The h Bar

Bernardo Meurer

00:00

Ah, I see

Well very neat taking the limit is

easier than hand-computing the partial derivative

Okay!

Next! Gradients, lemme check my notes

The gradient is just the partial derivative for each variable, right?

ACuriousMind

@BernardoMeurer basically, yes

Bernardo Meurer

@ACuriousMind Hmm, do you still have the problem set open?

ACuriousMind

Yes

Bernardo Meurer

2.a) How the heck am I supposed to do this

I mean

Okay, he gave me the gradient at a point

Do I

Wait

Hm

ACuriousMind

Chain rule.

Bernardo Meurer

00:07

$\phi(1,0) = f(1,1)$ right?

ACuriousMind

Just compute $\nabla\phi(1,0)$ by the chain rule, you'll see why you're given $\nabla f(1,1)$.

Bernardo Meurer

Okay, so $\nabla\phi(1,0) = (\frac{\partial}{\partial x}f(1,1), \frac{\partial}{\partial y}f(1,1))$

ACuriousMind

How did you arrive at that?

Bernardo Meurer

Probably through being silly

Well

Oh wait

no

It will be just the derivative of f

wrt x and y

because $f\colon \Bbb R^2 \to \Bbb R$

Wait, will it?

What am I doing

I thought

$f(x_1, x_2, \ldots, x_n) \implies \nabla f = (\frac{\partial}{\partial x_1}f, \frac{\partial}{\partial x_2}f, \ldots, \frac{\partial}{\partial x_n}f)$

@ACuriousMind No?

ACuriousMind

Yes

Bernardo Meurer

00:16

So

$\nabla\phi(1,0) = (\frac{\partial}{\partial x}f(1^3 + 0^2,1 + 0), \frac{\partial}{\partial y}f(1^3 + 0^2,1 + 0))$

ACuriousMind

Did you use the chain rule there?

Bernardo Meurer

No, why do I have to?

ACuriousMind

Because...well, where does that equation come from?

Bernardo Meurer

Ah

because $\phi$ and $f$ are composite

Well shit

Sigh

I should have listened to my father and become a lawyer

He was right

@ACuriousMind Okay, I don't even know how to apply the chain rule here

What am I even doing

ACuriousMind

@BernardoMeurer Yeah - $\phi$ is the composite of $f$ and the function $\mathbb{R}^2\to\mathbb{R}^2,(x,y)\mapsto (x^3 + y^2, x+y)$

Bernardo Meurer

00:24

So first i's going to be

$\nabla\phi(1,0) = (\frac{\partial \phi(1,0)}{\partial x}, \frac{\partial \phi(1,0)}{\partial y})$

But then I gotta figure those derivatives out with the chain rule

0celouvsky

00:36

I am back

what's poppin

Bernardo Meurer

God bless

Here's the problem statement:

0celouvsky

@ACuriousMind I thought you didn't do analysis?

Bernardo Meurer

$f\colon \Bbb R^2\to\Bbb R; f\in C^2$

0celouvsky

Just say $f\in C^2(\Bbb R^2)$

Bernardo Meurer

$\phi(x,y) = f(x^3 + y^2, x+y)$

a) Knowing that $\nabla f(1,1) = (2,3)$ determine $\nabla\phi(1,0)$

I wrote this

$$\nabla\phi(1,0) = (\frac{\partial \phi(1,0)}{\partial x}, \frac{\partial \phi(1,0)}{\partial y})$$

And then I got stuck :)

ACM said to use chain rule, but I don't know how to do that here

0celouvsky

00:39

what did ACM say after that?

Bernardo Meurer

Nothing?

He agreed with me on it being a composite function

ACuriousMind

@BernardoMeurer Don't you have a chain rule that looks like $D(f\circ g) = Df \circ Dg$? (The $\nabla$ is just a special case of the $D$)

Bernardo Meurer

@ACuriousMind Not that I could find, no

Now I do

0celouvsky

do you take notes in class?

Bernardo Meurer

Yes, a lot

0celouvsky

00:42

So you have the function $g(x,y)=(x^3+y^2,x+y)$

The Jacobian of that is a 2x2 matrix

Bernardo Meurer

@0celouvsky yeah

0celouvsky

Compute that

Then multiply on the right by the Jacobian of $f$ at $(1,1)$, which you don't need to do any work for

Jacobian matrix and determinant

In vector calculus, the Jacobian matrix (/dʒᵻˈkoʊbiən/, /jᵻˈkoʊbiən/) is the matrix of all first-order partial derivatives of a vector-valued function. When the matrix is a square matrix, both the matrix and its determinant are referred to as the Jacobian in literature. Suppose f : ℝn → ℝm is a function which takes as input the vector x ∈ ℝn and produces as output the vector f(x) ∈ ℝm. Then the Jacobian matrix J of f is an m×n matrix, usually defined and arranged as follows: J = [ ...

Bernardo Meurer

Wait, let me figure out how to compute the jacobian

0celouvsky

Use the picture there for the correct order of things

Bernardo Meurer

so I want $D\phi$

0celouvsky

00:44

No, you want $Df$ and $Dg$

Because $D\phi=DfDg$

Bernardo Meurer

Ah

I see

0celouvsky

matrix multiplication

Bernardo Meurer

$Dg$ is $2\times 2$ right?

0celouvsky

@BernardoMeurer The DA gave Kodak a plea for 8 years and he denied it. He's going away for a looooong time

I'm sure Fantano is happy

@BernardoMeurer yes

Bernardo Meurer

Jeezus

0celouvsky

00:46

they gave Bobby Shmurda like 117 years lol

Kodak seems like a pretty bad guy. He violated parole to go to a strip club and apparently assaulted a stripper. He also has an open rape case

He's poppin in the skreets tho

Bernardo Meurer

How do I make a matrix here?

0celouvsky

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

Bernardo Meurer

$$Dg = \begin{bmatrix}
3x^2 & 2y \\
1 & 1
\end{bmatrix}$$

0celouvsky

Do you know the final answer?

Bernardo Meurer

Nope

0celouvsky

00:49

that's not good

Bernardo Meurer

Is that right?

How do I find $Df$?

0celouvsky

I think so

You know $Df(1,1)$

Bernardo Meurer

Yeah

Oh

That's all I need?

0celouvsky

So $g(1,0)=(1,1)$

Bernardo Meurer

yeah

It's all I need

0celouvsky

00:51

So you need $Df$ at $g(1,0)$

which is $Df(1,1)$

Bernardo Meurer

Which is $(2, 3)$?

0celouvsky

yes

So multiply on the right by that vector

Bernardo Meurer

so I have:

0celouvsky

and plug in $x=1,y=0$

Bernardo Meurer

$D\phi = \begin{bmatrix}2 & 3\end{bmatrix}\begin{bmatrix} 3 & 0\\ 1 & 1\end{bmatrix}$

0celouvsky

00:55

looks right

Bernardo Meurer

$[9, 3]$

That's it?

The gradient is just the derivative?

0celouvsky

you need to transpose that to get a vector

but then you should be good, yes

Bernardo Meurer

Hm? Why must it be a column vector?

0celouvsky

gradient is a column vector, traditionally

Bernardo Meurer

Okay, got it

0celouvsky

00:58

idk if you're making the distinction between vectors and covectors

Bernardo Meurer

b) Determine $\frac{\partial^2 \phi}{\partial x^2}(x, y)$ in terms of partial derivatives of $f$

0celouvsky

maybe @ACuriousMind has a better idea

I never took multivariable calculus

@BernardoMeurer chain rule again

Bernardo Meurer

So I write $\phi = f\circ g$

0celouvsky

There's a formula for the double chain rule but it's horrible

ugh

why are they making you do this?

Bernardo Meurer

Because I study in hell

Okay, we can do a more instructive and less shitty one

Determine the critical points of $g(x,y) = \frac{1}{2}x^2 +xy+y^3$

I gotta compute the gradient, right?

0celouvsky

01:04

yes

ACuriousMind

Why did I have to look at the Portugese pdf and he gets the translated exercises? :P

Bernardo Meurer

@ACuriousMind Because you didn't ask :P

0celouvsky

because I actually help but that's my one condition

Bernardo Meurer

We already have it established that he won't learn my Spanish 3.0 ++

ACuriousMind

@BernardoMeurer lol, it's actually not so hard to figure out what the exercises want

0celouvsky

01:05

I can understand it

Bernardo Meurer

@ACuriousMind I said the same thing in Analysis I

0celouvsky

I just don't want to

Bernardo Meurer

^ Yep

0celouvsky

Supply and demand

He demands so he supplies

Bernardo Meurer

$\nabla g = (x+y, 3y^2 + x)$ ?

0celouvsky

01:07

@ACuriousMind Btw there's no tomfoolery in the Hodge theorem. That's the one bit of geometric analysis I feel pretty good about.

@BernardoMeurer sure

Bernardo Meurer

Now I gotta find the points where it's (0,0)

Can I use linear algebra for this? is that practical?

0celouvsky

it's nonlinear

you can use substitution though

Bernardo Meurer

PITA

0celouvsky

oh please

it's $3y^2-y=0$

ACuriousMind

pita bread is delicious

0celouvsky

01:09

that means kiss

Wtf is a smooth measure on a vector bundle

ACuriousMind

@0celouvsky it doesn't, and that's not very nice

0celouvsky

nvm

Bernardo Meurer

$y=0 \vee y= \frac{1}{3}$

0celouvsky

@ACuriousMind Sorry

@BernardoMeurer yes, and $x=?$

Bernardo Meurer

$x = 0 \vee x = -\frac{1}{3}$

0celouvsky

01:11

sure

ACuriousMind

apology accepted

Bernardo Meurer

@ACuriousMind Pita bread is pretty good yeah

@0celouvsky Okay, so I have 4 critical points

0celouvsky

why is this guy writing $\mathscr D(X,E)$ on a compact manifold :/

@BernardoMeurer what

you have two

Bernardo Meurer

$(0,0), (0, 1/3), (-1/3, 0), (-1/3, 1/3)$

0celouvsky

intersection of a line and a parabola has at most two points

Bernardo Meurer

01:13

Oh

I was being stupid

0celouvsky

No, it's $(0,0)$ and $(-1/3,1/3)$

Bernardo Meurer

Yeah

Now I need to compute the hessian, to classify them, right?

0celouvsky

yes

Bernardo Meurer

Okay, let me try

0celouvsky

@ACuriousMind Does modern algebraic geometry use this old Hodge-era analysis or have they found clever ways to avoid it?

Cows

01:18

6d(2,0) SCFT

hi

0celouvsky

that doesn't sound like analysis to me

Cows

heheheh neh it isn't lol

I had a question held

Bernardo Meurer

$$H_f(x,y) = \begin{pmatrix} 1 & 1 \\ 1 & 6y \end{pmatrix}$$

Cows

just lobbying on here for its release lol

0celouvsky

you pmatrix you scrub

Cows

01:19

@ACuriousMind

ACuriousMind

@0celouvsky I'm not sure what you mean, but "modern algebraic geometry" is rather broad and largely concerned with schemes and such where you can't really apply analysis at all

Bernardo Meurer

@0celouvsky There

0celouvsky

@ACuriousMind So it's rather disjoint from Hodge theory?

Cows

@ACuriousMind Are people still voting on the fate of my question?

Bernardo Meurer

@0celouvsky Does that look right?

0celouvsky

01:20

no

what was $Df$ again?

Cows

I was out for a walk so I could not address the comments in real time

ACuriousMind

@Cows What I wrote in my comment still stands - adding links that may or may nor explain what you're talking about wasn't the point, a well-informed reader should be able to understand what you're talking about without following any links.

@Cows It's not a real-time process

Bernardo Meurer

@0celouvsky Oh, my bad, I meant g there

$$H_g(x,y) = \begin{pmatrix} 1 & 1 \\ 1 & 6y \end{pmatrix}$$

0celouvsky

what was $Dg$ again?

Bernardo Meurer

$\nabla g= (x+y, 3y^2 + x)$

ACuriousMind

01:22

@0celouvsky Well, if you do complex geometry Hodge theory of course plays a large role, but algebraic geometry is much larger than that

Cows

@ACuriousMind awesome, I guess I just need to wait :D

0celouvsky

they're all the same to me

ACuriousMind

Alas Hodge theory and the like is alarge part of why complex manifolds are "nicer" than real ones

Bernardo Meurer

$g = (\frac{1}{2}x^2, 3y^2+x)$

0celouvsky

@ACuriousMind I don't understand the complex hodge theorem

it says something quite different than the Riemannian one if I'm not mistaken

ACuriousMind

01:24

@0celouvsky I don't know what exactly you mean by Hodge theorem and it's a bit late for me to have a technical discussion of things I'm only aware of in my peripheral vision

0celouvsky

ok, good night

Bernardo Meurer

@0celouvsky Is my Hessian right?

0celouvsky

no

wait

it is

Bernardo Meurer

:)

So for (0,0) it will be definite-positive right?

0celouvsky

for $y=0$ it looks like it has zero determinant

I'm stupid

negative determinant

so yes

wait, no

negative definite :D

is that right?

Bernardo Meurer

01:29

Yeah

-1

So (0,0) is a max

And for 1/13 it will also be negative

0celouvsky

don't actually trust me

I don't know what a negative definite matrix is

Bernardo Meurer

so also a max

0celouvsky

it has a positive and a negative eigenvalue

that doesn't make sense for negative definite

@ACuriousMind help

Bernardo Meurer

$h^t Hf(a)h> 0 \quad\forall h\in\Bbb R^n\setminus\{0\}$

0celouvsky

what's that

Bernardo Meurer

01:31

The condition for it to be definite positive

invert the inequality for negative

0celouvsky

so that matrix is indefinite

the eigenvalues are $1/2 (1\pm\sqrt 5)$

Bernardo Meurer

2

A: How to check if a matrix is positive definite

I don't think there is a nice answer for matrices in general. Most often we care about positive definite matrices for Hermitian matrices, so a lot is known in this case. The one I always have in mind is that a Hermitian matrix is positive definite iff its eigenvalues are all positive. Glancing ...

Nice

0celouvsky

too bad acm left

he likes this stuff

I tried very hard to not take calculus because of this stuff

Bernardo Meurer

Hmm

Both have positive and negative eigenvalues

I guess that makes them indefinite

So two saddle points

I hate this shit lol

@0celouvsky Oh, there's a proof in here, wanna try it?

0celouvsky

let's see it

Bernardo Meurer

01:39

Let $f\colon \Bbb R^2\to \Bbb R$ be continuous. Show that if there is an open set $U\subset\Bbb R^2$ limited by a closed curve $C$ such that $C$ is the level set of $f$, then $f$ has at least one local extreme

Papa bless

0celouvsky

let me track down the proof of this inequality, then we'll see

Bernardo Meurer

How do I verify continuity at a point in a multivariable function in $\Bbb R^n$ @0celouvsky?

0celouvsky

check $\lim_{x\to x_0}f(x)=f(x_0)$

@BernardoMeurer limited?

Bernardo Meurer

@0celouvsky Yeah "Um aberto U em R^2 limitado por uma curva fechada C"

Limited, contained by

bounded

0celouvsky

$\partial U=C$?

Bernardo Meurer

01:46

I dunno

0celouvsky

The point being that $\bar U$ is compact, I'm sure.

So $f$ attains a max or a min in $\bar U$

I'm sure you can take it from there

Bernardo Meurer

I don't even know what a level set is lol

Oh

Is this Weierstrass?

0celouvsky

$C=f^{-1}(c)$, for some $c\in\Bbb R$

@BernardoMeurer I don't know your shitty names for things :P

Bernardo Meurer

Weierstrass Theorem: $f\colon D\subset \Bbb R^n \to \Bbb R^n$ continuous with $D$ compact has max and min in $D$

0celouvsky

yes

Bernardo Meurer

01:48

That's it?

0celouvsky

No, not quite

Bernardo Meurer

It seemed to good to be true

0celouvsky

Your max can be on the inside, which is good, but it could be on the boundary

so you need an argument

you need to show that it's impossible for both the max and min to be on the boundary

that's where the level set thing comes in

because $f$ is constant on the boundary

Bernardo Meurer

Ah

And the max and the min can't be the same?

0celouvsky

so IF the max and the min are both on $C$, then max=min and that's either done or a contradiction

because then $f$ is constant on all of $U$ and every point is an extrememum

Bernardo Meurer

01:50

Why would it be a contradiction? The function can be constant, no?

Aha

0celouvsky

Yeah but if it's constant you're trivially done

Bernardo Meurer

Cool

0celouvsky

you assume it's nonconstant

Bernardo Meurer

V. nice

math.tecnico.ulisboa.pt/~gpires/CII/auto-aval/Tprep-nov-07.pdf

Working on this now

0celouvsky

this sans serif is killing me

Bernardo Meurer

01:52

What the hell is a serif anyway

0celouvsky

the tail on serif letters

Bernardo Meurer

Ah

Yeah, people here in portugal like shitty fonts

my mechanics prof will only use Comic Sans

0celouvsky

hahaaa

Bernardo Meurer

Really

ONLY

Okay

So check the limit thing

to find the vlaue of a so it's continuous

0celouvsky

yeah

Bernardo Meurer

01:58

Hmm that's going to be a tricky limit

This will give me 0/0

I'm certain

it's a shitty limit

It's gon' be 5

so $a=5$

@0celouvsky How do I calculate the derivative on a vector?

@0celouvsky Save me

0celouvsky

02:14

what do you need?

Bernardo Meurer

math.tecnico.ulisboa.pt/~gpires/CII/auto-aval/Tprep-nov-07.pdf

Look at 2

they give me $g$

and $Df(1,1,1)$

and ask for $Dg\circ f$ over $v$

0celouvsky

so $D(g\circ f)=Dg\circ Df$

what's the problem?

Bernardo Meurer

How do I do that "over $v$"?

0celouvsky

wait what is secundo o?

I think they just want you to apply the matrix to $v$

Bernardo Meurer

"Segundo o vetor" I would translate to over the vector, or according to the vector

Wait, before we return to this one, check this out:

$f(x,y,z) = 3x^2+2\sqrt{2}xy+2y^2+\frac{2}{3}z^3-2z$

$\nabla f=(6x+2\sqrt{2}y, 2\sqrt{2}x+4y, 2z^2-2)$

and

$$H_f(x,y,z) = \begin{pmatrix}6 & 2\sqrt{2} & 0 \\ 2\sqrt{2} & 4 & 0 \\ 0 & 0 & 4z\end{pmatrix}$$

@0celouvsky Correct?

0celouvsky

02:30

fuck

why are you making me do this

Bernardo Meurer

$$\nabla f = (0,0,0) \implies \begin{cases}6x+2\sqrt{2}y = 0 \\ 2\sqrt{2}x + 4y = 0 \\ 2z^2 - 2 = 0\end{cases}$$

@0celouvsky Because I love you

0celouvsky

@BernardoMeurer yes

@BernardoMeurer sure

Bernardo Meurer

I tried solving the system with a matrix

i failed

0celouvsky

because it's nonlinear

Bernardo Meurer

Ah

Duh

0celouvsky

02:32

you can solve the third one by hand

Bernardo Meurer

Wait

0celouvsky

the other two with a matrix

Bernardo Meurer

The third one is trivial, yeah

So I put it apart

and matrified the first two

$$\begin{bmatrix}6 & 2\sqrt{2} & 0 \\ 2\sqrt{2} & 4 & 0\end{bmatrix} \to \begin{bmatrix}1 & \frac{\sqrt{2}}{3} & 0 \\ 0 & 1 & 0\end{bmatrix}$$

0celouvsky

I'm not teaching you how to invert a matrix, sorry.

Bernardo Meurer

According to my calculator

Echelon reducing it

@0celouvsky Hm? Why do I have to invert the matrix?

0celouvsky

02:39

that's how you solve an equation I guess?

actually it's a homogeneous equation

idk dude

Bernardo Meurer

Huh? I don't invert it

0celouvsky

idk linear algebra

Bernardo Meurer

I make it into an extended matrix and use Gauss

0celouvsky

I don't know linear algebra

that's all I'm going to say

Bernardo Meurer

Alright

Let me use WA for the answer

0celouvsky

02:40

just subtitute

it's much easier than linear algebra shit

$x=-y/\sqrt 2$ from the second equation

idk something like that

Bernardo Meurer

wolframalpha.com/input/…

WA says only solution is (0,0)

0celouvsky

that's fine

Bernardo Meurer

I guess that makes sense if you continue to do Gauss-Jordan Elimination the non-augmented matrix gives you the identity

So we only have one critical point

(0,0,1)

0celouvsky

no

Bernardo Meurer

O.o

0celouvsky

02:43

solve $z^2=1$ again

Bernardo Meurer

Ah

+-1

two points

0celouvsky

yes

Bernardo Meurer

Ookay, for (0,0,1) we have $\sigma=\{8,2,4\}$, so it's a minimum

0celouvsky

sure

Bernardo Meurer

For (0,0,1) we have $\sigma = \{8,2,-4\}$ so it's indefinite and therefore a saddlepoint

Okay

4AM

heading home

@0celouvsky Thanks for all the help :)

0celouvsky

02:56

np

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