Mathematics - 2016-12-12 (page 1 of 6)

Kasmir Khaan

12:00 AM

but i dont see any difference between direnctional derivative and analytic

they seem to be the same thing

GFauxPas

hmm

well is this theorem true for directional derivatives?

Ted Shifrin

@Kasmir: That's the magic of the complex derivative. Having a complex derivative (which gives you the Cauchy-Riemann equations) makes the Cauchy Integral Formula work, and this gives you infinitely many derivatives and, in fact, a convergent Taylor series. Magic.

GFauxPas

If a complex function is continuously differentiable, it's infinitely continuously differentiable

which is absolutely amazing

Ted Shifrin

But the only way to prove that is with the Cauchy Integral Formula, @GFauxPas.

GFauxPas

Is that so? I'll take your word for it

Ted Shifrin

12:01 AM

LOL, that's generous of you :D

GFauxPas

But I don't know if you need to know about residues to understand the inuition of the complex derivative

:)

Ted Shifrin

Nothing to do with residues.

GFauxPas

Oh I'm thinking of the residue theorem

Kasmir Khaan

hmm it is still not clear to me to be honest

Ted Shifrin

There's some serious mathematics going on, @Kasmir. It is not intuitively obvious.

WDUK

12:03 AM

what is non serious math then ?

GFauxPas

bad math jokes

Akiva Weinberger

It's a matter of degrees

WDUK

got any examples?

lol !

GFauxPas

here's another way to think about complex differentiability

Kasmir Khaan

okay that helps :D i need to pay extra attention on class next 4 lectures.

Ted Shifrin

12:03 AM

Things like the Division Algorithm for integers are completely intuitively obvious.

Mike Miller

Non-serious math is what I do.

GFauxPas

remember how some functions have power series, and if you're lucky the power series converge to the function exactly , and everywhere

leo

@TedShifrin There are other involved ways see

Ted Shifrin

Even the Fundamental Theorem of Calculus is an important theorem, but it's completely "obvious" and believable to anyone who sees the picture.

Akiva Weinberger

I think you should try to work out why $\bar z$ is not complex differentiable (that is, not analytic), by using the definition to try to find the derivative at $0$ @KasmirKhaan

Kasmir Khaan

12:04 AM

thanks all ! :)

GFauxPas

if a function is complex differentiable, it has a power series that converges everywhere that it's complex differentiable

Ted Shifrin

@leo: That link is talking about something else.

Akiva Weinberger

You end up with $\lim_{h\to0}\dfrac{\bar h-\bar 0}{h-0}=\lim\dfrac{\bar h}h$

If $h$ is real and approaching $0$ then it's $1$,

but if $h$ is imaginary and approaching $0$ then it's $-1$.

Kasmir Khaan

intressting

GFauxPas

and you only have to find two paths that are different to ruin differentiability, because differentiability requires EVERY path to converge to the same limit

Akiva Weinberger

12:06 AM

Thus it's not differentiable at $0$, and a similar argument shows it's not differentiable anywhere

Kasmir Khaan

how can i save this conversation ?

GFauxPas

memorize it, then burn your computer

Kasmir Khaan

I think it will make sense after i go to the class

leo

I mean, in his book he proves that if a complex function is continuously differentiable, it's infinitely continuously differentiable without Cauchy's formula

Kasmir Khaan

-.-

Balarka Sen

12:06 AM

@MikeMiller If I understand the terminology right, I'd say leftist governments always used such fear tactics, at least from what I heard of the world.

GFauxPas

Another surprising thing about analytic functions

Ted Shifrin

@leo: OK, I don't know the book and so I can't argue. What I said is philosophically quite correct, even if it's not literally correct.

Akiva Weinberger

@KasmirKhaan You could make a permalink of a particular comment

Mike Miller

@Balarka You have no idea what you're talking about.

Akiva Weinberger

http://chat.stackexchange.com/transcript/message/34054377#34054377

Kasmir Khaan

12:07 AM

okay thanks @AkivaWeinberger

Ted Shifrin

I can discuss elliptic regularity without mentioning the Cauchy Integral Formula, too, @leo, but it is still a huge cannon to shoot a fly.

leo

@TedShifrin Indeed. I don't want to argue either. Just saying :-)

GFauxPas

is that, under certain assumptions, the definition of the function on a simply connected domain UNIQUELY define the function even outside the domain it's defined on

so if two analytic functions agree on some shared simply connected domain, they have to agree everywhere else they're analytic, barring something going wrong

Akiva Weinberger

@KasmirKhaan Apparently, on the transcript (what I linked to) you can bookmark a conversation by clicking that button on the right

Kasmir Khaan

All right thanks guys ! :) i should keep working on more porblems

Mike Miller

12:09 AM

@Ted I think about it in terms of elliptic regularity, but ultimately elliptic regularity is a general version of the Cauchy integral formula (an expression of the differential operator as convolution with some kernel)

Kasmir Khaan

I got that akiva :)

Ted Shifrin

I know, @MikeM. Or you can do it with pseudodifferential operators :P

GFauxPas

the book I use is by Brown and Churchill

it's dirt cheap on amazon

Mike Miller

Yes, we know the story.

GFauxPas

Complex Variables and Applications

Ted Shifrin

12:11 AM

@GFauxPas: If two analytic functions have the same domain and agree on any set with a limit point, then they have to be the same everywhere.

Kasmir Khaan

Saff and Snider: Fundamentals of complex analysis, Prentice-Hall.

this is the book they asked us to get

is it any good?

Balarka Sen

@MikeMiller I know some isolated stories from pre-WWII communist republics; I'd say the politics in there were founded upon fear.

Semiclassical

Fear is a tool of any authoritarian government, rightist or leftist.

Ted Shifrin

It's fine, @Kasmir. These are all pretty much the same sort of undergraduate textbook with some applications. They're more involved than a calculus course, but not as sophisticated as a graduate level text.

GFauxPas

@TedShifrin oh even if it's not simply connected? awesome

Ted Shifrin

12:16 AM

Simple connectivity is important for things like defining logs, @GFauxPas ... back to your branch issues :P

GFauxPas

oh bother

Semiclassical

The politics professed by such a regime seem basically meaningless; what matters is power and control.

Mike Miller

@Balarka You are completely misunderstanding what the term politics of fear in that article is, the use of the term "the left" as if it refers to a government in the US, the use of fear of other as political motivator vs fear of persecution by the state, conflating the USSR with "leftist government", ...

GFauxPas

guys this isnt math

Mike Miller

Any other observations?

GFauxPas

12:19 AM

plenty

Balarka Sen

@MikeM Ah, ok, fair enough. (I disagree on the interpretation of what I said at the 2nd and the 4th point, however, but that's pointless anyway).

Akiva Weinberger

The set of possible political views forms an infinite-dimensional metric space

^ this is math

GFauxPas

can you have an uncountably infinite dimensional space?

I guess that's what $\mathcal{C}^\infty$ is

Akiva Weinberger

Sure, set of functions from $[0,1]$ to $[0,1]$, supremum metric

Probably

Balarka Sen

direct product of R's.

or R over Q with Hamel basis.

Akiva Weinberger

12:31 AM

The Dehn invariant of the tetrahedron was ${\rm something}\otimes{\rm irrational~times~}\pi\in\Bbb R\otimes_{\Bbb Q}\Bbb R/\pi\Bbb Q$

Why are we sure this isn't $0$?

Ted Shifrin

DogAteMy: That question is a little vague (and I don't know the topology in question). Are you worried about why $\frac1\pi\cdot\pi\otimes_{\Bbb Q}\Bbb R/\pi\Bbb Q$ is nonzero? That really doesn't even make sense. :)

Kasmir Khaan

abs (z-a) = r how to see that is a circle

z is complex ?

Ted Shifrin

Think.

GFauxPas

start with $\vert z \vert = r$

Mike Miller

I have no idea what a Dehn invariant is.

GFauxPas

12:41 AM

and think about how that's a circle

Kasmir Khaan

thinking give me one second

Akiva Weinberger

Specifically I'm not sure why $6(72)^{1/6}\otimes\arccos\frac13\ne0$

Like, how we know for sure

Kasmir Khaan

abs(z) = x^2+y^2

Akiva Weinberger

I know $\arccos\frac13$ isn't a rational multiple of $\pi$

Balarka Sen

Niven's theorem

In mathematics, Niven's theorem, named after Ivan Niven, states that the only rational values of θ in the interval 0 ≤ θ ≤ 90 for which the sine of θ degrees is also a rational number are: sin ⁡ 0 ∘ = 0 , sin ⁡ ...

Akiva Weinberger

12:42 AM

@BalarkaSen I know that

Balarka Sen

oops, ok

Ted Shifrin

Balarka is supposed to be fast asleep.

Akiva Weinberger

But sometimes $a\otimes b=0$ even when $a,b\ne0$

Ted Shifrin

Well, that's what I was about to talk about. You're thinking of this as an element of the tensor product of what two rings?

Akiva Weinberger

I think I need to find some $\Bbb Q$-linear function from $\Bbb R\otimes_{\Bbb Q}\Bbb R/\pi\Bbb Q$ to $\Bbb R$ that doesn't map it to $0$

GFauxPas

12:43 AM

@KasmirKhaan yup

Kasmir Khaan

ahhhhhh the distance between z and a is r

makes it a circle :D

:)

thanks ! :)

Sure

which would first involve some linear function from $\Bbb R/\pi\Bbb Q$ that doesn't map it to $0$

Kasmir Khaan

12:44 AM

am supposed to show that the line integral dz/ (z-a)^k = 2pi*i

for k =1

Ted Shifrin

So you can only end up with a $0$ answer if you have a rational multiple of $\pi$.

Balarka Sen

@AkivaWeinberger Well $\Bbb R$ is free as a $\Bbb Q$-module, yeah?

Akiva Weinberger

I don't remember what free means

Ted Shifrin

no torsion

Akiva Weinberger

@TedShifrin That's what I'm trying to prove

meow-mix

12:45 AM

hi

Ted Shifrin

hi @meow ...

Akiva Weinberger

I think I would be done if I had a linear function from $\Bbb R/\pi\Bbb Q$ to $\Bbb R$ that didn't map $x\notin\pi\Bbb Q$ to $0$

meow-mix

hey, @TedShifrin, what was that geometry problem you said i should do?

Ted Shifrin

You have several, @meow. Why did you get parallel lines when you bisected sides? Why are you ending up with circles in those constant-angle constructions?

Akiva Weinberger

Don't I need Choice or something?

GFauxPas

12:46 AM

Are there functions with primitives that aren't analytic?

Ted Shifrin

You mean analytic functions, @GFauxPas?

Yes. $1/z$ ... depending on domain.

Akiva Weinberger

@GFauxPas 1/z?

GFauxPas

I meant, are there function that are not analytic but have primitives

Akiva Weinberger

Oh

I think not, it should follow from how differentiable functions have all derivatives

so the second derivative of the primitive should exist

Ted Shifrin

@GFauxPas: Huh? Then we're back to regular calculus?

Whoa, whoa. Slow down.

GFauxPas

12:48 AM

I meant

Like, the thm that f has a primitive iff the integral along a closed contour is zero

Ted Shifrin

I want a precise question.

What hypotheses on what?

GFauxPas

Is there a function that had the property that it's integral on every closed contour is zero, buy isn't analytic?

Ted Shifrin

Continuous function?

GFauxPas

Yeah sure

Ted Shifrin

I.e., no branch issues?

Then, no. This is called Morera's Theorem.

GFauxPas

12:51 AM

Right

Ted Shifrin

(Assuming you're still talking about complex functions on open subsets of the plane.)

meow-mix

@TedShifrin is complex projective space the same as the riemann sphere?

Kasmir Khaan

integral dz / (z-a)^k = 2pi*i for k=1 , 0 for other natural numbers

GFauxPas

Cool, Googling moreras thm now, thanks Ted

Kasmir Khaan

should I parametrize my curcve?

abs (z-a) =r

Ted Shifrin

12:52 AM

@meow: $\Bbb CP^1$.

meow-mix

oops

meant the complex projective line

Ted Shifrin

Yes. You'll get to some of that if you keep reading the chapter :)

Balarka Sen

Then yeah, just like RP^1 is a circle

GFauxPas

Circles centered at $0$ are trivial to parameterize if you know about the polar form of complex numbers

meow-mix

or a line with point at infinity ;)

@GFauxPas yes. $re^{i\theta}$

GFauxPas

12:53 AM

@meow-mix I haven't seen that

Kasmir Khaan

guys please help , how to parametrize abs (z-a) = r ? is it x=a+cost , y=a+sint ?

Ted Shifrin

The Riemann sphere is in that book, @GFauxPas. They might call it the extended complex plane.

meow-mix

@TedShifrin which chapter?

Ted Shifrin

@Kasmir: if you're doing complex stuff, use $e^{it}$.

GFauxPas

I didn't ask about the Riemann sphere

meow-mix

12:54 AM

also, after this you want me to read the next 2 chapters?

Kasmir Khaan

@TedShifrin but a could be any value , how would i parametrize that ?

Ted Shifrin

$a$ is in there, too, Kasmir.

@meow: You've asked me that a bunch of times, and I've said yes, if you're interested.

GFauxPas

Do a change of variables to put the circle centered st the origin; at least, that's what I do with conics

Ted Shifrin

Sections, not chapters. But meanwhile there's lots of exercises in section 2.

Kasmir Khaan

e^it is unit circle now I have to figure out where that "a" should fit

meow-mix

12:56 AM

indeed.

Ted Shifrin

Kasmir: Seriously. You need to think more and ask fewer questions.

You've done this a bunch of times already.

Adeek

hey @arctictern are the only subgroups of order 8 in $S_4$ $D_8$ ?

Kasmir Khaan

sorry ! it is just stressfull how many things am supposed to know to the exam

sorry again all

Ted Shifrin

Actually, the quaternion group doesn't embed as a subgroup of $S_4$, @Balarka, right?

Adeek

yeah

it doesn't

Balarka Sen

12:57 AM

That's why I deleted it

Ted Shifrin

So, Karim, the only nonabelian groups of order 8 are ... and the only abelian ones are ...

GFauxPas

Kasmir, just like in regular calculus, substitutions of the form $u=ax +b$ are very easy to do

$du = a dx $

So make the circle at the origin

Ted Shifrin

Karim: What are the Sylow 2-subgroups of $S_4$? you can see that geometrically.

Adeek

$Q_8$ and $D_8$ why can't it be abelian though ?

why can't the subgroup of order 8 be abelian ?

Ted Shifrin

Easy answer to this, Karim. Use Sylow, as I just asked you.

Adeek

12:59 AM

we have that $n_2 = 1$ so $n_2 = 1\ or\ 3$

Ted Shifrin

But what do you know about all Sylow p-subgroups?

And do you not know that $S_4$ is the group of symmetries of the cube? So you can see everything geometrically.

Adeek

they don't have trivial center.

Ted Shifrin

Most basic result, part of the Sylow theorems is that all Sylow p-subgroups of any finite group are ... what?

Adeek

oh really ?

conjugates ?

Balarka Sen

Bingo

Adeek

1:03 AM

oh

Ted Shifrin

But you can see the Sylow 2-subgroups of the cube group very geometrically. Think about how $D_4$ (which you call $D_8$) sits in there geometrically.

Adeek

let us say

oh I see it now. We have reflection

like if we draw a line cutting the middle of the cube

and we can reflect across that line and also rotate vertices as well right ?

Ted Shifrin

So you have symmetries of (pairs of) faces.

Adeek

yeah

cool

Ted Shifrin

And you see clearly that there are 3 such conjugate subgroups :)

Adeek

1:08 AM

yeah

Ted Shifrin

That's my algebra wisdom for you for the year.

Adeek

that is awesome :) haha

I am almost done preparing for algebra :D I feel like 90 % prepared

just have to review galois theory and rep theory then I am done

meow-mix

@TedShifrin am i considering this projection of $\Bbb P^1$ to be given by $x_0 = 1, x_1 = x, x_2 = y$?

Ted Shifrin

Huh?

meow-mix

like

hmm i don't know how to phrase this

so right now im in $\Bbb R^2$

Adeek

1:11 AM

@TedShifrin My exam schedule for next semester it looks I have 2 exams in the same day

** is wrong with this stupid thing system.

I should probably ask prof next year I can't have exam in same day

Ted Shifrin

People scheduling classes can't avoid all possible issues.

Grow up.

You're sounding like a whiny undergraduate.

haha

I'm serious.

yeah I guess lol

Ted, what's wrong with 1/z that you can't apply Korea on

Ted Shifrin

1:12 AM

LOL, Korea?

What's the integral around the unit circle?

GFauxPas

Ted, what's wrong with 1/z that you can't apply Morera on it

Ted Shifrin

That's exactly the whole point :)

GFauxPas

Also, autocorrect

Ted Shifrin

LOL, oh. I hate that thing. It happens on my phone but not on my desktop.

meow-mix

@GFauxPas some of its contour integrals have non-zero value?

Ted Shifrin

1:13 AM

@meow: You keep getting distracted !!

GFauxPas

Oh whoops, was mixing up sufficient and necessary

I'm on my phone

meow-mix

well im really not sure how im supposed to take $R^2$ and apply the projective stuff to it

GFauxPas

I guess the converse isn't a thm :)

Ted Shifrin

@meow: You did that for the medians problem. $\Bbb R^2\subset\Bbb P^2$, so you can take a figure that lives in the Euclidean plane, think of it in the projective plane, and deduce things.

meow-mix

adios!

oh

wow

my messages sent in the wrong order..

what i said was

" i have an essay due in 4 hours, adios"

anyways bye

Ted Shifrin

1:18 AM

Bye. :)

GFauxPas

@Ted come write for proofwiki with us

Ted Shifrin

Huh?

GFauxPas

It's a website of theorems and definitions

proofwiki.org

Mike Miller

The arXiv is similar but more expansive.

GFauxPas

No its not a repository for papers

It's a repository of theorems

Mike Miller

1:28 AM

yeah but what's a paper but a repository of theorems

Ted Shifrin

A lot more than a repository, I hope.

GFauxPas

Proofwiki catalogues theorems and proofs, you don't read a page in proofwiki, you follow a chain of links

It's a wiki

You don't read a paper I mean

CausingUnderflowsEverywhere

hey guys, Im the guy who asked how to do trig stuff without a calculator, I'm back with a new question.

When factoring a trinomial,
Why do we do a * c and then find the factors of that result that when added or subtracted equal the middle term?

How did we come up with a * c?

oh fail

GFauxPas

proofwiki.org/wiki/…

This one I wrote and I like it

Akiva Weinberger

@CausingUnderflowsEverywhere I think there's a question on the main site on that, let's see if I can find it

@CausingUnderflowsEverywhere I think the idea is that, if the first coefficient is $1$ (so we have $x^2+bx+c$),

and we want to factor it into $(x+P)(x+Q)$,

(By the way, do you have LaTeX enabled? It turns the dollar-sign things into cool rendered math formulas)

...then (by expanding the last thing) we essentially want $x^2+bx+c$ to be $x^2+(P+Q)x+PQ$.

And it's clear that we want numbers $P$ and $Q$ that sum to $b$ and multiply to $c$.

Mike Miller

1:38 AM

@GFauxPas I'm being disingenuous. In any case, I still don't quite understand the point of the website. I don't see anything it fixes that was broken before.

Akiva Weinberger

If the first coefficient isn't $1$, we end up wanting something of the form $(Px+Q)(Rx+S)=PRx^2+(PS+QR)x+QS$ to be equal to $ax^2+bx+c$.

GFauxPas

Oh it does a lot, but the mission statement there is more eloquent than what I could say. Part of its uniqueness is its accessibility without sacrifice of rigor

Akiva Weinberger

There we can notice, trying to generalize the version with $1$ as the first coefficient, that $b=PS+QR$ is the sum of two numbers that multiply to $PSQR=(PR)(QS)=ac$.

The equation becomes easier to factor once you have these two numbers $PS$ and $QR$, and now we know how to look for them.

Mike Miller

I can't find the mission statement but I doubt I would be convinced.

I'm going to play picross.

GFauxPas

proofwiki.org/wiki/ProofWiki:About

Well of you won't read it then that's all there is to it

Ted Shifrin

1:42 AM

reminds DogAteMy that he has a paper to write and submit :D

@GFauxPas: I've found enough stuff wrong on Wiki that I'm not going to worry about trying to proofread or check proofwiki.

Mike Miller