The h Bar

Yes

12:16 AM

So

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

Did you use the chain rule there?

No, why do I have to?

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

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

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

12:24 AM

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

12:36 AM

I am back

what's poppin

God bless

Here's the problem statement:

@ACuriousMind I thought you didn't do analysis?

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

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

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

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

12:39 AM

what did ACM say after that?

Nothing?

He agreed with me on it being a composite function

@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 $D$)

@ACuriousMind Not that I could find, no

Now I do

do you take notes in class?

Yes, a lot

12:42 AM

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

The Jacobian of that is a 2x2 matrix

@0celouvsky yeah

Jacobian matrix and determinant

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

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 = [ ...

Wait, let me figure out how to compute the jacobian

Use the picture there for the correct order of things

so I want $D\phi$

12:44 AM

No, you want $Df$ and $Dg$

Because $D\phi=DfDg$

Ah

I see

matrix multiplication

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

@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

Jeezus

12:46 AM

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

How do I make a matrix here?

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

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

Do you know the final answer?

Nope

12:49 AM

that's not good

Is that right?

How do I find $Df$?

I think so

You know $Df(1,1)$

Yeah

Oh

That's all I need?

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

yeah

It's all I need

12:51 AM

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

which is $Df(1,1)$

Which is $(2, 3)$?

yes

So multiply on the right by that vector

so I have:

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

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

12:55 AM

looks right

$[9, 3]$

That's it?

The gradient is just the derivative?

you need to transpose that to get a vector

but then you should be good, yes

Hm? Why must it be a column vector?

gradient is a column vector, traditionally

Okay, got it

12:58 AM

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

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

maybe @ACuriousMind has a better idea

I never took multivariable calculus

@BernardoMeurer chain rule again

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

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

ugh

why are they making you do this?

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?

1:04 AM

yes

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

@ACuriousMind Because you didn't ask :P

because I actually help but that's my one condition

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

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

1:05 AM

I can understand it

@ACuriousMind I said the same thing in Analysis I

I just don't want to

^ Yep

Supply and demand

He demands so he supplies

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

1:07 AM

@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

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

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

it's nonlinear

you can use substitution though

PITA

oh please

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

pita bread is delicious

1:09 AM

that means kiss

Wtf is a smooth measure on a vector bundle

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

nvm

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

@ACuriousMind Sorry

@BernardoMeurer yes, and $x=?$

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

1:11 AM

sure

apology accepted

@ACuriousMind Pita bread is pretty good yeah

@0celouvsky Okay, so I have 4 critical points

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

@BernardoMeurer what

you have two

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

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

1:13 AM

Oh

I was being stupid

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

Yeah

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

yes

Okay, let me try

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

1:18 AM

6d(2,0) SCFT

hi

that doesn't sound like analysis to me

heheheh neh it isn't lol

I had a question held

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

just lobbying on here for its release lol

you pmatrix you scrub

1:19 AM

@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

@0celouvsky There

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

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

@0celouvsky Does that look right?

1:20 AM

no

what was $Df$ again?

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

@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

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

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

what was $Dg$ again?

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

1:22 AM

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

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

they're all the same to me

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

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

@ACuriousMind I don't understand the complex hodge theorem

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

1:24 AM

@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

ok, good night

@0celouvsky Is my Hessian right?

no

wait

it is

:)

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

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?

1:29 AM

Yeah

-1

So (0,0) is a max

And for 1/13 it will also be negative

don't actually trust me

I don't know what a negative definite matrix is

so also a max

it has a positive and a negative eigenvalue

that doesn't make sense for negative definite

@ACuriousMind help

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

what's that

1:31 AM

The condition for it to be definite positive

invert the inequality for negative

so that matrix is indefinite

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

A: How to check if a matrix is positive definite

2

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

too bad acm left

he likes this stuff

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

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?

let's see it

1:39 AM

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

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

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

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

@BernardoMeurer limited?

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

Limited, contained by

bounded

$\partial U=C$?

1:46 AM

I dunno

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

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

Oh

Is this Weierstrass?

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

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

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

yes

1:48 AM

That's it?

No, not quite

It seemed to good to be true

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

Ah

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

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

1:50 AM

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

Aha

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

Cool

you assume it's nonconstant

V. nice

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

Working on this now

this sans serif is killing me

1:52 AM

What the hell is a serif anyway

the tail on serif letters

Ah

Yeah, people here in portugal like shitty fonts

my mechanics prof will only use Comic Sans

hahaaa

Really

ONLY

Okay

So check the limit thing

to find the vlaue of a so it's continuous

yeah

1:58 AM

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

2:14 AM

what do you need?

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 $v$

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

what's the problem?

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

wait what is secundo o?

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

"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?

2:30 AM

fuck

why are you making me do this

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

@0celouvsky Because I love you

@BernardoMeurer yes

@BernardoMeurer sure

I tried solving the system with a matrix

i failed

because it's nonlinear

Ah

Duh

2:32 AM

you can solve the third one by hand

Wait

the other two with a matrix

The third one is trivial, yeah

So I put it apart

and matrified the first two

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

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

According to my calculator

Echelon reducing it

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

2:39 AM

that's how you solve an equation I guess?

actually it's a homogeneous equation

idk dude

Huh? I don't invert it

idk linear algebra

I make it into an extended matrix and use Gauss

I don't know linear algebra

that's all I'm going to say

Alright

Let me use WA for the answer

2:40 AM

just subtitute

it's much easier than linear algebra shit

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

idk something like that

wolframalpha.com/input/…

WA says only solution is (0,0)

that's fine

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)

no

O.o

2:43 AM

solve $z^2=1$ again

Ah

+-1

two points

yes

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

sure

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 :)