Mathematics - 2018-06-20 (page 3 of 4)

But this splitting is exponentially small and therefore can’t be captured by perturbation theory. Instead you have to proceed from either the WKB approximation or use a saddle point approximation on the path integral

Rudi_Birnbaum

this is second order Jahn-Teller, isn't it?

Semiclassical

Couldn’t tell you. I don’t remember the various terminologies

Rudi_Birnbaum

5:19 PM

"this splitting is exponentially small and therefore can’t be captured by perturbation theory" sounds quite interesting

Semiclassical

The example I remember from textbooks is the level splitting of ammonia?

Rudi_Birnbaum

never thought that this could be an issue

Semiclassical

Closely related to this is the spectrum of a cosine potential

Rudi_Birnbaum

@Semiclassical: Yes a kind of strange example for a chemist, this is rather something we'd summarize by non-Born-Oppenheimer effect. But I think its true it could be a 2nd order Jahn-Teller case as well.

As essentially ANY symmetry breaking is somehow the Jahn-Teller effect.

Semiclassical

Sounds right.

Rudi_Birnbaum

5:22 PM

Because in NH3 tunneling is essential. So you need the coupling between nuclear and electronic wave function

And the cases where this coupling is essential often include the Jahn-Teller effect.

Semiclassical

All of my stuff is 1D, I should note. So nothing terribly realistic

Rudi_Birnbaum

The better!

Semiclassical

Lol

Rudi_Birnbaum

The stuff I am torchering people here with sice some days is actually also about some Jahn-Teller consequences.

Semiclassical

yeah, sounded like it

Rudi_Birnbaum

5:25 PM

though I am not sure if THAT kind of effect shows up in 1D as well. Let me think ...

nope unfortunately not, because you have only two irreps.

trivial one and "B".

Semiclassical

Well, there’s some interesting stuff in the cosine case

Rudi_Birnbaum

Yes?

Semiclassical

Since you have both reflection symmetry and a discrete translation symmetry

Rudi_Birnbaum

oh you got translation!

then there is chance - but go on

Semiclassical

Eh, that’s about all I remember

Rudi_Birnbaum

5:28 PM

OK :-)

So what I look at is what is the symmetry of the split states.

Semiclassical

The actual classification of eigenstates by symmetry is pretty easy though

Rudi_Birnbaum

after the distortion.

And then you see that there are two cases, one which includes some irrep of a magnetic transition and another case which doesnt.

Semiclassical

I should note that, in the case I’m describing, one only has the electronic energy levels

In particular, one doesn’t talk about how the lattice may deform in response

Rudi_Birnbaum

(in my case also) If you assume now that you have one patricle in the levels.

Semiclassical

If you do include that, that can lead to a Pierls transition which I think may be more directly related

Rudi_Birnbaum

5:31 PM

Ou I see, then thats something slightly different...

Peirls is the Physics name for Jahn-Teller

Semiclassical

Right

Rudi_Birnbaum

in principle

in practice the one is about periodic systems

and the other about "molecules" (subgroups of SO(3))

Semiclassical

Right.

Rudi_Birnbaum

But the rest is the same.

So you look at electronic symmetry breaking?

purely?

Semiclassical

To some extent, yeah. But I did a few things and it’s hard for me to sum it up

Rudi_Birnbaum

5:34 PM

is it a masters thesis?

Semiclassical

PhD

Rudi_Birnbaum

Oh sorry!

No offense indended ;-)

Semiclassical

Lol, np

Rudi_Birnbaum

Physics then I guess?

Semiclassical

Ya

Eric Silva

5:36 PM

lol wtf i thought the space force thing was a joke

i cant believe this is real

Rudi_Birnbaum

So does it happen then in your systems that the electron density localizes asymmetrically?

in a symmetric well

?

Semiclassical

Well, my interest was typically less to do with the form of the wavefunction than the level splitting itself

Rudi_Birnbaum

sure, but you will get the WF as well

Semiclassical

But in both the ground state and first excited state you have the electron being mostly confined to the two wells. The difference is whether it’s s symmetric or antisymmetric combination

Rudi_Birnbaum

Oh wait. It seems its really more kind a tunneling

yes its like the ammonia

now I see the connection

Semiclassical

5:40 PM

So bonding-anti bonding is maybr the better analogy

ÍgjøgnumMeg

physics sounds like magic to me

lol

Rudi_Birnbaum

Guess it would be VERY special case of bonding-antibonding, something we call near-degenrate multi-reference case.

does it sound familiar?

How does the method work you use to calculate the splittings?

Semiclassical

Basucally it’s just the WKB approximation

Rudi_Birnbaum

Thats some very new DFT stuff, isn't it?

Semiclassical

No.

Maximus

5:44 PM

in ZFC, Axiom of union just forces the existence of the set which we describe as union, correct?

Semiclassical

WKB is as old as regular QM stuff

Werner-Kramers-Brillouin I want to say

Rudi_Birnbaum

I just checked it on WP, I see never got in touch with that. How many electrons do you have?

or better: do you have inter-electronic interactions to account for?

Semiclassical

Typically you’re just doing one-electron stuff

Rudi_Birnbaum

@Semiclassical: That simplifies life a lot. So you basically need to solve an ODE?

Fargle

Alright, I was given a question you guys may be able to help out with. A friend wants to be able to rank kill-death ratios in a game such that:
-Returns on kills diminish (i.e. the difference in weight between n:m and n+1:m should be less than between n-1:m and n:m)
-If n:m and p:q have the same quotient, but n < p, then n:m is rated less than p:q.

I've tried a few heuristic solutions but none has worked for this yet. Any ideas?

Semiclassical

5:48 PM

Well, if I was solving an ODE it’d be the Schroedinger equation in which case I’m not doing anything approximate

Main thing is that the tunnel splitting is on the order of exp(-S/hbar) where S is the tunneling action

Rudi_Birnbaum

What is the tunneling action?

its the exponential decay rate outside the barrier right?

What is then the "classical" part, if you solve the SE for the potentials and get the tunneling action for different barrires from that.

Semiclassical

$\int_a^b \sqrt{2m V(x)}\,dx$ for a particle with energy E in a 1D potential, where $a$ and $b$ are the classical minima

Which is computed entirely from knowledge of the potential energy, which is classical

That’s how the tunneling action is defined and calculated

vzn

@Rudi_Birnbaum hi, interesting! googled that up & came up with this golem.ph.utexas.edu/category/2013/06/… ams.org/notices/200902/rtx090200212p.pdf

Semiclassical

The whole point of the WKB approximation is to be able to express QM results in terms of integrals in classical phase space

Rudi_Birnbaum

Hi @vzn: Yes that was the "birds and frogs" I was refering to (Freeman Dyson wrote that)

Semiclassical

5:58 PM

(well, classical configuration space )

Rudi_Birnbaum

@Semiclassical OK you calculate the action classically but the WF comes from QM and some "potentials" you get from somewhere.

Semiclassical

sure. This is all done under the assumption that you know what the potential is

It’s much more simple-minded than something like DFT

Rudi_Birnbaum

@vzn : ams.org/notices/200902/rtx090200212p.pdf cvery cool article: "Jokes of nature" that complex numbers & quasicrystals are out there, ...

Simply Beautiful Art

@Rudi_Birnbaum No idea, call me SBA, and I specialize in very fast growing functions.

Rudi_Birnbaum

@Semiclassical: Yes since you have only one electron than the Schrödinger eq is a charm ...

vzn

6:02 PM

@Rudi_Birnbaum gowers had a very similar essay on "birds vs frogs" using different terms... remarkable it seems dyson does not cite him or vice versa... gowers contrasted grothendieck vs erdos. dpmms.cam.ac.uk/~wtg10/2cultures.pdf

Rudi_Birnbaum

@SimplyBeautifulArt: I avoid specialization so its really hard to say what I do ... but my money comes from chemistry teaching and research in university...

Simply Beautiful Art

¯\_(ツ)_/¯

vzn

oops my memory may have been off, rereading, he did cite erdos, but maybe he didnt cite grothendieck, but a preeminent "bird" by the classification...

Rudi_Birnbaum

@vzn: I see :-) I have read some critics if or how useful it might be to classify 1D quasicrystals for the sake of understanding the Riemann $\rho$s. In case there is no big scepticism I'd be interested in that topic.

Semiclassical

I’m not sure the two traditions which Gowers described are quite the same distinction as birds vs frogs, though I agree they’re close

Rudi_Birnbaum

6:08 PM

I have bookmarked it and will read it soon.

Maximus

@LeakyNun So what did you mean by ZFC is just sentences in logic and why is it bad

vzn

@Rudi_Birnbaum the n-category cafe blog alludes to fractal aspects of Riemann which is a key interest of my own. have found a lot of fractal properties in Collatz conjecture over the yrs. are you a student? there are also some deep connections between Riemann and QM theory, have collected misc refs on that over the yrs...

Semiclassical

With bird vs frog, the distinction seems to be one of scope: look at lots of of problems at once vs focus on one at a time

Whereas the traditions that Gowers proposes are more to do with whether mathematics is a search for a systematic understanding vs building up a range of techniques and methods

vzn

@Semiclassical gowers calls it theory building vs problem solving & think a key part/ insight of his essay is that they are complementary & not mutually exclusive. like yin + yang :) ☯

Rudi_Birnbaum

@vzn: I'm docent. There is one recent one where a guy uses some $t^-1 xp -px t$ Hamiltonian to prove the Riemann hypothesis. One of a thousand to solve it, but I think the ansatz itself is regarded still ineteresting

RH is then to show that the operator is self-adjoined

Semiclassical

6:16 PM

Yeah, that was a Bender paper

Rudi_Birnbaum

because one can show that the zeros are its eigenvalues.

The Bender was the "serious" one. But then one followed from some "unknown" guy who claimed to have shown it - on second I'll pass the link ...

Semiclassical

There was a stack question on it at the time

Ah. I wasn’t aware of a follow up

Right, the MSE question:

52

Q: Riemann hypothesis: is Bender-Brody-Müller Hamiltonian a new line of attack?

There is a beautiful paper in Physical Review Letters [PRL 118, 130201 (2017), DOI:10.1103/PhysRevLett.118.130201] by Carl Bender, Dorje Brody, and Markus Müller (BBM) on a Hamiltonian approach to the Riemann Hypothesis. The paper is surprisingly easy to follow for a physicist. BBM define a H...

Rudi_Birnbaum

researchgate.net/publication/…

A Frederick Moxley

I am not sure if this can be called follow up, since he implies he solved the RH....

Semiclassical

The main reason for skepticism I’ve seen raised is that similar arguments apply to zeta functions which are known to not satisfy the RH

So those arguments can’t by themselves be sufficient

Rudi_Birnbaum

Ah

I once in while thought about how to construct Operators that have prime number eigenvalues. Then there was something which I still do not understand properly but which dampened my interest in prime somehow very strongly.

Leaky Nun

6:24 PM

@Maximus will you be here an hour later?

Semiclassical

The story I linked in the comments is a fun side story btw (though of course I think it’s intesting since I linked it)

Rudi_Birnbaum

It is that the 2D sequence $(m+1)(n+1)$ with $n,m\in \Bbb N$ produces all interges except any prime. that looks simple is simple and somehow demystifies the primes in my eyes a bit... dunno

vzn

@Semiclassical thx for digging that up, reminds me, am now organizing my riemann bookmarks + added that one... have been meaning to blog on this...

Rudi_Birnbaum

@Semiclassical: Oh great Belizzard himself replied.. I'll read that :-) Thanks

Semiclassical

@Rudi_Birnbaum Where?

Rudi_Birnbaum

6:34 PM

@Semiclassical: In the MSE post you linked : math.stackexchange.com/questions/2211278/…

Maximus

@LeakyNun yes

Maximus

6:54 PM

Is anyone able to explain this proof to me?

Rudi_Birnbaum

What is example 3.3?

Maximus

Here but it's kind of how they start the proof

I am assuming they are trying to use comprehension

Rudi_Birnbaum

From "Therefore" on it is clear, isn't it?

Maximus

nevermind they are using comprehension

we assume P is true, so x in P

so we let A = P to conclude there exists a B such that x \in B iff x in A and P(x, P, Q)

with B being the original set

ÍgjøgnumMeg

7:28 PM

Unless I'm mistaken, the isomorphisms in the category of $k$-vector spaces are just... the invertible $k$-linear transformations right?

Tobias Kildetoft

@ÍgjøgnumMeg right

ÍgjøgnumMeg

Okay lol, sanity check

learning the language

thanks

Semiclassical

@Rudi_Birnbaum sure, but where in there? his answer has been there a while

Ted Shifrin

We need more insanity checks around here ... :)

4

ÍgjøgnumMeg

7:43 PM

lol

A category needn't have infinitely many objects right?

Sorry if these are silly questions!

no of course not

a group is a groupoid with one object

So if I want to describe a groupoid that isn't a group I can take $\operatorname{Obj}(C) = \lbrace V_1, V_2\rbrace$ with $V_1, V_2$ $k$-vector spaces and then $\operatorname{Hom}_C(V_1, V_2)$ as the identity maps on $V_1$ and $V_2$ respectively and then an invertible linear transformation (and its inverse)? Or is this cheating?

(the linear transformation should map $V_1$ to $V_2$, not just any old linear transformation lol)

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Hi, @Alessandro. All done graduated?

Alessandro Codenotti

Nope, I have the number theory presentation on Friday

I tried it today and it's about 2 hours and a half long instead of an hour.... I decided that if the intermediate results need an hour to be explained they might as well be the topic of the presentation instead

Ted Shifrin

Well, is the presentation for your colleagues or for faculty?

Fargle

Heya @Ted.

Ted Shifrin

7:57 PM

heya @Fargle

Alessandro Codenotti

@TedShifrin colleagues

Ted Shifrin

Aha. So the question is whether the intermediate results are interesting enough in and of themselves. Sometimes it's better just to quote some black-box results to prove more interesting things.

valer

Someone please i have another question. When/if an infinite sum can go inside a function: like infinitesum ln(f(x)) = ln ( infinitesum f(x) ) ?

Ted Shifrin

But it's a standard phenomenon that people underestimate the time by a factor of 3 or so ... and you should allow for interruptions/clarifications/questions.

@valer: That's a big vague a question. But uniform convergence of the series might be the answer.

@Fargle: You got my recent answer?

Fargle

Yes, I did. I appreciate it.

Alessandro Codenotti

8:00 PM

Nah they're pretty interesting, I'm going to show how factorization of primes works in cyclotomic extensions, but I'll also prove a more general result about factorizing primes first. I'll also show why in the Galois extension case the factorizations have a simpler form

Ted Shifrin

Oh, OK, that sounds not just technical but interesting.

Make sure you intersperse plenty of examples.

ÍgjøgnumMeg

@Alessandro Nice!

Alessandro Codenotti

We dealt with quadratic extensions in class so it's not completely new either for my classmates but a generalisation

Leaky Nun

@Maximus i'm here now

Ted Shifrin

Just make sure you have examples ... and maybe examples illustrating why hypotheses are needed, etc.

One of the standard mistakes I've seen — even in graduate presentations and seminar talks — is dry theorem proofs without concrete examples illustrating the ideas and the need for hypotheses.

Alessandro Codenotti

8:04 PM

That's good advice, thanks, I'll make sure to have examples ready

Ted Shifrin

I try to give only bad advice :P

Leaky Nun

@ÍgjøgnumMeg that's ok

or just construct the category abstractly by giving the 2 objects and 4 morphisms

ÍgjøgnumMeg

@LeakyNun As in, a satisfactory answer to "describe a groupoid that is not a group" would be "a category with $2$ objects, the identity morphisms on these objects, and an isomorphism/its inverse"?

Leaky Nun

yes

ÍgjøgnumMeg

ha nice, thanks

Maximus

8:08 PM

Awww, I waited, but now I have to brb myself. In case I am not back on time we can talk tomorrow or some other time since I come here often @LeakyNun

Leaky Nun

sorry @Maximus

Maximus

no worries

Leaky Nun

@ÍgjøgnumMeg You would say there are two objects $X$ and $Y$, and $\operatorname{Mor}(-,-)$ is always a singleton, and you could also give the composition table but that's quite enough

ÍgjøgnumMeg

Fair, the exercise is labelled as "Unimportant exercise" anyway

Ted Shifrin

LOL ... I don't think I'd ever write such a label. I mean, I did have an Exercise 0, which was a standard joke, but ... really!

Leaky Nun

8:11 PM

@ÍgjøgnumMeg recall that a category is a set of objects, and for every two objects a set of morphisms, satisfying some axioms

so you just have to give the objects and morphisms

user193319

Is the collection of all sums of commutators in $B(\mathcal{H})$ topologically closed, where $\mathcal{H}$ is some Hilbert space? Closed with respect to the strong operator topology would be nice, but if it's closed with respect to any of the standard topologies on $B(H)$, I would be interested in knowing.

Leaky Nun

oh and the composition function @ÍgjøgnumMeg

user193319

By the way, by commutator I mean ring commutator; i.e., $[X,Y] = XY-YX$ for $X,Y \in B(\mathcal{H})$.

ÍgjøgnumMeg

@LeakyNun which is something like Hom(B, C) x Hom(A, B) --> Hom(A, C) right?

Leaky Nun

well, Mor instead of Hom

ÍgjøgnumMeg

8:13 PM

What's the difference? lol

Leaky Nun

we use Mor to denote the set of morphisms (used in the definition of category, but not really much afterwards)

and for Hom, it's a functor

ÍgjøgnumMeg

Ah I see

I was under the impression that the distinction was purely based on preference

Leaky Nun

I've never seen Mor except in the definition

ÍgjøgnumMeg

ha fair, that's good because Hom looks nicer to me

Leaky Nun

@ÍgjøgnumMeg we also sometimes equip Hom with other structures, e.g. in an abelian category, the Homs are abelian groups

Peter Sheldrick

8:22 PM

@Semiclassical so why was there a problem with (f1g)(x):=int_0^1(t-1/2)g(x-t)dt? i agree that for g(x)=x it should be -x^2/2 + x/2 - 1/12 and not just -1/12, but how do we get that?

Ted Shifrin

@user193319: You might ask Demonark (@Daminark) if he's thought about your question.

Semiclassical

@PeterSheldrick because $g(x)\neq x$ on $[0,1]$, at least not if $g$ is supposed to be expressible as a Fourier series

ÍgjøgnumMeg

@LeakyNun Right, for instance the hom sets of $R$-modules are abelian groups

GFauxPas

i'm not convinced by my own argument here, if anyone wants to check the reasoning

ÍgjøgnumMeg

and the category of $R$-modules is an abelian category if I'm not mistaken

Semiclassical

8:23 PM

facecpalm i edited that into nonsense

Leaky Nun

@ÍgjøgnumMeg indeed, it is an abelian category

GFauxPas

context: $k, h$ are mappings on well ordered sets, $k$ is order-preserving

Leaky Nun

so we have a functor $\mathcal C^{op} \times \mathcal C \to \mathbf{Ab}$

Semiclassical

$g(x)=x$ on $[0,1]$ but not outside of it. in particular, $g(x-t)\neq x-t$ for all $t\in [0,1]$

GFauxPas

Thus $k(a) \prec \min \left({E \setminus h\left[ {S_a} \right]}\right)$.

This implies that $k(a) \notin h\left[ {S_a} \right]$.

As $k$ is order preserving, we must have that $k(a) \succeq \min \left({E \setminus h\left[ {S_a} \right]}\right)$.

doesn't feel right to me

Ted Shifrin

8:26 PM

heya @GFauxPas.

GFauxPas

hi Ted

seeking a contradiction here

not sure I have one

ÍgjøgnumMeg

@LeakyNun Nice, thanks

Semiclassical

You'll have $g(x-t)=x-t$ for $t\in[0,x)$. But for $t\in (x,1]$ you'll have (by periodicity) $g(x-t)=g(x-t+1)=x-t+1$ since $x-t+1\in [0,1]$ when $t>x$

Hence you have to divide up your integral into two parts, one for $t\in[0,x)$ and the other for $t\in (x,1]$

Ted Shifrin

Is the argument just what you have here or was there something before? What you have here seems to assume I know what you're talking about.

Leaky Nun

@GFauxPas can you give me the full context?

Ted Shifrin

8:28 PM

Oh, good, better for @Leaky to think.

GFauxPas

I guess that's not enough context. proofwiki.org/wiki/User:GFauxPas/Sandbox , "proof of lemma"

Leaky Nun

@GFauxPas the lemma is crap

surely there's an order-preserving mapping from $2$ to $1$

ÍgjøgnumMeg

so if $A \cong B$ in the category then $\operatorname{Aut}(A) \cong \operatorname{Aut}(B)$ under the map $f \mapsto f^{-1}$ (now in $\textbf{Grp}$)

GFauxPas

It's an exercise from Munkres, and I haven't found a mistake in Munkres before

Ted Shifrin

No, Munkres does not make mistakes.

I don't even remember any in the typewritten version of his book we had before it got published.

Not that I'm paying attention ...

GFauxPas

8:32 PM

do you know him personally Ted?

Peter Sheldrick

@Semiclassical okay how about x-floor(x) - how do we calculate the integral of that?

Ted Shifrin

Yes, @GFauxPas. I took topology from him as an undergraduate (and diff. top. from Guillemin and algebra from Artin). I was a lucky boy.

Now you know why I've made my students suffer so :D

GFauxPas

nice :)

Ted Shifrin

@PeterSheldrick: Draw a picture!

ÍgjøgnumMeg

@Ted Your students will be telling people on Math.stack 3.0 Chat that they took lessons from Ted

Fargle

8:33 PM

@TedShifrin Munkres is still one of my favorite books to this day. I'm a bit envious of the fact that you got to learn from the man himself. And from both of those others.

ÍgjøgnumMeg

lol

Leaky Nun

@GFauxPas well the lemma would be correct if the mapping is injective

Ted Shifrin

Munkres was a bit stiff and formal, but for that course when I was a second-year student it was perfect. I wouldn't have liked it so much for an advanced course like algebraic topology. But he was great for me then. And we interacted some when I went back to MIT as a postdoc. He was always kind.

Leaky Nun

@ÍgjøgnumMeg surely $f$ isn't sent to $f^{-1}$...

ÍgjøgnumMeg

bleh why not?

Leaky Nun

8:37 PM

because $f^{-1}$ is still a member of $\operatorname{Aut}(A)$

ÍgjøgnumMeg

oh right hahaha

makes sense

I guess

$f$ maps to something like $f \circ \varphi$ where $\varphi : A \to B$ is the isomorphism

dunno if that even makes sense I'm just "typing out loud"

Leaky Nun

you're almost there

@GFauxPas why $h(a) \ne e_0$?

Peter Sheldrick

@Semiclassical, @TedShifrin hmm im almost there but there is a sign problem
if g(x)=x-floor(x) then int_0^1(t-1/2)*g(x-t)dt=int_0^x(t-1/2)*g(x-t)dt+int_x^1(t-1/2)*g(x-t)dt=int_0^x(t-1/2)*(x-t-0)dt+int_x^1(t-1/2)*(x-t-1)dt=x^2/2-x/2-1/12 but it should be -x^2/2+x/2-1/12

Leaky Nun

mathematics is always correct up to indexing error and sign error

so treat i as i+1

and then by induction...

GFauxPas

order-increasing mappings are always injective I thought

Leaky Nun

8:43 PM

@GFauxPas ok

GFauxPas

because if $h(a) \succ \text{ anything }$, then $h(a)$ can't be minimal

Leaky Nun

@GFauxPas also, $k(\alpha) \prec \min(E \setminus h[S_\alpha])$ actually implies that $k(\alpha) \in h[S_\alpha]$

Peter Sheldrick

okay if t>=x then g(x-t)=(x-t)-floor(x-t)=x-t + 1 and not x-t - 1

thanks everyone

struggle ;)

GFauxPas

hmm

ÍgjøgnumMeg

Right $f\circ \varphi$ can't be an automorphism of $B$ because $f$ is an automorphism of $A$

unless $B = A$ I guess

GFauxPas

8:46 PM

even if $h$ isn't given as order-preserving?

Leaky Nun

@GFauxPas because $\min(..)$ is the minimum item that isn't in $h[S_\alpha]$

GFauxPas

It would be nice if $h[S_a] = S_{h(a)}$ but I don't know if that's ture

Leaky Nun

i.e. everything below it is in $h[S_\alpha]$

@GFauxPas I believe it's true

GFauxPas

proofwiki.org/wiki/… I'm looking at this, I did this yesterday

but now I'm second guessing what I did yesterday

Leaky Nun

@GFauxPas just use the product trick

the one you felt unjustified last time

idk if you have convinced yourself of its veracity now

GFauxPas

8:55 PM

i dont remember what that is

Leaky Nun

given wosets $A$ and $B$, consider the set $\{(a,b) \in A \times B \mid \exists~\text{order preserving bijection}~S_{A,a} \to S_{B,b}\}$

GFauxPas

oh, well, let me finish it this way first, in Munkres's exercises

Leaky Nun

claims:
1. if (a,b1) and (a,b2) are in the set, then b1=b2
2. if (a1,b) and (a2,b) are in the set, then a1=a2
3. the set exhausts either all of A or all of B
4. in the first case, it is an injection A->B with connected image
5. in the second case, its inverse is an injection B->A with connected image

GFauxPas

oh I don't need equivalence between $h[S_a]$ and $S_{h(a)}$, it's enough to have subsets I think

Leaky Nun

this is found in Jech's Set Theory

GFauxPas

8:58 PM

let me try that

okay fixed I think

Maximus

9:56 PM

Anyone here familiar with the convolution theorem of the Fourier Transform?

Leaky Nun

@Maximus hi

Maximus

Hey, you can write stuff now

but I have to do work so I will read it when I can

Mohammad Areeb Siddiqui

10:21 PM

consider $\int \Psi^*\dfrac{\partial^3 \Psi}{\partial x^3}$. If i do integration by parts on this by taking $u = \Psi$ and $v = \dfrac{\partial^2 \Psi}{\partial x^2}$. The integral should become: $-\int \dfrac{\partial^2 \Psi}{\partial x^2}dx$

right?

assume limits $(-\infty, +\infty)$

And ignoring the boundary term

Now consider the integral -$\int \dfrac{\partial^2 \Psi^*}{\partial x^2}\dfrac{\partial \Psi}{\partial x}$ I can't figure out how to apply by parts on this?

After by parts the resulting integral should be the same as the previous result but with a changed sign

Will I have to apply by parts twice?

nvm got it

Symposium

10:59 PM

Is it actually possible to receive send a text message then delete it?

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$Mathematics$ Mathematics