numerical-analysis

[neon snow]

Or wait

I see

or no

idk

[wide spear]

First, can you discretize the boundary conditions?

[neon snow]

WRite these as finite differences?

[wide spear]

Yeah

[neon snow]

Hmm forwards?

[wide spear]

Well, at the left side it will need to be forwards

[neon snow]

Yeah, and right side backwards

[wide spear]

And at the right side it will need to be backwards

Yeah

[neon snow]

so

[
N\left(u_{j+1}(t)-u_{j}(t)\right)=0
]

[pine jetty]

Victor H

[neon snow]

For the left side

[wide spear]

Yes

[neon snow]

But is this with respect to x now

[wide spear]

Well, this also holds for t+1 right?

[neon snow]

Yeah you're right

[wide spear]

How can you incorporate this into your update routine?

[neon snow]

Well since this is the first one

I need

[brave crypt]

So that's the compressible mass conservation, incompressible is just $\nabla\cdot(u)=0$. The answer to number 2 in this picture about the pressure was that it enforces mass conservation. I definitely did not get that right at the time, still don't understand it now totally. But pretty sure that was it and physically it makes sense somewhat. Like in a pipe that is thinner downstream than upstream flow speeds up according to mass conservation. How would the individual fluid particles know to speed up? Because of the pressure, which is still a wave with finite speed. But very large compared to the velocity of the fluid such that it is infinite in the mathematical analysis.

[pine jetty]

Abdul Aziz

[neon snow]

Right?

[wide spear]

Yes

[neon snow]

So I need u_j=1 and u_j=0 to be the same every iteration

[wide spear]

Sure, you're talking about Bernoulli's principle or whatever similar concept applies here

[wide spear]

Yes

[neon snow]

But I can't just set them to whatever value

[wide spear]

I would not call this a Lagrange multiplier

No, but how did you update them in the past?

[neon snow]

What do you mean?

[wide spear]

How did you compute u_1(t+1)?

[neon snow]

u(2:end-1,m+1) = u(2:end-1,m) + dt * udot(:,m+1);

[wide spear]

Ok

So this is how you get u_1(t+1)

Then you can just set u_0(t+1)=u_1(t+1)

I think

[neon snow]

oooh

so I still step inside only

but I also update the boundary

YES OK

[wide spear]

[neon snow]

for
u(2:end-1,m+1) = u(2:end-1,m) + dt * udot(:,m+1);
u(1,m+1) = u(2,m+1);
u(end,m+1) = u(end-1,m+1);

?

[wide spear]

Something like this, yes

Also obligatory Matlab sucks

[neon snow]

why haha

[wide spear]

Python is faster

Open source

Free

Etc....

[neon snow]

Comparing Python to MATLAB is weird

MATLAB is fast when you use MATLAB properly, i.e using vectorized instructions and utilizing it's perfect implementation of LAPACK

[wide spear]

Python calls LAPACK via Num/Scipy

[brave crypt]

Python is definitely superior

[neon snow]

Yes ofc I know that

But one actually pays for MATLAB because of it's really good documentation and robust functions that work out of the box

It's like a big ass library

But Python is good too imo

I wouldn't pay for MATLAB personally 😄

Academic license

[brave crypt]

Yup the libraries are very nice

[wide spear]

I would hesitate to say that Matlab has good documentation

[neon snow]

??

I find MATLAB's documentation to be one of the best I've ever seen

[wide spear]

I've tried working with some things

I prefer the Scipy style of documentation

[neon snow]

The von Neumann BC doesn't work

It shouldn't turn around

Why the fuck is the video 3 min

It should reflect without inverting the wave

[wide spear]

First, I'm going to suggest that you replace all of your first order approximations with second order approximations

This should help with the oscillations

[neon snow]

other than this

[wide spear]

Yes

Second order approximation

Also for the endpoints

[neon snow]

sorry I'm back

I just had to commit seppuku

From this piece of shit assignment

but we back

[wide spear]

[neon snow]

So I looked in my book

And found something similar

Dog shit book btw

And the author uses second order method for derivatives

Then he solves some system of linear equations in the for-loop

[wide spear]

Ok that makes sense

Second order method will reduce oscillations

System of linear equations at the endpoints, yeah

[neon snow]

But he only uses second order method for the

first derivatives

ie. the boundary cond.,

[wide spear]

Yeah yeah

[neon snow]

But I don't know how to fix it with symplectic Euler

So the derivative approximations are

[wide spear]

Oh you don't need second order for the main symplectic Euler part

[neon snow]

For the left end-point:
[
-3u_{j=0}(t)+4u_{j=1}(t)-u_{j=2}(t)=0
]

[wide spear]

Yep

[pine jetty]

Victor H

[neon snow]

For the right endpoint:
[
-u_{j=N-2}(t) +4u_{j=N-1}(t)-3u_{j=N}(t)=0
]

[pine jetty]

Victor H

[neon snow]

How the fk do I do this

[wide spear]

Yes these are correct

[neon snow]

Ok so

[wide spear]

So

[neon snow]

Ok my try broke matlab

It grew a little bit

Ok I don't know what to do

[wide spear]

What did you do?

[neon snow]

Honestly I don't know

[wide spear]

[neon snow]

The book changes the matrix A to this

To reflect the end points somehow

[wide spear]

Is this what you did?

[neon snow]

Yeah

And I did a random \ in MATLAB and hoped for the best

[wide spear]

For u(1,m) right?

Not u(2,m-1)

[neon snow]

But I don't understand where to put it

[wide spear]

Put what

[neon snow]

The \

What system am I solving?

[wide spear]

Why do you want to solve a system?

This is the matrix A

[neon snow]

for m = 1:M
    udot(:,m+1) = udot(:,m) + dt * c^2 * A*u(2:end-1,m);
    u(2:end-1,m+1) = u(2:end-1,m) + dt * udot(:,m+1);

This is what I have now

With the "normal" A

[wide spear]

You don't need to solve a system

udot(:,m+1)=udot(:,m)+dt * c^2 * A * u(:,m)

Where A is the modified matrix you have

[neon snow]

and change A to correct size too?

[wide spear]

Well you add the two rows on top and bottom so that it's the correct size right

[neon snow]

Yes it's N-1 x N-1 rn

So I did N+2 x N+2

[wide spear]

No it should be N by N

[neon snow]

u = zeros(N+1,M+1);
udot = zeros(N-1,M+1); %
e = ones(N-1,1);
A = N^2*spdiags([e -2*e e],-1:1,N-1,N-1);
x = linspace(0,1,N+1)';

But u is N+1

[wide spear]

Oh wait yeah N+1 by N + 1

[neon snow]

Yeah!

[neon snow]

omg Im retarded

[wide spear]

Not N+2 by N + 2

[neon snow]

yeah

I meant I added +2

xD

[wide spear]

Yeah

[neon snow]

from n-1

[wide spear]

Ok

So the first row is 1 -4 3 0s

And then the next row is 0 coeff coeff coeff 0s

And so on

[neon snow]

u = zeros(N+1,M+1);
udot = zeros(N+1,M+1); %
e = ones(N-1,1);
A = N^2*spdiags([e -2*e e],-1:1,N+1,N+1);
x = linspace(0,1,N+1)';

A(1,:) = [-3 4 -1 zeros(1,N-2)];
A(N+1,:) = [zeros(1,N-2) -1 4 -3];

Still wrong

😦

Still inverts

[wide spear]

The right hand signs are wrong

1 -4 3

[neon snow]

These?

[wide spear]

Yeah

I think

[neon snow]

Other ones correct?

[wide spear]

Yeah left hand side is -3 4 -1

[neon snow]

Literally no difference

or am I interpreting shit wrong?

[wide spear]

Hmmmm

Changing the signs is not supposed to not change anything

[neon snow]

But why would this even make sure it's zero

I never say it should equal zero

[wide spear]

Oh ok I think I see what you need to do

You keep the matrix A as a N-1 by N-1 matrix

You update u(2,m-1)

Then you know that -3u(1,m+1)+4u(2,m+1)-u(3,m+1)=0

So u(1,m+1)=(u(3,m+1)-4u(2,m+1))/(-3)

And you do the same at the other endpoint

Does this make sense?

Sorry for the late reply

[neon snow]

I can try

[neon snow]

Why u(2,m-1)?

[wide spear]

u(2:n,m-1)

Whoops

[neon snow]

Why the one previous?

[wide spear]

u(2:n,m+1)

My bad

What a mess

[neon snow]

for m = 1:M
    udot(:,m+1) = udot(:,m) + dt * c^2 * A*u(2:end-1,m);
    u(2:end-1,m+1) = u(2:end-1,m) + dt * udot(:,m+1);
    u(1,m+1) = (4*u(2,m+1)-u(3,m+1))/3;
    u(end,m+1) = (4*u(end-1,m+1)-u(end-2,m+1))/3;
end

Let's see what happens

It's probably gonna suck fucking ass

Yes, it sucked ass, more than usually actually

Looks even more Parkinsony than usual

But wait a minute

For the g(x)=sin(3pi*x)

Look how the end point at x=0 is moving

[neon snow]

Yo

This shit can't possibly work

It cant

Because we need to pick these 3 end points closest to each side to fulfill the equations

but they're all coupled to everything, arent they?

So we need to solve a huge system of eq.

???

[wide spear]

Oh yeah do that

[neon snow]

yeah do that

[wide spear]

Usually when I get here, to something that is almost correct, I randomly switch +/- signs, add +1/-1 to things, etc...

[neon snow]

xddd

i laughed irl

the proper desperation

[brave crypt]

Kek

[wide spear]

Try solving the system u(:,m+1)=u(:,m)+c^2 * A * u(:,m+1)

Maybe?

[neon snow]

with the N+1 matrix with the -4 3 -1 shit?

[wide spear]

Yeah

[neon snow]

But how do I update it with that?

What's the update scheme

iteration scheme

I solve that system first?

Sorry if I'm dumb but I'm legit on the verge of death

[wide spear]

Yeah you solve the system

(I-c^2*A)u(:,m+1)=u(:,m)

Solve for u(:,m+1)

[neon snow]

I can just use backslash in matlab

[wide spear]

Yeah

[neon snow]

So I do

u(:,m+1)=(I-c^2*A)\u(:,m)

What about udot and everything?

[wide spear]

udot(:,m+1) = udot(:,m) + dt * c^2 * A*u(2:end-1,m);

In this line

Shouldn't it be udot(:,m+1) = u(:,m) + dt * c^2 * A*u(2:end-1,m);

I don't know

Nothing makes sense anymore

[neon snow]

I know man

[wide spear]

This is when debugging becomes a spiritual experience

[neon snow]

My spirit is legit zero

nada

[wide spear]

udot(:,m+1) = udot(:,m) + dt * c^2 * A*u(2:end-1,m);

What is this line doing

Why udot(:,m) and not u(:,m)

[neon snow]

Well it's just the scheme isn't iot??

First I do

udot(:,m+1) = udot(:,m) + dt c^2 Au(:,m)

then

u(:,m+1) = u(:,m) + dt*udot(:,m+1)

[neon snow]

I just can't figure out what system I'm solving

lmao

but it's like close

[wide spear]

Is that close?

It seems like you're off by a sign

[neon snow]

I mean it's not close at all

But the end-points are mving

[wide spear]

I would guess

[neon snow]

Dude

I'm just bruteforcing stuff

Testing every possibility

for m = 1:M
    udot(:,m+1) = udot(:,m) + dt * c^2 * A*u(:,m);
    u(:,m+1) = u(:,m)+dt*(eye(N+1)-A)\udot(:,m+1);
end

eye() is identitiy matrix

[wide spear]

Don't you need c^2*A?

[neon snow]

c is one anyways

[wide spear]

Oh lol

Haha

[wide spear]

Oh my

[neon snow]

Now it's just sperging out

[fleet sail]

any hint to show divided difference is the coefficient of lagrange interpoly

[neon snow]

difference = zeros(2,M);
for i = numel(M)+1
    difference(1,i) = -3*u(1,i)+4*u(2,i)-u(3,i);
    difference(2,i) = -u(end-2,i)+4*u(end-1,i)-3*u(end,i);
end

Only two values

Probably round off values

So it's like correct

[wide spear]

Ummmmm

Your signs are still off

End should be 1 -4 3

[neon snow]

It doesn't change anything

[wide spear]

Well you want it to be correct

[neon snow]

Honestly, I see your point but

[wide spear]

And it should change something after it stops randomly flipping

[neon snow]

for m = 1:M
    udot(:,m+1) = udot(:,m) + dt * c^2 * A*u(:,m);
    u(:,m+1) = u(:,m)+(eye(N+1)-dt*c^2*A)\udot(:,m+1);
end

Is still even slighty

like

this is 100% nonsensical

Like legit

[wide spear]

Oh wait in the second line you don't need the first u(:,m)

I think

Wait nevermind

[neon snow]

But this doesn't even make sense in the slightest

[wide spear]

Nothing makes sense anymore

[fleet sail]

how in za warudo

any hints

[fleet sail]

cant seem to use induction

pain

also I dont want to assume they are permutation invariant as i prove permutation invariant assuming this fact, but if its really necessary then its ok

[wide spear]

How did you define Lf?

[wide spear]

[neon snow]

lmao

speaking of neumann boundary conditions

I solved it

[wide spear]

Nice

[brave crypt]

Excellent

[neon snow]

Thanks for all the help yesterday btw

Really appreciate it

[wide spear]

I'm not sure how much I actually helped though

[neon snow]

Well you were part of the process

Who cares who drove it across the finish line

[wide spear]

[fleet sail]

Lf is the unique interpolating poly of degree m for m data points

the lagrange form is ez, but the professor just took it for granted that the recursive def of divided difference will give the coefficient w.r.t the other basis for the same interpol

[pine jetty]

Anticipation

[wide spear]

Is this how you defined Lagrange interpolation? This is Lagrange interpolation with divided difference coefficients right?

I think that can induct on k, the number of interpolating points

[fleet sail]

yea

but i tried inducting and failed

wait

yea i think i failed

[wide spear]

Well what did you try

[fleet sail]

so assume its true for n ≤ m

[wide spear]

Ok

[fleet sail]

i can't show its true for n = m+1 because everything is a mess

[wide spear]

Well I agree that it will be a mess

[fleet sail]

wait either im wrong yesterday or im wrong today

[wide spear]

[fleet sail]

OK so I suppoed

$\sum_{k=0}^mf[x_0,\cdots,x_k]\omega_k(x) = \sum_{k=0}^ma_k\omega_k(x)$ where RHS is the interpolating polynomial

[pine jetty]

Anticipation

[fleet sail]

we know it exists cuz the wks are a basis

now i need to show $\sum_{k=0}^{m+1}f[x_0,\cdots,x_k]\omega_k(x) = \sum_{k=0}^{m+1}a_k\omega_k(x)$

and using inductive assumption thing

[wide spear]

Are the w_k the same on both sides?

[fleet sail]

yea

[pine jetty]

Anticipation

[fleet sail]

which is same as showing $f[x_0,\cdots,x_{m+1}]\omega_{m+1}(x) = a_{m+1}\omega_{m+1}(x)$

[wide spear]

Right

[pine jetty]

Anticipation

[wide spear]

So it's showing that f[x]=a_{m+1}

[fleet sail]

wait but i need to show that

[wide spear]

Well omega_{m+1}(x) is on both sides so you cancel it out

So you need to show this

Yes I agree

[fleet sail]

no i mean

[wide spear]

Expand out the definitions

[fleet sail]

i need to show this thing

and by showing this thing im done as u said

ok so another way to phrase this is I need to show

[pine jetty]

Anticipation

[prime kraken]

i'm under the impression it is a telescoping sequence if you look at partial sums

do a few smlal ones and look for a pattern

[wide spear]

Wrong channel?

[fleet sail]

i think not but it doesn't work 😦

[fleet sail]

like it gets messy once i start doing f[x0,x1,x2,x3]

[wide spear]

This is not a telescoping series

[prime kraken]

yeah mb, i didn't mean telescoping, but self similar

[brave crypt]

I'm trying to understand why fixed point iteration is at least quadratic convergence when g'(x) = 0.

[wide spear]

Which fixed point iteration? Just x_{n+1}=f(x_n)?

[brave crypt]

yeah

[wide spear]

Or g I guess

Ok

[brave crypt]

I get to this point and dont know how to get a constant out of this?

[wide spear]

What does quadratic convergence mean?

To get a constant, try taking the Taylor expansion of f(x_n) about r

So f(x_n)=f(r)+f'(r)(x-r_n)+....

[brave crypt]

one sec

[brave crypt]

it means the picture above is equal to some constant

[brave crypt]

yeah that works

[wide spear]

I should hope so, I don't have the applied math role for no reason

[rose turret]

can someone give me like a eli5 for EM GMM? like how do we initialize mean, covariance and pie with respect to the number of clusters? In python, given m x n matrix, would the implementation of expectation just be scipy.stats.multivariate_normal(array[:, n], mean[cluster_num], cov[cluster_num]) for all n?

[wide spear]

Wikipedia suggests that the parameters can be randomly initialized

[rose turret]

@wide spear between [0,1], right? because normal/gaussian distribution

[wide spear]

(Also disclaimer I don't know any stats but I am in close contact with someone who does so they're answering via proxy)

Wait why between [0,1]?

[rose turret]

oh good question... umm

[jolly wadi]

This guy works out an example:
https://stephens999.github.io/fiveMinuteStats/intro_to_em.html

I guess that's the wrong language...

[light rover]

my homework asks me to do this

for

but they are not circles?

and im reading the textbook and it only has stuff for circles

can someone pls explain

[wide spear]

Newton's method to find the root? Or Newton's method for finding minima

[light rover]

uhh

im assuming root

[wide spear]

Well

Make sure which one it is

[light rover]

you see, I would if my professor actually responds to his emails

but does the first two steps differ based on which it is?

[wide spear]

Well

They're doing two completely different things

So.......

[light rover]

can I know both?

[wide spear]

First, can you explain what the textbook has about circles?

[light rover]

this

[wide spear]

Ok

Where is the Gauss-Newton method defined

[wide spear]

Ok

Why does being a circle matter

[light rover]

I wish I knew the answer to that question LOL I'm just trying to understand this thing

because my prof doesnt lecture :)

[wide spear]

Can you write down the r_1 and r_2 for example 3?

[light rover]

uh

[wide spear]

r_1(u,v)=....

r_2(u,v)=....

[light rover]

yeah but isnt R1 the radius

so like

[wide spear]

Don't think about that

Do you understand how the r_1,r_2, and r_3 were derived in example 4.21?

[light rover]

I understand the parts before the radius yeah

isnt that just the distance

[wide spear]

The distance from what

[light rover]

x to y?

[wide spear]

No

[light rover]

x, x1?

[wide spear]

(x1,y1) to (x,y)

Do understand what you're trying to do?

[light rover]

noppe

[wide spear]

Did you read the chapter?

[light rover]

it doesn't really make sense but it's okay sorry for bothering you

[wide spear]

No

I'm going to make sure you understand

For example 4.21, we are trying to find a point (x,y) that minimizes the distance to the three circles with centers (x_1,y_1), (x_2,y_2), and (x_3,y_3) with radii R1, R2, and R3

r_1 is the distance from (x,y) to the 1st circle

[light rover]

okay

[wide spear]

r_2 is the distance from (x,y) to the 2nd circle

r_3 is the distance from (x,y) to the 3rd circle

[light rover]

oh my gosh

so sorry about that

[wide spear]

[light rover]

i don't understand why we are finding a point that minimizes the distance

[wide spear]

Well

[light rover]

and that entire thing you typed

[wide spear]

Why is another question

But that's the goal of this section it seems

Which thing I typed

[light rover]

the three circles and the centers

[wide spear]

Ok

What don't you understand

[light rover]

what this method exists for

and how do I apply it to this equation

[wide spear]

First, do you understand the example they give?

4.21

[light rover]

why they used what they used? no
how to use it on another circle? maybe

[wide spear]

Well, it turns out that finding minima of these types of functions is pretty important in applied math

[light rover]

okay I did not know that thank you

sorry, I am just a very confused being but I will try my best here

so is this related to finding the minima?

[wide spear]

Well, you have the three distance functions

(x,y) to each of the three circles

You combine these to get the function you want to minimize

Then you use the Gauss-Newton method to minimize it

[light rover]

okay is the radius relevant or irrelevant in the u^2 + 4v = 4

[wide spear]

Well, stop thinking about the radius

[light rover]

why?

[wide spear]

Instead, think about how you can find an analogous distance function

It won't be in exactly the same form

[light rover]

okay

[wide spear]

So how can you find the distance from a point to the curve you were given?

[light rover]

the abs value of a (point on the curve) - (the point we're given)?

[wide spear]

If you aren't sure you should ask in #geometry-and-trigonometry

I think

[light rover]

d(x,y) = |x - y|

or thats the distance I know

[wide spear]

Well that's the distance between x and y

You want the distance between (x,y) and the curve u^2+4v^2=4

So you'll need to find another equation

[light rover]

okayy

[wide spear]

Find a formula for this distance and then come back

[light rover]

okay gotcha

[wide spear]

You'll also want to find the formula for the distance between (x,y) and the curve 4u^2+v^2=4

[light rover]

okay

thank u

[light rover]

so I finally found my profs "notes" on this as I was on the hunt to find if he posted this equation somewhere, and I think I am even more confused than when I started

but I think I found what he wants out of this question (?)

[light rover]

nvmm I figured it out

[wide spear]

[light rover]

his definition of newtons method is very different

it seems like

what is this

is this also newtons method?

[wide spear]

Well, Newton's method refers to many different things

[light rover]

such as?

[wide spear]

Newton's method for finding zeros of functions

Newton's method for finding minima of functions

[light rover]

ohhh

okay this makes more sense to me now

thanks for all ur patience ^^

even though I asked some pretty irrelevant and dumb stuff I hope I didnt take too much of your time

[wide spear]

[light rover]

wait can I ask one more question?

[wide spear]

Ok

[light rover]

are you a math major?

[neon snow]

I see you're using Sauer numerical analysis book

Dog shit book

[light rover]

oh is there a better book you would recommend?

[neon snow]

Nah it's the only book I've tried

[light rover]

LOL

[neon snow]

Course recommended

So maybe it's the best but then all must be trash

[light rover]

yeah but imo its better than trying to decipher my professor's chicken scratch

when he decided to give up lecturing all together

[neon snow]

Oh yeah anything is usually better than the professor's parkinson

[light rover]

yep

ngl

i thought that was wiqqles

for like 2 hours

[neon snow]

Does his g go the other way

[light rover]

no idea

[neon snow]

Why did he reflect his g around the horizontal axis

[light rover]

LOL IKR

[wide spear]

Yes I am a math major

There are no good numerical analysis books

[light rover]

wow... I have much respect

I can't handle that

[wide spear]

[light rover]

are you undergrad?

[wide spear]

Yes

[light rover]

ohh ok!

[wide spear]

[hollow ingot]

michael heath’s introductory survey is alright

wait angetenar is an undergrad

[wide spear]

It doesn't really discuss numerical methods for pdes though?

Just looking at the table of contents

[hollow ingot]

not in any sort of depth

[wide spear]

Also very little about FFT and signal processing

[wide spear]

What does this mean

[hollow ingot]

you seem pretty knowledgeable in this stuff and i thought you were a grad student at least

[wide spear]

Well

I'll probably be a grad student next year

And I've done a fair amount of classes about numerical analysis

[hollow ingot]

sounds great

i had like, one numerical analysis class that was offered in my undergrad

[wide spear]

Ah

I go to a big school

With a lot of applied mathematicians

2 undergrad numerical analysis classes, 3 graduate ones

Some other applied math classes that touch on the topic

[light rover]

Wait how does that happen

I thought usually people take a year off before going to grad school?

[wide spear]

Ummmm

Gap years are of varying popularity

Depending on circumstances

[light rover]

Wow

I applaud u

[wide spear]

Thanks

Grad school admissions are hard

[light rover]

Which school are you aiming for?

[wide spear]

Well

Of the schools that I haven't been rejected from yet, Caltech would be cool

[light rover]

:0

thats very ambitious of you

[wide spear]

Well, I still have dreams

I haven't given up all of them yet

[light rover]

whOa that sounds sad

but not sad at the same time

[wide spear]

Becoming an adult means giving up on your dreams

[light rover]

not necessarily true

[wide spear]

[light rover]

eerr i meann

what is your dream?

[brave crypt]

testing testing

[wide spear]

[fleet sail]

my dream is to taylor expand every day

[fleet sail]

this proof valid?

feel like its pretty unjustified for some reason

[light rover]

Proofs

Heh the bane of my existence

I'm surprised I even lived thru my real analysis class

[light rover]

question - how do I find d degree of polynomials that pass through four points?

I could just be really dumb and this could just be something super simple, but I'm going to accept my dumbness

where does this come from

[random hornet]

Lagrange interpolation could give you a 4th degree polynomial, right?

I'm not sure if that's the lowest possible you can get

[light rover]

uhh is that okay?

I can do that?

cus the question asks for d degree

but never specified 3rd or 4th

[random hornet]

Okay so a quick web-search reveals you'd get a polynomial of least possible degree from Lagrange interpolation

For k given points, it will be at most k-1

So in your case, at most 3

Once you plug in the actual values of the points, you can possibly obtain a polynomial of a smaller degree than 3

[light rover]

oh I see

alright, thank you

😄

[random hornet]

I'd still suggest you read about polynomial interpolation methods

[dark sinew]

just try remembering that you need two points to get a polynomial of degree 1 (a line)

[random hornet]

My knowledge about this is too superficial

[dark sinew]

that gives you an idea of the relationship

[light rover]

yeah I'll probably hunt the textbook

[dark sinew]

d+1 data points for polynomial of degree d

[light rover]

oh okay thank you for the pointer

[dark sinew]

yeah whenever you're trying to remember whether its like plus or minus one

think about the edge cases

[light rover]

ohhh ok

[wide spear]

Ultimately this is because a polynomial of degree n has n+1 coefficients

[light rover]

i see

[pine jetty]

davidW

[wide spear]

What does triangular support mean

[pine jetty]

davidW

[glad mantle]

yeah so trying to see if there's a slick way to do this without induction lol, maybe something involving Schur Complements? hm, thinking

[wide spear]

No induction seems like the best way to do this

Schur complements will indeed work, but I don't think it will give you the triangular support condition

[brave crypt]

currently i parsed the input file into these:

1. nu of faces (cells)
2. nu of edges (Using Eulers Identity V - E + C = 2)
3. coordinate of each vertex
4. cell centroids
   and made this image

could one suggest an algorithm to enumerate the edges of such graph?

[light rover]

Imagine knowing the answer to this... I can't lol

https://tenor.com/view/cute-dragon-omocat-charmander-pokemon-gif-16022601

[brave crypt]

it should be sth like

create edge center array
for each cell loop around the edges
if mean of any 2 is not in edge center array
fill center array

if i do not mess up with logic :\

[wide spear]

Have you considered breadth first graph traversal?

[brave crypt]

wasnt it for trees only

[wide spear]

Wait how is your graph stored

Do you not store it as a set of vertices and a set of edges?

[brave crypt]

just coords

x array
y array
triangle connectivity array

[wide spear]

Can you convert this into an adjacency matrix

[brave crypt]

had to work in google sheets also :\

[wide spear]

Triangle connectivity array?

[brave crypt]

size of 3 by nu of triangle regions

[wide spear]

Oh my god why is this in google sheets

Turn your triangle connectivity array into an adjacency matrix

[brave crypt]

😐
okay

with adj mat it would be very tidy ikr
but moving connectivity into adj mat is another nightmare

[wide spear]

Write some code

[brave crypt]

ye, kinda doing it rn

nvm it should be ez to create adj matrix

run over all triang
put + 1 for each combination of 3 vertices in my file (or +0 if 1 already)

boom, made it

strange, nu of edges in adj matrix does not coincide with eulers identity formula >_>

nvm, had to fill ij and ji, then take half

[brave crypt]

all done! thank you @wide spear

[wide spear]

[fleet sail]

i rly want to make the last part of this pf better

[fleet sail]

heres the updated i think its better but maybe its still flawed

[light rover]

how did they get 1???

for f[x1x2x3]?

is it not 2 - (-1) / 1 - 3?

ohhhh sike

nvm

[wide spear]

[light rover]

haha my math isn't the greatest

i respect the mathematicians

I am a part of the 10% of chinese ppl that can't math

[wide spear]

Well

I don't really believe that there are people who can't do math

[light rover]

wait can I ask a question why is -1 and 2 both f[x1, x2]

[light rover]

haha then you've yet to meet my math skills

[wide spear]

f[x1,x2]=2

Oh I see

That's a typo

The second one should be f[x2,x3]

And the third should be f[x3,x4]

[light rover]

ohh okay that makes more sense

thank u

i've taken a lot of math classes and still learned nothing

as you can see

[wide spear]

So?

Maybe math classes aren't the right way for you to learn math

[light rover]

i mean watching yt videos aren't really like amazing either

or reading stuff from textbooks

unless you have a different way

although patrick JMT - great dude

[wide spear]

Well I have no experience as an educator so I don't think I'm the right one to make recommendations

But something that I've found very important for me when learning is motivation

Learning math for the sake of doing things with it

[light rover]

such as?

(besides everyday life things)

[wide spear]

Well I'm interested in fluid dynamics

I think it's a really cool field

https://www.outpan.com/app/44bdd9869c/interactive-fluid-simulation

See this

Fluids are very pretty

And behave in very cool ways

[light rover]

that is indeed very cool

[wide spear]

And understanding this is a very difficult mathematical question

[light rover]

I would imagine so

I respect ^^

[fleet sail]

divided difference is just pascal triangle

[wide spear]

[light rover]

oh oke

I mean thanks

how

[turbid jay]

hey, has someone done some convex optimisation? We proved that gradient descent for a convex and Lipschitz $f$ with stepsize $\gamma = \frac{R}{B\sqrt{T}}$ where $||x_0-x^|| \leq R$ and $\nabla f(x) \leq B$ is in $O(\frac{R^2B^2}{\varepsilon^2})$. $T$ is the number of total steps. However, we don't know $B$ and $R$ (only that they exist) so we can't actually use this algorithm. Suppose I know $R$ and also know that some $B$ exists, I need to find an algorithm to find the optimum $x^$ up to some arbitrary error in the same complexity as before. I'm not sure how I should go about that tbh. Should I just try different stepsizes that only depend on $B$ and $T$ and try to prove that this has the correct complexity, or is there a better way? I feel like I should somehow use the fact that $f$ is Lipschitz

[wide spear]

What did you prove about gradient descent

[pine jetty]

lyinch

[wide spear]

If you know R and that B exists, why can't you just run gradient descent until the error is small enough?

In practice you never know what R and B are

You just run gradient descent

[turbid jay]

it's an exercice in a convex optimisation lecture 🙂

I proved that if R and B exist, then I can find the optimum in with the above given steps

where epsilon is f(x)-f(x*) < epsilon

[wide spear]

Sure

[turbid jay]

but I can only reach this for the correct gamma, and to chose that I need to know R and B. Which is, as you said, not feasible

[wide spear]

Gradient descent still works if gamma is not chosen optimally

You just converge slower

[turbid jay]

yes

but I need to prove how slow

[wide spear]

Ah ok there we go

[turbid jay]

now I assume that I know R and that B exists but is unknown. I need to find a gamma such that the convergence has still the same complexity above and it's actually feasible

(or better complexity of course, it's big O)

[wide spear]

8da is typing

[brave crypt]

I don't think B is knowable, or boundable, right? No matter how finely you know f, there can be some crazy shit happen in some small neighborhood

[wide spear]

Hopefully you are optimizing on a convex compact set?

[turbid jay]

it's an assumption here, we start with convex differentiable Lipschitz functions

and a minimum exists

[wide spear]

Even for something like f(x)=x^2

B does not exist

f'(x)=2x is not bounded

[turbid jay]

yes, we've also treated this. To reason about such functions we need different assumptions such as strong convex functions

but that's not part of this problem

strong convexity will be treated in the next 2 exercises, don't worry 😄

(note strong convex is different than strict convex)

[wide spear]

Ok so what happens if you do the error analysis with an arbitrary step size gamma

[brave crypt]

Oh, I guess if it's convex nothing too weird can happen

[turbid jay]

yes I think that's the starting point. One second

that's for an arbitrary gamma

[light rover]

i have a question

nvm

wait I do

wait nvm

[wide spear]

[brave crypt]

[wide spear]

"Waifu's Region"

[light rover]

is this

not

right?

[wide spear]

It's fine

[light rover]

then whats up with this

my brain

is not comprehending

so which one is right and which one is wrong?

[wide spear]

They're the same

[light rover]

what????

they are???

I'm sorry

my brain doesnt understand that

[wide spear]

Do ask your algebra questions in #prealg-and-algebra

[light rover]

oh sorry

[wide spear]

In the first one you can divide 12 and 15 by 3

[light rover]

https://tenor.com/view/anime-cute-shocked-zoom-surprised-gif-15198304

sorry sorry I instinctively come here bc this is for my computational math class

but I will note that next time

[light rover]

um I'm not sure if I'm starting this question right

it was the newton's multivariate method thing

but like my prof's notes look like this:

and thats kind of what I got out of it?

but I'm not sure if i'm even on the right track ...

[hollow ingot]

looks okay i think? you have the jacobian and the function evaluated at x, now you can do a linear solve for the step delta x.

[turbid jay]

For anyone wondering about this question, I got the solution today. Since we know that the gradient is bounded by some B (we don't know the value), the algorithm is simply to guess the correct B. Let's call the guess B' . We initialise this by B'=epsilon/R (not sure yet why, I need to derive the proof myself) and then for every iteration, we check if the gradient <= B' . If it's not, then we double it: B' = 2B' . This means that in the worst case, our guessed bound B' is at most 2B, the real but unknown bound. Now we know B' and also know R (by assumption) and this gives us the stopping criterion T >= 4(RB'/epsilon)^2 and we have in bigO the same number of steps as mentioned above

[prime kraken]

looks like the Armijo rule

[brave crypt]

interesting

[turbid jay]

I don't know the Armijo rule but it looks indeed similar

[light rover]

Okay thanks!

[wide spear]

[brave crypt]

I would like to compute first and second derivatives of scattered 2D/3D data using fast fourier transforms, are there any good sources for this if it's even possible?

[wide spear]

Ok so you know how Fourier transforms turn derivatives into polynomials right

If you have data, look at Poisson summation formula or Shannon-Whitaker sampling to reconstruct the Fourier transform

This will at least be a starting point

[lofty bone]

I have a very good reference but itis a untranslated french reference

[wide spear]

Oh Anatole are you attending Sylvie's talk

[lofty bone]

this is right now ?

Oh f*ck

[wide spear]

[lofty bone]

thanks bruh

[wide spear]

Ah I see you here

[lofty bone]

who are you ?

[wide spear]

[brave crypt]

I took french in high school, I could give it a shot

[wide spear]

Way to ignore my suggestion

[brave crypt]

Oh sorry I'm on tablet and just missed it, reading it now

[lofty bone]

https://cel.archives-ouvertes.fr/cel-00573975/document

It's free in the open archive

HAL

the link is legal

[brave crypt]

Ok I will look at those formulas and that reference thank you

[empty umbra]

Hello

I need help

[wide spear]

Ok

With what

Did you read #❓how-to-get-help

[empty umbra]

Yes

[wide spear]

Ok

I will say that based on the quick glimpse I saw of the image you sent, this is not the correct channel

[empty umbra]

Ohh which channel should i go to 🥺

[wide spear]

Ask in one of the math help channels

[empty umbra]

Okay thank you

[glad mantle]

@wide spear do you have thoughts on how to continue, given that I can express transpose as matrix mult?

[wide spear]

Ok so you have A'=LPL'

[glad mantle]

right

wait no

[wide spear]

P is your permutation that we use as a transpose

Wait no that's not legal

[glad mantle]

well, maybe you're using different notation, but I originally had that we have A_n = A' + (LPL')_n

where A' is non-invertible

[wide spear]

Oh yeah

[glad mantle]

and we have that (LPL')_n -> A' uniformly

[wide spear]

You should look more into the properties of the Bruhat decomposition

Is all I can say

[glad mantle]

I think there is still room for it to work

since I know we have LUP decomp for any square matrix

and any triangular matrix is the limit of invertible triangular matrices

[wide spear]

Yes triangular matrices are closed

[glad mantle]

Ok, so then aren't we done?

[wide spear]

We are?

[glad mantle]

(LPL')_n = L_n P_n L'_n

and triangular matrices and permutation matrices are closed

well, maybe not permutation, is limit of permutation mat even well defined or smth

[wide spear]

Would you like to explain how you've written a general matrix as the product of two lower diagonal matrices and a permutation?

[brave crypt]

There are only finitelyany permutation matrices, so certainly closed, right?

[wide spear]

Yes finite number of permutation matrices

[glad mantle]

well that's why I wrote what I did originally, but what is the limit of a permutation matrix

[wide spear]

I don't think the limit needs to exist

[brave crypt]

I guess we must pass to a subsequence for convergence

[wide spear]

It can keep spinning around

Or something

[glad mantle]

exactly my point

[glad mantle]

but yeah, maybe that's not a problem for here

[glad mantle]

I just used Bruhat decomp, and a neat proof from https://mathoverflow.net/questions/15438/a-slick-proof-of-the-bruhat-decomposition-for-gl-nk

[wide spear]

Sure, but it is not true that you can write a general non-invertible matrix as the product of two lower triangular matrices and a permutation

[glad mantle]

yes, that's why I'm asking n the first place lol, since this is only for invertible

but I know that square matrices have LUP decomp, so there should be a way to extend

[wide spear]

Well you'll need to turn L' into an upper triangular matrix

[brave crypt]

Is there such a thing as LUL decomposition?

[wide spear]

No

For spd matrices you have Cholesky

Which is A=LL*

[glad mantle]

sure just take P' s.t. LPL' = L P (P' L' P'^{-1}) = L (P P') (L' P'^{-1}) = L P'' U

[glad mantle]

you have LDL

for D diagonal

[brave crypt]

Sadly LDL doesn't sound as lulz as LUL

[glad mantle]

which is basically cholesky like ang said

[wide spear]

Some indefinite matrices have D with negative elements

Which is definitely not a permutation matrix

[glad mantle]

I'm aware

[glad mantle]

do you agree with this though

[wide spear]

So you have L'=P'L'P'^-1

Why is P' invertible

Ok it's a permutation

[glad mantle]

it's the inverse permutation

[wide spear]

Sure

Then you say L'P'^-1 is upper triangular

Why is this true

[glad mantle]

that's how I chose P'

[wide spear]

Can you elaborate further

[glad mantle]

it's the permutation matrix that makes L'P'^{-1} upper triangular

[wide spear]

I don't know why you can choose P' in this way

Why does it exist

[glad mantle]

isn't a permutation matrix just a reordering of the rows and columns

can't I reorder the rows and columns of a lower triangular matrix into an upper triangular one

[wide spear]

If you have U=LP then you're saying that UL^-1=P

[glad mantle]

no I'm not

[wide spear]

I'm still not sure this P exists

[glad mantle]

I'm saying L = UP^{-1}

I never claim L' is invertible

[wide spear]

Sure

But you're writing transposition as a permutation

Which I'm not sure can be done

[glad mantle]

is this not a valid argument

[brave crypt]

I think it surely exists right? We can prove it by drawing a triangle and reflecting it twice

[wide spear]

Can you?

Let's say L is diagonal

Then it is lower and upper triangular

So P is the identity

Clearly the identity will not take the transpose of any non-diagonal matrix

[brave crypt]

At least assuming permutation matrix is composition of row and column interchanges, and not just row interchanges

[glad mantle]

I don't understand how what you just wrote doesn't show what I want

then P is the identity

so you found such a P

[wide spear]

AP only permutes columns

You want the P to be the same for all of them

[glad mantle]

no I don't

[wide spear]

So that you have convergence

Anyways multiplying L'P only reorders columns

[glad mantle]

sure, for convergence we will need the sequence of Ps

[wide spear]

It doesn't reorder rows

[glad mantle]

the sequence of P' s s.t. each L' P'^{-1} is U

but a P' will always exists

[wide spear]

No

It doesn't

You need two permutation matrices to turn a lower triangular matrix into an upper triangular one

[glad mantle]

oh

is that your point? hm

[wide spear]

One to rearrange rows and one to rearrange columns

[glad mantle]

sure

[wide spear]

Your P' doesn't exist

You would need to multiply P_1LP_2

[glad mantle]

wait I have two Ps here

I have P'' = P P'

I can also write L = P_1 L P_1^{-1} if we want

so we'd have like

[wide spear]

You don't know that P_1^-1 is P_2

[glad mantle]

$$LPL' = (P_1 L P_1^{-1}) P (P_2 L' P_2^{-1}) = P_1 L (P_1^{-1} P P_2) L' P_2^{-1} = P_1 L U P_2^{-1} $$

[pine jetty]

davidW

[glad mantle]

maybe you're right we're missing another matrix, or does this fix anything

I'm not sure

[wide spear]

Why is there a P_2 inverse at the end

[glad mantle]

did you see the latex I put above

[wide spear]

Yeah

[glad mantle]

we get to pick P1 and P2

[wide spear]

It should be in U

[glad mantle]

well, I hesistated to write that since then you have

P_1^{-1} P (P_2 L' P_2^{-1}) = P_1^{-1} P L'

so if we needed to choose 2 matrices to be able to permute the rows and columns then you'd want to group (P_1^{-1} P P_2) so it acts on L' to make it U

[wide spear]

You misunderstood the 2 matrices

You need to multiply one matrix in front and one matrix after

Multiplying before permutes rows, multiplying after permutes columns

[glad mantle]

Ok sure

$$LPL' = (P_1 L P_1^{-1}) P (P_2 L' P_2^{-1}) = P_1 L (P_1^{-1} P P_2) L' (P_2^{-1}) = P_1 L U$$

[pine jetty]

davidW

[glad mantle]

so now the two parenthasized permutation matrices act on L' before and after, respectively

[wide spear]

Yes

What's next

[glad mantle]

I'm not sure, did you have any thoughts?

[wide spear]

Well, do you want the individual matrices to converge?

This is hard

Because you only had bounds on the product of the matrices

[glad mantle]

yes

[glad mantle]

I'm not sure what you mean by that

actually now I see

(sorry :p)

[wide spear]

Well, as the condition number of A_n goes to infinity, these 3 matrices can behave wildly

[glad mantle]

sure so let's add some error mats

yep

[wide spear]

They will behave wildly probably

[glad mantle]

but we know PLU is indeed a way to represent a square mat

[wide spear]

Yes

[glad mantle]

we shouldn't need any other conditions on these matrices

[wide spear]

But we don't know that the PLU coming from the Bruhat decomposition is

[glad mantle]

?

[wide spear]

I don't know do we

Maybe we do

[glad mantle]

why would we

[wide spear]

Why does PLP^-1=L

[glad mantle]

we know A is square, and L is any triangular matrix and U is any triangular matrix, and P is a permutation matrix

[wide spear]

Right multiply by P gives you PL=LP

But there's no reason for these to be the same

[glad mantle]

where does this show up

[wide spear]

LPL' = (P_1 L P_1^{-1}) P (P_2 L' P_2^{-1}) = P_1 L (P_1^{-1} P P_2) L' (P_2^{-1}) = P_1 L U

[glad mantle]

where did I use the equality you wrote

[wide spear]

You wrote L=P_1LP_1^-1

[glad mantle]

yes

we get to pick P_1

oh

did I just use the identity matrix lmao

[wide spear]

Yes

So this doesn't work

[glad mantle]

good catch

[glad mantle]

wait sorry, isn't this always true

by definition of permutation matrix basically

[wide spear]

Well

No

P_1P_1^-1L=L

And P_1^-1 exists because permutations are a group

But when you are inverting group elements you need to act from the same side

[glad mantle]

right, but if I first permute the entries then I move them back why is that not what I started with

[wide spear]

When you do P_1LP_1^-1 you have P_1^-1 permuting the columns and P_1 permuting the rows

How can a column permutation and a row permutation cancel each other out?

[glad mantle]

ah ok, sorry for dumb question lol

[wide spear]

[glad mantle]

@wide spear so now I think I see why you brought up cholesky before, since bc A is invertible so we have cholesky, and we're gonna be able to choose L = L', with P being the inverse ordering of the rows

[wide spear]

Invertible does not mean psd

[glad mantle]

whoops positive definite

[wide spear]

Invertible does not mean positive definite

[glad mantle]

touche

[wide spear]

[glad mantle]

@wide spear so why did you think this related to cholesky then?

if we're only demanding invertible and never anything psd

[wide spear]

Burhat feels like Cholesky

[glad mantle]

it do 😩

[wide spear]

But it isn't

[glad mantle]

it's cholesky but made worse by algebraists

[wide spear]

Yes exactly