anyone reading this who doesn't know what I mean by regulator, read this.

Oh really? Is it "nonsense" when a pro mathematician doesn't write out every single step of a proof because parts of it are well known and/or obvious?

Give me a break.

I abandon this conversation, sorry.

No. But @IceLord isn't a pro mathematician.

@0celo7 Yes, you need it to prove the Fourier transforms make sense.

I have to go for now.

I'll look at this problem again later.

In a couple hours.

bye

@IceLord We can fix it.

@DanielSank Yeah but you have to admit that the people that do this stuff don't usually know how to do it 4realz

Either way, it's fine

Ok, maybe you can do the integral. But the answer, as given on the problem sheet, is a divergent sum.

I don't see how that can be fixed.

If we're accepting a divergent sum as an answer, then yeah, switch limits. We've abandoned mathematics at that point.

I never really looked down upon what physicists do with divergent sums and get meaningful answers. That's intriguing and all. All I said was without a complete mathematical framework, writing integral of a divergent sum is mathematically nonsense.

I hope I didn't give the former impression.

@Secret I'm not convinced by the regulator business, but the answer given also diverges.

There's no getting around that.

If the manipulations are meant to be purely formal, then I guess one can switch limits.

I cannot really give a sensible comment n that issue unless I knew more about regularisation and renormalisation, thus icelord have to figure this out himself for now using danielsank's advice

Mathematically as balerka said it is divergent and nothign will save that, but physicists have tricks to extract physics stuff from it that I don't understood well

TIL using the word 'deceleration' for acceleration that's opposite to velocity is wrong because it's not proper physics, you should call it 'decrease in velocity' or 'negative acceleration'. studyphysics.ca/newnotes/20/unit01_kinematicsdynamics/…

It would be funny if people it didn't make people go around the interwebs saying you can't say that sudden deceleration from 20 m/s to 0 m/s may kill you.

Hey guys, I know this is a really basic problem but I can't for the life of me figure out why flipping switch in this 'http://i.imgur.com/Zr8XpaF.png' circuit would change the output of voltmeter. Isn't the voltage equal when connected in parallel and as such it shouldn't make any difference if the additional 'branch' is connected as the voltage from the supply stays the same. Thanks.

@IceLord We can make you into a mathematiker.

@MAFIA36790 cool post/ "question" by 't Hooft, thx for bringing it to attn, touching on area of huge personal interest/ research, but not sure what you are referring to in particular, do you have any suggestion? are you pointing to a particular answer?

@IceLord We begin with a simple problem.

Every compact, simply connected 3-manifold is homeomorphic to a sphere.

We will try to solve this problem.

@IceLord This is not a problem we can solve in a day.

correct

Because the Riemann tensor is zero.

Gaussian curvature makes sense in $2$-d.

@IceLord To solve this problem, we need: Bredon, Hatcher, Lee, do Carmo, Petersen, Cheeger-Ebin, Evans, Jost Analysis, Jost Riem Geo, Helgason, Chow-Lu-Ni, Morgan-Tian.

I think that's enough.

Oh, Kobayashi-Nomizu (both volumes).