Mathematics - 2018-10-22 (page 1 of 2)

@LeakyNun Sorry for the late response, I figured that one out. However this one bugging for quite some time now. How do I show that a harmonic function has partial derivatives of all orders?

@user330477: Because complex differentiable functions have partial derivatives of all orders.

12:22 AM

@LeakyNun How do you get this? But how do we prove this?

@user330477: Write out the Cauchy integral formula for a circle in terms of polar coordinates and real and imaginary parts.

@LeakyNun Cauchy Integral formula or just Cauchy's Theorem?

hlo

no, Cauchy Integral Formula, @user330477.

heya @Eric. Just reading.

oh ya ty!

12:31 AM

As part of an MIT alumni thing, I just spent the afternoon doing what turned out to be 4 separate "classes" 20-25 minutes long with middle school kids. I thought I was only doing 1. :P

i still havent because im awful

haven't what? @MikeM

o wut that’s wild

My topic was a bit too hard for most of 'em. :(

what was the topic

12:33 AM

But I made it work, mostly.

The moral of the story was that you need 3D to solve some 2D problems.

Given a bunch of rectangles of length $>2r$ and whose widths sum to $<2r$, is there any way you can use them to cover a disk of radius $r$?

@MikeMiller u isn’t awful bruv

What is Mike talkin' about?

i assumed the statement i sent u

i sent him 2

Ohhhh.

I'll send some comments.

many thank

12:38 AM

Very well written.

ty i beat myself up over it a lot

hi chat

hi ShmoJoe

whatsup? i see u yaught a few classes today?

pseudo-classes ... yeah ... my 2-hour calculus class with Art of Problem Solving and then the afternoon with an MIT-sponsored thing for middle school students.

So I'm pooped.

12:47 AM

dont worry. im not here to ask questions :) i was doing problems all of yesterday and today. so im equally pooped

You don't count. You're still a young'un.

so there are always a few kids in class, every semester, who seem to know all the right answers to all the homework problems, and the exam problems (as I inevitably find out after the fact), but as soon as they open up their mouth in class, or you ask them to explain their solutions, it's pretty evident they don't know what they're talking about. what am i to think? that they get their answers from students who'd taken the class previously?

some people aren't good at explaining stuff they actually know ... witness what goes on in here.

we're all just kids

I would hope that exams change from year to year, teacher to teacher.

12:59 AM

that point of view is also affirmed by the fact that they get pretty angry if i pry, when i ask them to explain their solution

No, getting angry would suggest defensiveness because they've cheated. Are you sounding accusatory?

not at all. i ask simple things like "how did you get this?" even (especially) if i know how

I'm just saying I wasn't nearly as good at explaining stuff I'd figured out when I was in college as I am now. ... But I was better than average.

and watch them crumble

That's not encouraging. But your tone of voice could be upsetting.

1:01 AM

no no. im not asking in an obnoxious way at all

I know when I was in my 6th grade math class my teacher thought I was a know-it-all. she would yell at me to teach the class "since I know everything" and I would just sit there silently. Obviously I don't think that's what Joe Shmo does, but tone can mean a lot.

I find that I have a decent amount of difficulty in explaining my thought process for any particular math problem to others, at least in person.

My 11th grade physics teacher got upset at me a lot because he really didn't know what he was doing and I tried to explain things for the other kids in the class.

I think we've seen that a lot of young folks in here have worked hard on improving their on-line explanations and pedagogy. In person is even harder.

I once had an English teacher who had a specific case of "principle/principal" wrong, and I got in trouble when I pointed out--by asking, rather than correcting!--the error.

and hi @Fargle and @CaptainAmerica

1:03 AM

Heya @Ted.

@JoeShmo engage with the possibility that you are, because it's difficult for one to be objective about themselves

@TedShifrin Lol, yeah. Later in the year she made me tutor the other kids in class.

Wassup Ted

i am absolutely aware that i could be, even though i don't mean it. and i make an active effort not come off that way.

Well, that's a reasonable, constructive response, @CaptainAmerica.

I'd have to be a fly (or flea) on the wall to judge, @JoeShmo.

and i dont think that thats the case here

1:05 AM

We realize that you don't. But I would hope that the teacher isn't just reusing somebody's (perhaps his own) tests over and over.

if you were a fly on the wall, it would be a nifty trick for exam-taking

I did cheat once on an exam ... 8th grade history. Damn teacher believed in stupid memorize names/dates exams, and I rebelled at that.

History is the worst CLASS on the planet. I just study what I want to in my own time. At school, I learn what I need to learn and then forget it all. I kind of feel bad about it though.

That's bad teaching. As is math just about formulas bad math teaching.

i hope not. i like the guy this semester. he gave us a speech about academic honesty and how the school is aware of the problem, and is taking measures to crack down on the problem so if anyone even thinks about cheating the department is going to bring down the hammer

1:08 AM

Most faculty say that and do nothing in the face of cheating.

and im sitting there thinking its about time. but what can they do? these kids are paying crazy money to be there, and they can't just turn them away.

exactly. ill be shocked if they even get a slap on the wrist

How do I show that $z=0$ is not a local maximum of the power series $|p(z)|=|a_0+a_1z+\ldots+|$ which converges for $|z|<R$ where $R>0$?

I wouldn't want to pay money to not learn anything. That's one thing that worries me about college. Cramming and then forgetting stuff.

and how do you prove someone cheated anyway. unless its blatantly obvious

well thats the beauty of it. they can teach you, but that doesn't mean youll learn

Daminark

Why can't they? Seems they don't but like, when you go to college you don't pay for a degree, you pay for the opportunity to learn and prove that you learned. If you squander it with your own hands then you squandered your money

1:10 AM

for me learning happens when i read a book, or bang my head against the wall on a problem. not sitting in on lecture.

Some students take challenging classes to get the most out of their experience. Others just want the highest grades for the least work. Lots of different students out there.

sometimes^

I learned a tremendous amount from lectures. Different students have different learning styles. And teaching styles affect those.

I take the most challenging classes in stuff I'm interested in. My CS club leader said I have to be more well-rounded. That's why he says me taking an AP English class will be a good experience.

As long as you take it seriously and engage, sure, @CaptainAmerica.

1:12 AM

I learn best visually. So after class, I have to work out examples myself.

@TedShifrin I'm trying to.

Good, CaptainAmerica. Although I always tried to teach more visually than average ... I stress to people learning to teach that there are different learning styles and one shouldn't assume one's students learn the same way the teacher does.

i like lectures where the lecturer really knows his stuff. and on top of that is a good teacher/speaker. and thats two rare ingredients

last semester was a disaster, for instance

this semester is a lot better

The first shouldn't be rare at a quality university. The second should be a bit rarer, but not rare.

as far as lecture goes

I don't like when teachers repeat the same thing over and over again. I can see if it's really important and has to be stressed, but my calculus teacher seems to do it for no reason sometimes.

1:15 AM

true. i mean to say that both taking place at the same time is rare. but yes, in a quality university it is more common

OMG and my American Gov. teacher.

Sometimes teachers repeat things because they have only one way to say/explain something, @CaptainAmerica. Ideally, one should have different ways to do so.

Oh, yeah. I'm going to finish the problem @TedShifrin. I just ended up taking a break yesterday and then went to bed. I looked up projective geometry today and found out that there's no parallel line axiom in it. I think it sounds really interesting. Non-euclidean geometries in general.

@CaptainAmerica: For that probability problem, projective geometry isn't so good, but it's affine geometry that is the key thing.

@TedShifrin That's why when I learn something new in math. I write it on my whiteboard from memory and then explain it to someone I know. (They don't always listen lol.)But it's really helped me retain stuff and understand.

1:21 AM

Yes, explaining something to someone else is a good way to learn.

Ooh, I'll definitely look into that.

I can email you the chapter from my book on affine/projective geometry if you're interested, @CaptainAmerica. Just drop me an email (email in my profile).

That would be great! I'll just let my parents know :D

Sure.

thanks for your wisdom @Ted !

1:26 AM

Not much wisdom.

If a power series converges for $|z-c|=R$ must the absolute value of the power series be bounded above some $M>0$?

that's the kind of thing a wise dude woud say

[Random]

@user330477: Make that question very precise.

This is not a question, but a random thought that is caught from daydreaming
$$\int_0^a \frac{\Gamma (e^{\sin x})}{\cos^2 x +1} dx$$

1:29 AM

@Ted Shiffrin This is in context of this question:

How do I show that $z=0$ is not a local maximum of the power series $|p(z)|=|a_0+a_1z+\ldots+|$ which converges for $|z|<R$ where $R>0$?

Oh, that's the open mapping property for complex differentiable functions.

It's false, by the way, if $a_1=a_2=\dots=0$.

@user330477 yes to your first question, i think

sP_

Hi all, I haven't touched mathematics in a long time so stuck with this simple question. I want to find x for which f(x) = z(e^x - e^-x) + e^-x is minimum. Can treat z as a constant.

@JoeShmo: I believe it's false as stated.

Did you take the derivative and set it equal to $0$, @sP_?

snipped

sP_

1:32 AM

@TedShifrin Yes, I'm kinda stuck at how to get x when I have both e^x and e^-x after taking the derivative.

@TedShifrin We are also given that $a_i>0$ for some $i>0$.

OK, @user330477, then it's the open mapping property or maximum principle.

Or you can do it bare hands (were you the one doing this with Akiva a few days ago?).

@sP_: Substitute $u=e^x$ and you get a quadratic equation to solve.

@TedShifrin Yes, but this is not possible using bare hands as we have an infinite series. This is a different question.

@user330477: so, the problem with just announcing questions like this is that we don't know what you know and what you don't. I've given you two theorems which solve it.

sP_

@TedShifrin Ty! I think I got it.

1:36 AM

@TedShifrin I don't know open mapping property. I do know maximum modulus principle, but it is in a section afterwards.

@user330477 you wouldn't happen to have a midterm coming up would you

@JoeShmo What midterm?

complex analysis

@JoeShmo No, I have already had a midterm in that class.

not a but the

@user330477: You can bound the series by a geometric series (e.g., using the root test), and then reduce to an argument like the one you and Akiva did.

But this is a very clunky way to do it, IMHO.

You could prove the maximum principle by writing down the Cauchy Integral Formula.

1:42 AM

@TedShifrin The problem is that the series has complex coefficients and so I am not sure how we should go about doing this?

Who cares about complex coefficients?

How do you get $R$ in terms of the coefficients?

@TedShifrin We know that for $|z|<R$ we have $|p(z)|$ converges. I don't understand your question?

How do you get radius of convergence from the coefficients?

@TedShifrin Is it not just $\sup{z}$ over $|z|<R$ as the series converges for this and thus is $R$?

Go do some reading, @user330477.

1:47 AM

@TedShifrin Can you be a bit more explicit?

I already was, in my preceding question.

Pig

hi Ted

hi Piggy

Pig

random question: do you still go to any conferences these days?

Nope.

Pig

1:51 AM

ah i see

@TedShifrin Ok, sorry for the previous post. But I just saw that $R=\frac{1}{\limsup_{n \rightarrow}\sqrt{n}{|a_n|}}$? How does this help now?

What you typed is incorrect, but I assume you have the correct thing on your paper. It will allow you to bound the series by a geometric series and do a similar argument as the one that worked for a polynomial.

I'm not going to do it for you.

@TedShifrin Ok. Thank you so much.

@TedShifrin did we need Liouville's theorem or the entire theory of complex numbers to prove the fundamental theorem of algebra? it seems like we are hardly using much with the $\frac{1}{|P(z)|}$ argument

in particular, why did it take Gauss an entire doctoral thesis?

2:07 AM

@JoeShmo The question which was answered by Akiva a few days earlier gives an alternative proof of Fundamental Theorem of Algebra.

what was the question?

there are many proofs of the FTA

@JoeShmo If $a_1 \neq 0$, then $z=0$ is not a local maximum of $|a_0+a_1z+\ldots+a_nz^n|$.

ok, so if $n=1$ the claim is straight forward. does the rest follow by induction?

uh yes, i think so

3:01 AM

how to count how many automorphism on $\mathbb Z_n\times Z_m$ when m,n not relatively prime?

and I need a hint on what are prerequisites of this problem

3:49 AM

Currently I only know $|\textrm{Aut}(G)|=\phi(m)$ where $G$ is cyclic of order $m$, and in short I want to find how many automorphism on $Z_2\times Z_2\times Z_2$.

What's considered too long for an email?

4:18 AM

ping me if anyone can help, thanks!

Corellian

6:56 AM

so I was playing with a 3d plotter

Let $0\le\theta<\pi$ and $-\frac{1}{2}\le t\le\frac{1}{2}$. Set a function $f:[0,\pi)\times [-0.5,0.5]\to\mathbb{R}^3$ defined by $(\theta, t)\mapsto (x,y,z)$ parametrically: $x=(1+t\cos\theta)\cos 2\theta,\; y=(1+t\cos\theta)\sin 2\theta,\; z=t\sin\theta$

on the other hand...

Now let $0\le\phi<2\pi$ and $0\le r\le 1$. Set a function $g:[0,2\pi)\times [0,1]\to\mathbb{R}^3$ defined by $(\phi, r)\mapsto (x,y,z)$ parametrically: $x=(r\cos\phi-1)\cos 2\phi,\; y=(r\cos\phi-1)\sin 2\phi,\; z=r\sin\phi$

Their image is the same, a Mobius band of unit thickness sitting about the origin. However, I think $f$ is a bijection while $g$ does a weird "double winding" thing

Oops. For $g$, it's supposed to be $z=r\cos\phi$.

Corellian

7:50 AM

Coming back, still left a few mistakes... For $g$, $0\le r\le \frac{1}{2}$ and for $x$ and $y$ the inside $\cos\phi$ should be $\sin\phi$. NOW $f$ and $g$ have the same image, and $f$ is one-one (not a bijection oc) while $g$ is not, e.g. $g(0,0)=g(\pi,0)$

Riddle me this, topologists

Can a compact hypersurface be non-orientable?

(assuming the original manifold is orientable)

Q: Every compact hypersurface in $\mathbb{R}^n$ is orientable

9:13 AM

1

Show that every compact hypersurface in $\mathbb{R}^n$ is orientable. HINT: Jordan-Brouwer Separation Theorem. This is an exercise from Guillemin and Pollack. So hypersurface means smooth hypersurface. Jordan's Theorem says that if $S$ is some hypersurface in $\mathbb{R}^n$, then $\mathbb{R}...

9:30 AM

is that a single element both chain and anti-chain?

11:11 AM

@IsanaYashiro there are 168 automorphisms of $Z_2 \times Z_2 \times Z_2$

do you know linear algebra? $Z_2 \times Z_2 \times Z_2$ is a vector space over $\Bbb F_2$ and any group automorphism is $\Bbb F_2$-linear

I like to think you spent the 8 hours from his message counting them by hand

lol

it didn't take my long, I just entered chat

@Slereah Hypersurface meaning dimension $n-1$?

Q: Is the functor that assigns to an algebra its algebraic group of units fully faithful?

I know, I suppose there is either some neat trick or a huge online database

abenthy

0

Let $A$ be a $d$-dimensional algebra over a field $K$. One can naturally assign to $A$ the linear algebraic $K$-group $\mathbf{GL_1}(A)$ that represents the functor $B \mapsto (A \otimes_K B)^\times$ for $K$-algebras $B$. My question is: Is the functor $A \mapsto \mathbf{GL_1}(A)$ from finite...

abstract-algebra category-theory algebraic-groups

11:13 AM

$168=(2^3-1)(2^3-2)(2^3-2^2)$, that's just combinatorics

(this is not an exclusive or)

we're looking for the order of $GL_3(\Bbb F_2)$

Oh, right, it's a vector space

so for the first row, there are $2^3-1$ possibilities, for the second there are $2^3-2$ possibilities, since it can't be a scalar multiple of the first row and for the third one, there are $2^3-2^2$ possibilities since it can't be contained in the span of the first two rows

this generalizes to a formula for the order of $\mathrm{GL}_n(\Bbb F_q)$

@Slereah Isn't ${\rm\Bbb RP}^3$ orientable?

So take the subspace ${\rm\Bbb RP^2}$

11:33 AM

The noun form of ominous is ominousness

ÍgjøgnumMeg

ew

I'd like ominosity

abenthy

@MatheinBoulomenos Hi, may I ask you a follow up question to my question on the unit group functor?

My original motivation for the question was the following: If $A$ is a $4$-dimensional central simple algebra over a field $K$, is every isomorphism $SL_1(A) \to SL_2$ of algebraic $K$-groups coming from a $K$-algebra isomorphism $A \to M_2(K)$?

11:49 AM

@AkivaWeinberger The correct general statement is "Every 2-sided hypersurface in an orientable manifold is orientable"

One can show that intersection number with a one-sided hypersurface defines a nonzero map pi_1(M) -> Z/2, and so every hypersurface is orientable as long as H^1(Y;Z/2) = 0

Should be true that your oriented manifold has a one-sided hypersurface iff H^1(Y;Z) -> H^1(Y;Z/2) is not surjective

12:15 PM

(^@Slereah)

4 hours ago, by Slereah

(assuming the original manifold is orientable)

Yes

${\rm\Bbb RP}^3$ is orientable

Is it?

Dang

But I wanted to point you at Mike's post there^

Q: Uniform Limit of Integrable Functions

Thx

user193319

12:47 PM

0

If $\{f_n\}$ is a sequence of bounded measure functions on $[a,b]$ and $f_n \to f$ uniformly, then $f$ is integrable. This is actually proved in the book I am working through, but I came up with a proof that looks quite different. Naturally, I was hoping someone could critique my proof and...

real-analysis integration sequences-and-series uniform-convergence riemann-integration

Gabriele Petrioli

1:44 PM

Hey everyone,
trying to solve a problem and do not know how to name/search for the related algorithm.
I need to find the Y value which will result in the minimum sum of two functions that accept it as a parameter.

In case what i describe is a XY problem, the full problem trying to go from point A to B where while in the -X the speed is S1 and while in the +X the speed is S2. I need to find the fastest route between them (so find the Y, the X will be 0, which will provide the fastest route)

Mats Granvik

1:56 PM

$\zeta \left(\frac{1}{2}\right)=\lim_{k\to \infty } \, \left(\sum _{n=1}^k \left(\frac{1}{\sqrt{n}}-\frac{2}{\sqrt{k}}\right)\right)$

2:18 PM

Is there any significance behind the fact that $\left(\dfrac fg\right)'=-\dfrac1{g^2}\det\begin{bmatrix}f&g\\f'&g'\end{bmatrix}$?

well, if $f=\lambda g$, then that determinant vanishes because the two functions are linearly dependent

And conversely.

Huh - that's the Wronskian.

yep

that's what I had in mind, to be clear

in turn you can probably link this up with how second-order ODEs work, since Wronskians usually show up there

Is there a connection to the version with more than two functions, I wonder?

I doubt it tbh. When you've got more than two functions, the condition for linear dependence isn't that a certain ratio is constant

2:29 PM

Yeah. I can generalize the RHS but not the LHS

Not even, 'cause of that $1/g^2$.

rschwieb

2:43 PM

wave

Liad

Hi!

i need to find an example of a morphism in the category of groups (Grp) that is has a right inverse but not a left inverse. someone can help?

@Semiclassical Through a pointless calculation, I have determined that $\left(\dfrac{(f/h)'}{(g/h)'}\right)'=-\dfrac h{(g'h-gh')^2}W(f,g,h)$

So it kinda is generalizable

But that's also really ugly

@AkivaWeinberger Huh. This seems to detect linear dependence decently well (e.g., try $f(x) = x^3 + x^2$, $g(x) = x^3$, $h(x) = x^2$), but I can't tell if it always does

Start with $Af+Bg+Ch=0$, divide by $h$ and differentiate

Oh! Duh. That makes sense.

2:52 PM

You get $A(\frac fh)'+B(\frac gh)'=0$

Divide by $(\frac gh)'$ and differentiate again

Hm, is $W(f,g,h)$ equal to $W(W(f,h),W(g,h))$?

The Wronskian is linear in its arguments, so the same thing from above works (kinda)

I think it equals $hW(f,g,h)$, actually

'cause doubling $h$ will quadruple it

Conjecture: $W(f,g,h)=\frac1hW(W(f,h),W(g,h))$

I dunno what to do with this information

yuck

...yikes

It doesn't fit in my screen

yeah same

The calculation checks out though, at any rate

Your conjecture holds

The two $fgh'h''$ terms cancel out and the remaining six terms have an extra $h$

3:03 PM

Indeed

Does that generalize in any nice way to, say, a 4d Wronskian being decomposable as a 3d Wronskian of Wronskians (or a Wronskian of Wronskians of Wronskians)?

At a guess, $W(f_0,f_1,f_2,g)=\frac1{g^2}W(W(f_0,g),W(f_1,g),W(f_2,g))$

which is a 3-input Wronskian with three 2-input Wronskians inside it

and in general I conjecture that $W(f_0,\dots,f_n,g)=\frac1{g^n}W(W(f_0,g),\dots,W(f_n,g))$.

To verify this, of course, you will need to take the $n$th derivative of a (two-input) Wronskian.

I don't wanna :(

Yeah, let's not

(Alternatively, there might be some identity involving determinants in general, but I dunno)

Or maybe there's a unique polynomial in those variables that detects linear dependence and which is linear in those terms (up to scalar multiplication)

Er, we don't know if the RHS is even a polynomial

StammeringMathematician

3:22 PM

math.stackexchange.com/questions/2964434/… Please have a look at this question of mine. It is related to complex analysis. I think I am missing something obvious here

3:33 PM

Hey, if $f(x,y)$ is a real function, what's the notation for $f_{xx}+f_{yy}$?

Also: Is this the same in any reference frame (aka if I choose any two orthonormal vectors)?

Should be

How to find volume of a beam whose base is 1 inch and height is .25 inch, its a regular beam like in a shape of a rectangle, wouldn't the volume just be (.25*.25*1) bc the height and width would be the same right?

Ik this is a really dumb question and def not in the level of what i should be asking here but this is confusing me cuz the answer is supposed to be 9 inches and idk why?

@AkivaWeinberger $\Delta f$

@MATHASKER Are you sure it's supposed to be 9? I don't see how that could be

@MatheinBoulomenos Ah, right, Laplacian

$\nabla\cdot\nabla f$

@AkivaWeinberger yes like they went over the problem in the class and it was 9 inches we were calculating the moment of intertia for like a regular beam

9 inches can't be a volume

3:40 PM

Inches cubed*

I just don't know how they got 9 inches cubed

It sounds much smaller than a 1x1x1 cube, which would have volume 1

so I think you probably misheard something in class

hey guys, does anyone know why in perturbation theory, it makes sense to expand our solution as a series, of the form $f_0+\lambda f_1+\lambda^2 f_2+\dots$, where $\lambda$ gives the 'strength' of the perturbation and $f_i$ is the $i$-th order correction to the solution? I can only find sources online that just take it as an ansatz, without really any justification, that would help with intuition

@MatheinBoulomenos What does $\Delta f=0$ represent for a function, visually

Hmm ya that could prolly be the case thanks anyways

@ShaVuklia I dunno anything about perturbation theory but that's just a Taylor series in $\lambda$, right?

3:46 PM

@Akiva I guess so.. would that mean that $f_i=f^{(i)}$? up to some constants maybe

I would guess, yeah

$f^{(i)}/n!$

What's the Laplacian of $\ln(x^2+y^2)$?

$0$

What does that mean for the function

Q: The riemannian metric of a neighborhood of the boundary of a compact manifold

it means that the value at a point is well approximated with the average value on a small square around it

Anderson Felipe Viveiros

0

Let $M$ be a compact riemannian manifold with boundary $\partial M$. We have that $\partial M$ is also compact and I was able to show that there is some $a>0$ such that the map $F:[0,a]\times \partial M \to M$ given by $F(r,p)= \exp_p(r\nu(p))$ is a diffeomorphism onto its image, say $U$ (here $\...

riemannian-geometry manifolds-with-boundary riemannian-metric

Up to a degree-two error I guess? @mercio

4:02 PM

the point is that the degree two error should vanish i believe

Ah, so degree-three.

oh maybe i forgot to integrate and then its the degree one

4:18 PM

Is there a natural interpretation of $f^2\Delta(\ln f)$?

@Akiva it is true that we can generally (aka, with the right assumptions) write $f(x,y)=\sum \dfrac{\partial^i f(x,y)}{dy^i}y^i/i!$?

I believe so

ah cools, then I think that's what they're doing with perturbation theory

oh, wait actually no:(

oh right, so we get $f_i(x)=\sum \dfrac{\partial^i f(x,0)}{\partial\lambda^i}\lambda^i/i!$

(without the summation sign)

and without $\lambda^i$:p

Alternatively, is there a natural interpretation of $\frac1f\Delta(\ln f)$?

4:33 PM

Delta = second derivative?

Laplacian @Semiclassical

$f_{xx}+f_{yy}$

How do I show that a harmonic function has partial derivatives of all orders? I know by a theorem that this function is the real part of some analytic function. Also, an analytic function is infintely differentiable.

What to do next?

5:02 PM

If $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, then Cauchy–Riemann says that $u_x=v_y$ and $u_y=-v_x$, right?

@user330477

I claim that $f'(z)=u_x+iv_x=v_y-iu_y$.

That is, $f'(z)=\frac\partial{\partial x}u(x,y)+i\frac\partial{\partial x}v(x,y)= \frac\partial{\partial y}v(x,y)-i\frac\partial{\partial y}u(x,y)$.

@AkivaWeinberger Yes. I see why it is true. Just use definition of analyticity and take two different paths.

So, we know that the analytic function $f$ is infinitely differentiable

and we want to show that that means that $u$ has all partials

I'll claim that $u$ and $v$ both have all partials

@AkivaWeinberger This claim is what I don't see.

I just showed that $u_x$, $u_y$, $v_x$, and $v_y$ all exist, right?

Because they're real and imaginary parts of $f'$ (up to a sign).

What are the real and imaginary parts of $f''$?

If I differentiate $f$ both times along a horizontal path, I get $u_{xx}+iv_{xx}$.

If I differentiate $f$ first along a horizontal path and then along a vertical path, I get

$f'=u_x+iv_x$ and so $f''=v_{xy}-iu_{xy}$.

What if I differentiate vertically and then horizontally?

@MatheinBoulomenos: Thank you sir!

5:11 PM

@IsanaYashiro you're welcome (and you don't need to call me sir)

Also: I didn't actually know that this theorem was true. I wonder if there's a direct proof that doesn't go through analytic functions?

@AkivaWeinberger Yes, this good upto second derivative. But my problem is how do you show existence from third derivative onwards. This must somehow follow from the fact that $f$ has complex derivatives of all orders.

Also, it only works for 2D functions, right? Is it true in other dimensions?

it's true in other dimensions as well

$\Delta f = 0$ implies $f$ is analytic

@user330477 To get $u_{xyx}$ differentiate $f$ horizontally then vertically then horizontally

5:12 PM

@MatheinBoulomenos Although I haven't understand it(because I just at the beginning about group theory), but I will spend time thinking your words

@MatheinBoulomenos ? $f$ is a real function

Oh, analytic

Not holomorphic

this is an instance of elliptic regularity, you should ask Eric or Mike when they're around about it, they know this PDE stuff

@MatheinBoulomenos Could you recommend some book about abstract algebra?

I did search some, but I would like to know your idea~

@IsanaYashiro I liked "Aluffi - Algebra Chapter 0", but I didn't read it without knowing some algebra beforehand, so it might be difficult as a first book on abstract algebra

I would like to learn something called burnside theorem(lemma?) and polya, that's the reason I start to learn group theory, and I spent a lot of time understand quotient group, then I found the idea of coset so beautiful

Daminark

5:19 PM

Hello Sir @Mathein!

@AkivaWeinberger yes, there's a proof that avoids complex analysis, you show the Poisson integral formula (similar to Cauchy's) and then derive analyticity from that

@MatheinBoulomenos It's ok thank you!

Hello memelord @Daminark

So in any case I was curious if there was a formula for the curvature of a Riemannian surface given a conformal map

since I already derived a formula for the geodesics in that case

(This is the case where the Tissot ellipses are circles)

The usual elliptic regularity arguments get you smoothness

But for elliptic PDEs with analytic coefficients, you do indeed get analytic solutions

5:24 PM

If the metric is $g_{ij}=g\delta_{ij}$ then the radius of the Tissot circle is $s:=1/\sqrt g$

But the argument is different and I don't know it off the top of my head

and the curvature is, apparently, $s^2\Delta(\ln s)=-\frac1{2g}\Delta(\ln g)$

and I dunno if there's any intuition for that

($\Delta$ is Laplacian)

The standard citation is a paper by Morrey which has very general results

But there's also an approach called "microlocal analysis", in which you can study the "wave front set" at each point, which sort of tells you the directions in which non-analyticity propagates - this will make sense if you ever see pictures of solutions to the wave equarion, it's mostly fine but gets screwed in certain cones

For an elliptic PDE the wave front set is just seen to be empty

i heard elliptic where the party at

Interesting

5:33 PM

I got really interested in this a couple years ago and took a class on microlocal analysis but didn't end up finding the time to get to this theorem

While the analyticity is not so useful in practice (just interesting), note that it implies strong identity theorems for solutions to elliptic PDEs

Can't agree at a point to infinite order, can't agree on an accumulation set

Ehhh drop the latter, I am thinking of the complex analysis case

But these sorts of identity theorems are usually true as long as your equations are well-behaved elliptic and I constantly need them

There is a baby on this flight which has been screaming almost the entire time; normally I have sympathy because it's them expressing pain because their ears are popping or whatever

Toddler

But it has become clear it's just throwing a 5 hour temper tantrum because it's being asked to sit still

I am impressed it has so much energy and unimpressed by the parents

5:49 PM

5 hour temper tantrum? Wow, what a baby

Nothing more horrifying to a toddler than five hours of boredom

Honestly, big mood