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6:02 PM
Wait, is $\Delta(\ln(x^2-y^2))$ also zero?
In addition to $\Delta(\ln(x^2+y^2))$
No; I just messed up.
 
hmm I'm gonna see a lot of $\nabla$ in tomorrow's test
any advice?
 
Remember that the real part of a holomorphic function is harmonic. You're locally just applying that to ln f, whose real part is ln |f|
 
@MikeMiller is there anything in common to the product rules for grad curl div?
 
So if the metric is conformal (i.e. $g\delta_{ij}$), and the log of that metric is harmonic, then the surface is Euclidean.
 
Idk
 
6:11 PM
Similarly for the log of the radii of the Tissot circles.
 
For forms it's just the Leibniz rule
So translate there I guess
 
I mean scalar multiplication by a scalar field
 
6:30 PM
How do I show a harmonic function has partial derivatives of all orders? All the proofs suggested earlier somehow assume that the partials exist without actually proving it.
 
I only assumed the complex derivatives of $f$ exist!
Where the harmonic is the real part of $f$
 
@AkivaWeinberger Ok, that is correct. But your proof only shows that u and v have first and second order partials which is obvious by definition of harmonic functions.
For n greater than 3, you have somehow assumed that the partial exists without actually proving it.
 
Why?
We know $f'''$ exists, right?
 
@AkivaWeinberger You said "To get $u_{xyx}$ differentiate $f$ horizontally then vertically then horizontally" without actually showing $u_{xyx}$ exists.
 
To show that $u_{xyx}$ exists, differentiate $f$ horizontally then vertically then horizontally.
 
6:40 PM
@AkivaWeinberger Ok. But how would write this for all such partials.
 
@user330477 just take the harmonic conjugate
 
Induction
 
@LeakyNun Can you elaborate?
@AkivaWeinberger We have $2^n$ paths for each $n$ degree partial for $u$. How do you do induction?
I mean to say how can I make this rigorous.
 
How do we even notate the general case
Let $\alpha$ be a sequence of $x$s and $y$s
Let $n$ be the length of that sequence, and let $r$ be the number of $y$s in it
I claim $f^{(n)}(z)=i^{-r}(u_\alpha+iv_\alpha)$
 
we claim that for every partial derivative of $u$ is either $\plusminus \text{Re}(f^(n)}}$ or $\plusminus \text{Im}(f^(n)}}$
 
6:52 PM
\pm $\pm$
And the other part is the corresponding partial derivative of $v$
 
@AkivaWeinberger Still now sure how I should go about proving this?
 
any hints for proving that the series $\sum_{n=1}^\infty \sqrt{x_n}/n$ converges when $\sum_{n=1}^\infty x_n$ converges where $x_n\geq 0$?
 
@user330477 Do you remember the proof of Cauchy–Riemann?
 
@AkivaWeinberger Yes. I know that since derivative exists and is well defined we can take in the derivative in the desired direction.
But still I am not sure how to go about doing this?
 
Induction step: In my notation, assume $f^{(n)}$ is either $\pm u_\alpha\pm iv_\alpha$ or $\pm v_\alpha\pm iu_\alpha$.
We need to show that $f^{(n+1)}$ is either $\pm u_{\alpha x}\pm iv_{\alpha x}$ or $\pm v_{\alpha x}\pm iu_{\alpha x}$
and we need to show that $f^{(n+1)}$ is either $\pm u_{\alpha y}\pm iv_{\alpha y}$ or $\pm v_{\alpha y}\pm iu_{\alpha y}$
By $\alpha$ I just mean some sequence of $x$s and $y$s
like, if $\alpha=xxyx$ then $u_\alpha=u_{xxyx}$
 
7:13 PM
Does every automorphism of the linear algebraic $\mathbb{R}$-group $\mathbf{PGL_2}$ come from an $\mathbb{R}$-algebra automorphism of $M_2(\mathbb{R})$?
 
7:38 PM
what is $\omega_n^{n/2}$ where $\omega_n$ is the nth root of unity?
 
@Anush: I guess it depends whether $n$ is even. And if $n$ is odd, what does this even mean?
 
$n$ is even
I should have said
 
Have you drawn pictures to see some examples?
 
is it just -1?
 
Yup.
 
7:40 PM
thanks! So this means that $\omega_n^{k+n/2} = -\omega_n^k$
which is very helpful indeed
 
Sure.
 
thanks.. I was trying to understand the derivation of the FFT algorithm
 
sup chat
 
Ah.
heya Eric.
 
how goes it
 
7:42 PM
Bumbling along ... and you?
 
trying to write up a little statement about a geometry/PDE problem
 
Hi everyone
 
hi demonic :)
 
apparently there's cool geometry that comes up when u study the asymptotics of the Allen-Cahn equation that isnt well understood
so i think im gonna be writing about this for my nsf proposal
 
7:45 PM
I have never heard of that, Eric. But there's all sorts of interesting geometry/algebraic geometry that comes out of all sorts of nonlinear PDE.
 
i think it's a science-y phase transition thing
one of Neves' postdocs worked on it
 
Is it a KdV type thing?
 
i know nothing about KdV so i couldnt teell u
 
I don't think so
I think it's a very nonlinear gradient flow type equation
 
I just googled.
Well KdV is super non-linear, but just works out cleverly ...
cf. solitons
 
7:49 PM
Allen-Cahn is only a little nonlinear
 
Oh my bad
 
it's a diffusion type boi whos nonlinearity is double well potential thing at order 0
 
I decided 5 courses per semester was too much for me so I dropped one, my weekly schedule looks way better now
 
@Eric: I can't even figure out the equation on Wiki. What the hell is $\epsilon_\eta^2$?
 
the wikipedia equation is gibberish to me
 
7:51 PM
I'm unsure what to do with this question. I generally object to answering questions I think shouldn't have been posed. But there are also interesting things to say (that are only vaguely related).
 
@Alessandro: By graduate school, 5 is way too much. Typically, 3 Ph.D.-level courses is plenty.
 
Lmao 5 courses
I can barely learn one thing at a time
 
@EricSilva That makes me feel better.
 
And I struggle with that
 
the form im working w is like $u_{t} - \alpha K \Delta u + \alpha f(u) = 0$
 
7:52 PM
I'm doing 4 now but one is only 2 hours per week without problem sets
 
I solved that, @MikeM. I quit learning. :)
 
where f is the thing u get from the double well potential
 
That looks linear to me, @Eric, except for the inhomogeneity.
 
-perk-
 
Well it's not like I do anything else interesting
 
7:53 PM
and $\alpha$ and $K$ come from science
 
I did 7 courses in one semester in my undergrad, but they are way harder now of course
 
Learning is all I've got
 
Double well potential is my bread and butter, lol
 
Leaves @Semiclassic and @Eric to commune.
 
Mostly because the way you compute the level splitting in a double well potential is via the semiclassical WKB approximation
 
7:54 PM
@TedShifrin the $f$ is the nonlinearity
 
Right, @Eric. I was thinking of it as making the homogeneous problem inhomogeneous.
So the basic operator is very linear.
 
so the pde is semilinear
 
Aren't $\alpha$ and $K$ constants?
 
Oh, I see what you're saying.
 
7:55 PM
(I only sorta know about KdV tho.)
 
I brought up KdV. It's irrelephant here, Semiclassic.
 
mmkay
it links up with the scattering theory of the linear Schrodinger equation, which is how you usually make sense of those soliton solutions
 
@Semiclassical it means nothing to me except that it has 2 minima and that's what makes the magic geometry happen
 
asymptotically what happens is the solutions converge to one of the two minima locally uniformaly and the interface is a surface moving by a curvature flow
 
7:57 PM
In physics terms, you can write down two approximate solutions, each with support on only one of the two wells
but these aren't good solutions since they don't respect the symmetry of the two wells
you fix that by taking symmetric and antisymmetric combinations of them
(blah blah blah)
 
yeah idk what the hell ur saying
 
all math-y stuff that isnt what ive immediately studied is gibberish 2 me
 
sits in the corner
 
@Ted it looks like Sid will be leave behind hodge and talk to us about Kähler geo now
exciting
 
8:02 PM
That's good ... I thought you guys were doing the reading, rather than his doing talking.
 
both
 
I'm ready to talk curvatures and linear systems, etc., whenever you get there.
 
as soon as i finish nsf app hopefully today or tomorrow...
 
Lemme try to not do it in physics terms, then. You've got a spectral problem $(-\hbar^2 \Delta + V)u=\lambda u$ where $V$ is the double well potential
 
8:05 PM
gotta throw in Planck's constant in order to talk about asymptotics eventually
If you just had a single well $V\propto x^2$, then the eigenvalues would be of the form $\lambda_n = \hbar(a n+b)$ where $a,b$ are constants that I'd have to work out more carefully
and the eigenfunctions would be stuff like Gaussians * polynomials
tedious stuff, but exactly solvable
(blah blah blah I don't feel like continuing)
 
8:24 PM
@Semiclassical cool
 
8:46 PM
$\varepsilon_{ijk}\varepsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$
I feel like a master in tensor notation now
 
lol, yeah
that one gets used a ton in physics calculations
for instance, if you contract both sides with $A_j,B_k,C_l,D_m,$ you get $$(\mathbf{A}\times\mathbf{B})\cdot(\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot \mathbf{C})(\mathbf{B}\cdot\mathbf{D})-(\mathbf{A}\cdot \mathbf{D})(\mathbf{B}\cdot\mathbf{C})$$
i.e. the vector quadrupole product
 
@Semiclassical do you have tricky problems involving tensor notation?
 
not so much off the top of my head, i'm afraid
I guess one would be: show that $\epsilon_{ijl}\epsilon_{jkm}\epsilon_{ikn}=C \epsilon_{lmn}$ for some constant $C$
You can see how the rule you just wrote would facilitate that, though it's still tedious
 
9:05 PM
$\varepsilon_{ijl} \varepsilon_{jkm} \varepsilon_{ikn} = (\delta_{lk} \delta_{im} - \delta_{lm} \delta_{ik}) \varepsilon_{ikn} = \varepsilon_{mln} - 3\varepsilon_{iin} = \varepsilon_{mln} = -\varepsilon_{lmn}$?
 
ah, i guess you can get away with just one use of the rule.
The fact that it's C=-1 is easy to confirm by taking $(l,m,n)=(1,2,3)$ and observing that the LHS vanishes except for the terms $(i,j,k)=(2,3,1)$
 
man i've no time to do unification
 
though the algebra of course shows that directly
as you can imagine, you can construct a bunch of rather silly exercises like that
for instance, you could also do $\epsilon_{ijm}\epsilon_{jkn}\epsilon_{klp}\epsilon_{ikq}$
 
a slightly trickier thing is $\delta_{ij}\delta_{jk}\delta_{ki}$
 
not so tricky: first two give $\delta_{ik}$, and then you're just doing $\delta_{kk}$
so just the number of allowed $k$
 
9:12 PM
yeah but you just apply the reduction rules you eventually get $1$ lol
 
eh, should be $\delta_{ij}\delta_{jk}\delta_{ki}=3$ in 3D
 
sure
@Semiclassical do you know de Rham cohomology?
 
Not well enough to be any help
 
do you think that's the easiest route for physicists to learn algebra?
 
Learn algebra, or learn algebraic topology?
Easiest route for physicists to learn algebra is how group theory shows up in physics
 
9:16 PM
(co)homology
 
I think when it comes to algebraic topology, cohomology is the one which requires the least introduction of new objects
however, a typical physicist takes certain things for granted, e.g. $d\alpha=0\implies \alpha=d\beta$
which is true in a local sense, but needn't be true globally
 
Tensor notation is garbage
5
Add it to the heap
 
@MikeMiller what can I do, I have a test on it tomorrow
@Semiclassical I thought physicsts know that the cartesian form of gradient only works for R^n
 
sure. that's why physicists doing E&M rely on their tables of differential formulae in various coordinate systems
But for instance you'll see stuff like "$\nabla\cdot \vec{E}=0$, so there exists a scalar function $V$ such that $\vec{E}=-\nabla V$"
which is not true without either some qualification or some conditions
 
really
 
9:25 PM
@LeakyNun for someone ranting about that: unapologetic.wordpress.com/2012/02/18/…
 
> So what’s missing? Full formality demands that we justify that the first de Rham cohomology of our space vanish. Now, I’m not suggesting that we make physics undergrads learn about homology — it might not be a terrible idea, though — but we can satisfy this in the context of a course just by admitting that we are...
:P
 
I mean, in physics you'll typically go from there to writing the potential as $$V(\vec{r})-V(\vec{a})=-\int_{\vec{a}}^{\vec{r}} \vec{E}\cdot d\vec{l}$$
in which case the point is largely that the RHS only makes sense so long as that integral is path-independent.
And that, in turn, means that you require $\oint \vec{E}\cdot d\vec{l}=0$ for any closed path
The real point, imo, is that an intro text takes for granted that $\nabla\vec{E}=0$ is valid everywhere in space, i.e. that that's true by definition in electrostatics
and once you assume that, then your domain is just R^3 and therefore contractible
 
10:01 PM
Suppose we are given that every real harmonic function is locally the real part of a analytic function. Then how do we deduce that every harmonic function has continuous partial derivatives of all orders?
 
What do you think?
 
@MikeMiller I don't know why the second should follow from the first. But I think it should involve that if the function whose real part is our harmonic, then it has continuous derivatives of all orders.
 
Yes, that does seem important. What does analytic mean?
 
@MikeMiller Analytic means that the function is complex differentiable and its derivative is continuous. Another word use for this is holomorphic.
 
10:17 PM
Hmm, I remember another rephrasing of this. Maybe in terms of power series?
 
@MikeMiller Power series is next chapter. Can we not prove this without using that?
 
10:33 PM
how about we don't prove this at all?
we clearly already told you how you can prove it, and you keep saying such and such hasn't been taught.
 
@LeakyNun You never told me anything. It was just Akiva and Ted Shifrin the other day who guided me. Stop acting so bossy.
 
Hey guys
how do you like my new name
I was studying adjoints in CT today
@Daminark, it's you! Am I busted?
 
10:52 PM
if $f=u+iv$ is infinitely complex differentiable on a domain $D$, how do I show that $u$ and $v$ are also infintely differentiable on $D$?
 
Probably the Jacobian matrix has something to do with it
 
:0
(Not sure quite who you are, perhaps because of the username)
 
hey @Daminark
 
Yo!
 
11:12 PM
lattice of propositions over one variable up to equivalence in intuitionstic logic
 
Hiya Ted.
 
hi Demonark, @CaptainAmerica, Leaky
@user330477: Continuity, differentiability, etc., all work component by component.
Demonark: Who is it?!
 
I learned about the Hairy ball Theorem today. I understand the basic explanation, but I didn't really get the math part. I need to learn what a vector field is.
 
Great theorem, @CaptainAmerica.
You can find it in one of my video lectures. The proof uses lots of stuff you don't know yet.
 
I think it's really cool. It'll probably be a while before I really get why it works though.
 
11:27 PM
That's really rather deep topology.
The proof I gave in my book/lectures is based on Stokes's Theorem.
 
hi @CaptainAmerica16
 
@LeakyNun Hey :D
Stokes theorem...is that the multivariable calculus or differential geometry? @TedShifrin
 
The one in multivariable calculus is a special case of the general one.
You'll eventually need to learn differential forms :)
BTW, you got my geometry chapter?
 
Hey Ted!
 
Yeah, I've read through it a bit. Then I did some school work, but now's math time :D
 
11:34 PM
sup chat
 
OK, just checking. You only need the first section for the stuff we were talking about.
 
If $(a_n)$ is a sequence of positive numbers such that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}=0$, what can I say about $\lim_{n \to \infty} a_n$? Is it $0$?
 
What do you think, @user193319?
 
I think it is $0$, but I'm currently brain dead.
 
11:35 PM
Write down the $\epsilon$-$N$ definition.
 
You know, I was a little surprised when you said it was from your abstract algebra book.
 
You're allowed to be surprised.
 
@MikeMiller u wound me
 
Lol, yeah. I guess I was more than a little surprised. I don't know much about Abstract Algebra so I don't really know how it fits in.
 
There's group stuff that appears in later sections ...
 
11:38 PM
Oh, I haven't read it all yet.
 
@Eric ADD IT TO THE HEAP
 
But there's some introductory algebraic geometry going on in later sections, so it makes more sense than you think.
Is Eric beheaped?
 
Did someone say algebraic geometry?
 
ya i guess im in the heap along w all my tensors and christoffel symbols
 
11:39 PM
Back to frame bundles :P
 
@Daminark Ted did.
 
Good
 
@CaptainAmerica16 I think you might like this video
 
@MikeMiller we agree the bourgeois belong in the heap tho
 
@LeakyNun Ooh, thank you. That does look interesting.
 
11:44 PM
@TedShifrin I'm looking at the definition, and it doesn't look like one can conclude that $a_n \to 0$.
 
Tensor notation is lumpengeometrie
 
antirev scum
 
So you have $\left|\dfrac{a_{n+1}}{a_n}\right|<\epsilon$ for $n\ge N$, @user193319?
 
@TedShifrin Yes.
 
This means $|a_{N+1}|<\epsilon|a_N|$, etc., so $|a_{N+k}|<\epsilon^k|a_N|$.
 
11:50 PM
Oh, you are very right. Thank you.
 
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