Mathematics - 2018-05-02 (page 1 of 4)

0celo7

00:00

the parametrization definition is the odd one out

Akiva Weinberger

50 mins ago, by Balarka Sen

Like, take a loop and twiddleschloopkinfloop it to floop it all into $\Bbb R^n$ and slurpslorp

0celo7

can't type

Sha Vuklia

right, so we get $Df(\phi(x))D\phi(x)=0$, which we can write as $\nabla f(\phi(x))\cdot D\phi(x)$. So we see that all vectors of $D\phi(x)$ are perpendicular to the gradient of $f$ at a point of the manifold, and since they span the tangent space, we know that all those vectors are perpendicular to the gradient. Also, the inclusion goes both ways because of the dimensions

Akiva Weinberger

@BalarkaSen I feel like there's a 60% chance you actually had something in mind when you typed that ^

GFauxPas

Akiva, that was us proving the fundamental group of $S^n$ is $\mathbb Z^n$

Ted Shifrin

00:00

Superb, @Sha. Kudos.

Sha Vuklia

yay

Ted Shifrin

I hope not, @GFaux.

0celo7

@GFauxPas $\pi_1(S^n)=\Bbb Z^n$?

Balarka Sen

@GFauxPas $0$ you mean lol

0celo7

that's...not good

Balarka Sen

00:01

@AkivaWeinberger I did

GFauxPas

errr I was thinking $T^n$ sorry

Akiva Weinberger

sigh of relief

Daminark

Lel

Balarka Sen

It was perturbing the loop a little to make it non surjective

Akiva Weinberger

Wait, no, relief is from topography, not topology

0celo7

00:01

@BalarkaSen how u gonna do that on $S^1$

Balarka Sen

then bloop it floop

@0celo7 I wont

0celo7

that's the $S^n$, $n\ge 2$ proof

Balarka Sen

not "the" but yes

0celo7

The proof.

The only proof.

GFauxPas

The proof to end all proofs

Secret

00:07

While I don't know how is power defined in "there exists to the power there exists to the power for all", googling "exponential logic" lead me to something called linear logic:

en.m.wikipedia.org/wiki/Linear_logic

GFauxPas

"as a refinement of classical and intuitionistic logic," immediately loses interest

LEM 4eva

Poptart

00:25

Hi everyone! Can someone please explain the intuition behind the definition of the complex-valued function $g(z) = \frac{f(z) - f(a)}{z-a}$ for $z \not = a$ and $g(z) = f'(a)$ for $z = a$ where $f$ is an entire function. Bak and Newman use this function frequently throughout their text and show that the rectangle theorem (integral of a function $g(z)$ around the boundary of a rectangle $R$ and the closed curve theorem apply to $g$, later showing that if $f$ is analytic at $a$, then so is $g$).

It seems to me that as you take the limit as $z \to a$, then $g(z)$ is the definition of the complex derivative of $f$ at $a$; so are Bak and Newman using $g$ to show that the complex derivative of $f$ at any point where $f$ is analytic is also analytic and inherits all properties of the analytic functions?

Akiva Weinberger

What's the rectangle theorem?

Is it Cauchy's integral formula?

Poptart

Bak and Newman define the rectangle theorem as if $f$ is entire and $\Gamma$ is the boundary of a rectangle $R$, then $\int_\Gamma f(z) dz = 0$.

Akiva Weinberger

Ah

$\Gamma$ doesn't need to be a rectangle, by the way, turns out

They'll probably prove that soon, 'cause it follows from the rectangle case

but $\Gamma$ can be any closed curve

I think they're gonna prove the Cauchy integral formula, then, assuming they haven't already

Poptart

Right, I read that it can be a smooth closed curve! I guess I'm just a bit unsure of the motivation behind the definition of $g$!

Akiva Weinberger

Say $\gamma$ is some curve around $a$, and $f$ is analytic

GFauxPas

00:33

pointwise smooth*

Akiva Weinberger

@GFauxPas We need that?

GFauxPas

rectangle

Akiva Weinberger

I mean, can't it be any curve?

Co-shi's integral formula lets us define $f(a)$ in terms of an integral, which will turn out to be useful later (we can get bounds on the value of $f$ using it)

So let's say we know that $\dfrac{f(z)-f(a)}{z-a}$ is analytic

(defined to be $f'(a)$ at $a$)

GFauxPas

well you want it to be differentiable with respect to the parameter so you can change of coordinates in the integral and therefore parameterization doesnt matter yadda yadda

Akiva Weinberger

Then $\displaystyle\int_\gamma\frac{f(z)-f(a)}{z-a}dz=0$

GFauxPas

00:35

$f(\gamma)\gamma'$

Akiva Weinberger

But also we know the value of $\displaystyle\int_\gamma\frac1{z-a}dz$

Poptart

Is $\gamma$ a clsoed curve around $a$? Then that would be $2\pi i$ right?

Akiva Weinberger

That's the derivative of $\ln(z-a)$, so we can try to use the Fundamental Theorem of Calculus. The natural logarithm can't be defined on the whole complex plane, but there's a way around that

and we end up with $2\pi i$ times the winding number of $\gamma$ around $a$

If it's just a counterclockwise circle then yeah it would be $2\pi i$.

Or any loop of winding number $1$.

So $\displaystyle\int_\gamma\frac{f(a)}{z-a}dz$ would be...

($f(a)$ is a constant)

$f(a)2\pi i$.

So we can take that $\displaystyle\int_\gamma\frac{f(z)-f(a)}{z-a}dz=0$ that we had earlier

and suddenly we realize we can do something to that, if we split up the fraction

$\displaystyle\int_\gamma\frac{f(z)}{z-a}dz=\int_\gamma\frac{f(a)}{z-a}dz=2\pi if(a)$

$\displaystyle\frac1{2\pi i}\int_\gamma\frac{f(z)}{z-a}dz=f(a)$

and now we have Cauchy's integral formula, a way to express $f(a)$ in terms of just the values of $f(z)$ on a loop around $a$.

Poptart

Oh wow, I see the importance of the expression $\frac{f(z) - f(a)}{z-a}$ now.

Akiva Weinberger

It's kinda motivated by the result, tbh

Poptart

00:42

Thanks Akiva!

Akiva Weinberger

We know $\displaystyle\int_\gamma\frac C{z-a}dz=0$ when $C$ is a constant, and we can wonder what happens when $C$ is a function

and we can make a conjecture, and then the expression helps us prove that conjecture

is kinda how I see it

Poptart

Gotcha.

Akiva Weinberger

'Cause it turns out that $\dfrac{f(z)}{z-a}$ is just an analytic function away from something of the form $\dfrac C{z-a}$.

And so their integrals are the same.

@Poptart Here's a neat application of this

Say that $f$ is bounded

For example, let's say we know that $|f(z)|<1$ for all $z\in\Bbb C$.

Note that $\sin$ does not satisfy this!

$\sin(i)\approx1.17520$.

$\sin(iz)=i\sinh(z)$.

So, here's the question: Can we find any functions that actually do satisfy $|f(z)|<1$ for all $z$, real or complex?

Well, constant functions, for one. $f(z)=\frac12$ is a boring function, but it works. Are there any others?

We know that $\displaystyle\frac1{2\pi i}\int_\gamma\frac{f(z)}{z-a}dz=f(a)$.

GFauxPas

$\operatorname{Arg}(z)$?

Akiva Weinberger

Analytic functions, I mean

GFauxPas

00:48

;)

Akiva Weinberger

That's not even continuous!

GFauxPas

lol

Akiva Weinberger

And also, it goes from $-\pi/2$ to $\pi/2$...

Let me actually see if I can remember how to do this

Let $\gamma$ be a HUGE circle

GFauxPas

omg so big circle

Akiva Weinberger

Let me first spell out an important consequence of this formula that I didn't mention before

$\displaystyle\frac1{2\pi i}\int_\gamma\frac{f(z)}{z-a}dz=f(a)$ means that, if we just know the values of $f$ at points on $\gamma$, we know the values of $f$ everywhere in $\gamma$

because we can just calculate them from the formula

(In fact, it will turn out that more is true, but I can't prove that now)

GFauxPas

00:50

because holomorphic functions are ~magic~

Akiva Weinberger

^This is the main lesson from complex analysis

Analytic functions are magic

So, uh, say $\gamma$ is a circle of radius $R$ around $a$.

Wait

GFauxPas

Akiva, you want to find $f'$

so differnetiate both sides wrt a

Akiva Weinberger

...I do

GFauxPas

wait, is that right

you want $2 \pi i f' = \int f/(z-a)^2$ thing

Akiva Weinberger

I can do this another way, I think?

If I show that $f(a)-f(b)=0$ for all $a$ and $b$, I'm still good

and $\dfrac1{z-a}-\dfrac1{z-b}=\dfrac{a-b}{(z-a)(z-b)}$

And so $\displaystyle\frac1{2\pi i}\int_\gamma\frac{f(z)(a-b)}{(z-a)(z-b)}dz=f(a)-f(b)$ if $\gamma$ contains both $a$ and $b$

But this is more complicated than I originally thought it would be

I had forgotten the details

@Poptart When you see this: Sorry

Lemme finish this anyway

GFauxPas

00:56

wikipedia/proofwiki lol

that's what I do

Akiva Weinberger

So $\gamma$ is a big circle around $a$ and $b$ of radius $R$

The left-hand-side, we know that the the integral of a function around a loop is at most the max value of the function times the length of the loop

If $|f|<M$ and $\ell(\gamma)=L$ then $|\int_\gamma f|\le LM$

There's a name for that, I forget it

0celo7

@Daminark yo are you here?

Akiva Weinberger

But it's the important thing that makes Cauchy's formula useful here

0celo7

@AkivaWeinberger basic L^\infinity estimate?

Konformist Liberal

ncatlab.org/nlab/show/simple+object can someone give a simple counterexample to prop 2.1 when the field is not algebraically closed?

GFauxPas

00:59

Rouche's Lemma or something?

wait no thats different

Akiva Weinberger

So $\displaystyle\int_\gamma\frac{f(z)(a-b)}{(z-a)(z-b)}dz\le\frac{1(a-b)}{{|z-a|}{‌|z-b|}}\cdot2\pi R$

GFauxPas

estimation lemma

Akiva Weinberger

'cause $|f(z)|\le1$ by hypothesis, and the length of $\gamma$ is $2\pi R$

and if $z$ is on the big circle, then $|z-a|\ge{}$uh, $R-|a|$ or something?

If $\gamma$ is centered around the origin, then sure, yeah

The point is, we're bounding this thing by something that's approximately $\dfrac{\rm constant}{R^2}\cdot{\rm constant}R$

which is approximately $\dfrac{\rm constant}R$

which goes to $0$ as $R$ goes to infinity

But $\displaystyle\frac1{2\pi i}\int_\gamma\frac{f(z)(a-b)}{(z-a)(z-b)}dz=f(a)-f(b)$

So as I make $\gamma$ bigger and bigger, the left-hand side goes to $0$

meaning the right-hand side, which doesn't change with $\gamma$, has to equal $0$ the whole time

which means $f$ is constant

so boom

0celo7

@Daminark I have some crap complex analysis that I'm stuck on

Akiva Weinberger

Also, I could have started this with $|f(z)|\le M$ instead of${}\le1$

so all bounded analytic functions are constant

Why? 'Cause we can write them as an integral, and then we can show that the integral can't go big enough

I have no intuition for this fact that doesn't go through the Cauchy integral formula

So the Cauchy integral formula lets us prove magic

Poptart

01:10

Are $a$ and $b$ two arbitrary points contained in the region contained in $\gamma$?

Akiva Weinberger

Two arbitrary points... at all

$\gamma$ is a big circle containing $a$ and $b$

We want to prove that, for any $a$ and $b$, we have $f(a)=f(b)$

Poptart

Oh i see! that's a cool result!

@AkivaWeinberger thank you again :)

Akiva Weinberger

There's another way, where instead of trying to show $f(a)-f(b)=0$ for all $a$ and $b$, you try to show $f'(a)=0$ for all $a$. But for that you need a different result, namely $\displaystyle\frac1{2\pi i}\int_\gamma\frac{f(z)}{(z-a)^2}dz=f'(a)$.

GFauxPas

yeah that's the way I was thinking of Akiva

Akiva Weinberger

Which your book will probably prove soon.

But it's probably annoying proving a second integral identity (even if it follows from the first)

You can get it by differentiating both sides of Cauchy's integral formula with respect to $a$, but I was worried about the rigor

GFauxPas

01:15

you need to show you're allowed to do $\mathrm d \int = \int \partial$

or, alternatively, that the thing minus the other thing goes to 0 in modulus, which is probably easier

Akiva Weinberger

Or consider the definition of the derivative, and replace each instance of $f$ in it with a whole integral

It's doable

You actually can get a formula for $f''$, $f'''$, etc., also

In fact, it turns out you can use the integral formula to prove that those all exist

All analytic functions are infinitely differentiable

In other words, if a function in the complex plane can be differentiated once, it can be differentiated infinitely many times

which is very not true for functions on the real line

So that follows from the integral formula as well

Again, analytic functions are magic

GFauxPas

:O

0celo7

that's just elliptic regularity theory

Akiva Weinberger

Also, all analytic functions have antiderivatives

GFauxPas

Also, all holomorphic functions are analytic

meaning that theyre equal to their power series

Akiva Weinberger

01:20

(Well, $\frac1z$ is defined on $\Bbb C\setminus\{0\}$, and it doesn't have an antiderivative defined on $\Bbb C\setminus\{0\}$.

GFauxPas

because it'

Akiva Weinberger

But if something's defined on $\Bbb C$, it has an antiderivative defined on $\Bbb C$.)

GFauxPas

s not holomorphic on any nbhd of 0

Akiva Weinberger

(And that integral is analytic. So you can integrate it infinitely many times also.)

0celo7

@AkivaWeinberger You mean functions defined in simply connected domains have primitives.

Akiva Weinberger

01:21

Right, you can strengthen it to that

Simply connected means it doesn't have holes, and primitive is a synonym for antiderivative

Nicholas Roberts

01:45

Hi all. I am tasked with writing $\sum_{i, \space j, \space k \space distinct} x_i^2x_j^2x_k^2$ as an elementary symmetric polynomial. Any ideas?

GFauxPas

hmm

Nicholas Roberts

In other words, I want to write this thing in terms of $s_1 = \sum x_i, \space ... \space s_n = \sum x_1 \cdot \cdot \cdot x_n$

skull

02:01

Proof that part of higher mathematics is salesmanship:

in Homotopy Theory, Apr 23 at 1:20, by Tyler Lawson

writing that out, i feel like i may have tried to make this sales job to the chatroom before

Secret

Bernoulli differential equation

In mathematics, an ordinary differential equation of the form: y ′ + P ( x ) y = Q ( x ) y n {\displaystyle y'+P(x)y=Q(x)y^{n}\,} is called a Bernoulli differential equation where n {\displaystyle n} is any real number and n ≠ 0 {\displaystyle n\neq 0} and ...

Nicholas Roberts

Lol Bernoulli diff eq, i remember those days....

Akiva Weinberger

02:25

@NicholasRoberts If you get rid of the "distinct" bit, it becomes $(\sum_i x_i^2)^3$

So from that you have to subtract out things where two are equal

and add back some thing where all three are equal probably? Maybe with some constant in front

Nicholas Roberts

Ok so youre saying to start with the sum of (x_i)^2 and then try and subtract things to get the result

seems like a lot of trial and error, but i'll try

Akiva Weinberger

I would write it out with some example (i.e. $n=3$)

@NicholasRoberts I don't have a better idea at the moment

Nicholas Roberts

Ya but n = 3 its already a symmetric polynomial

its simply s_3

Akiva Weinberger

Oh

True

Nicholas Roberts

(s_3)^2

Akiva Weinberger

02:27

5, then, I guess?

This all sounds annoying

Nicholas Roberts

Yeah i suppose. LOL yes it is very annoying

Akiva Weinberger

but if you write it out neatly maybe you'll get ideas for less annoying things

DarkRunner

02:59

I'm stumped on this geometry problem

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