is it the frequency that differs?

the derivative, for a start

how?

Also the value at every other value of $x$ :p

They have a phase difference.

@Phase yeah

I went out so wasn’t playing the whole time

Just got to Alexandria

Nice

Is Caesar in the game?

What period is it actually set in

~50 BC, so no

Well, @parvin just use some small but nonzero value

less than pi/2

and you'll see pretty marked differences

@0celo7 dang

wait

Yeah I dunno why they didn’t do it during Caesar time

What are you on about @parvin

Maybe the next one

they dont start from a maximum on the y axis

Cos(x) does but 1 + sin(x) doesnt

I came here not to praise Caesar but to bury him!

Ave Caesar

@0celo7 did you play ac2?

Because I just had a pretty cool idea

AC2 had the armour of brutus right? Imagine if you played as him during the relevant time

and assassinated Caesar

that shit would be cool

Yeah it would be

But we already had a Rome game

wait

50BC?

Headshotting people with the bow is so satisfying

Julius caesar isn't dead yet

A timeskip of about 5 years and you'll be near caesar's assassination

Maybe they had that planned afterall?

Oh maybe he will be in it

Idk no spoilers pls

it aint spoilers dude

Its random guessin

Hang on

this is probably a really dumb question but

Take a polynomial centred on x

Centered?

yeah like

it's 0 at x = 0

i defined that terribly

Ignore that entire point actually

I just mean take a polynomial where all it's roots are at x = 0

so if you take any polynomial with an odd power in the leading term, it's just a cubic looking equation

and the gradient is constrained to be the same sign either side of x = 0 as the derivative's leading power will be even

So how do you add an infinite number of odd powered polynomials

and get a periodic function with varying gradient signs

@Phase so it’s just x^n?

I guess Im confused why something like y = x - x^3/3! + x^5/5!... will always have all its roots at 0 for finite numbers of terms

but will be periodic and have nonzero roots for infinite terms

nah a sum of polynomials of odd powers of x

since function addition works by pointwise addition and all that

I honestly have no idea what you’re talking about.

Im confused why sin(x) is equal to an infinite number of the terms of the sum i gave above

but for any finite number of terms all it's roots are zero

wait

what the fuck

am I talking about

AAAAAAAAAAAAAAAA DELETE ME FROM SE

this is humiliating.

for some reason in my head I didn't register that the signs weren't all the same.

Does the Hamiltonian formalism actually give anything the Lagrangian formalism doesn't in the classical setting? Is it just a more convenient framework to consider?

@Phase Sounds like you might need to go to sleep my dude

I agree but I wont go to sleep yet

@0celo7 im tempted to get ACO now

You've tainted me

I just overclocked

Haven’t had to do that in a while

Alexandria destroys CPUs

Yeah i just read

apparently its the DRM

it basically makes it so you're almost emulating large parts of the game

Which is why it's so resource hungry

Think I'll give ACO a pass for now. Maybe if they fix their DRM

Good god I’m doing 4Ghz on 6 cores and it’s using 90%

idek how to overclock safely

I'm a bit scared to try it, don't wanna fry my components especially because it's a laptop, and that makes replacing components a bit.. trickier

Hidden blade! Finally.

wow spoilers

Cheeky fuckers

"Our teams have optimised Assassin's Creed: Origins to run as smoothly as possible on various systems, but if you experience performance issues (for example low framerate, screen freezing for a very short time, action slowed down) the following info may help you understand the root cause of the issue and correct the situation."

what is it

What is the info

My game runs fine but I’ve got a beast rig and shouldn’t have to use 100%

$L$ is lagrangian, $\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot{q_i}}=\frac{\partial L}{\partial q_i}$. How?

im just annoyed they pretend its on the players fault

@user104729 if that equation is true it reflects that the path is an extrema of a functional

the functional in this case being the action of a path

iirc

@user104729 $\frac{d}{dt}$, not $\frac{\partial}{\partial t}$

Apologies, the total derivative appears

and to calculate this you have to take the Frechet derivative of the action

or the variation, as physicists say

Thank you, I'll look into that

God

Anything rigorous is so terrifying to look at without 100% of the prereqs

I just googled Frechet derivative

If you don't know it (meaning anything), it is terrifying. If you know it, it is beyond trivial.

@Slereah you mean Gateaux

One of these yeah

whatever

the variational derivative

there's ANOTHER name for it but I forget

>french

>"Gateaux"

what a delicious derivative

@0celo7 PVAL just came up with a construction that resolves the issue like a magic trick

@BalarkaSen how do?

So?

@0celo7 We wanted to approximate -ydx + xdy on a perturbed nbhd of the circle by exact forms, yeah?

sounds right

We're doing this in the C^0 norm. So choose for a given $N > 0$ a $C^0$ isotopy $h_1$ of the unit circle such that $h_1(S^1)$ has length $> N$ wrt metric in $\Bbb R^2$

Ok

Define the volume from $\omega$ on $h_1(S^1)$ induced from $\Bbb R^2$

Notice that as $N \to \infty$, $\omega$ goes to $0$ in the $C^0$ norm

(Eg big circles have small volume forms - circle of radius r gets vol. form multiplied by 1/r^2)

@0celo7 I didn't mean $C^0$ isotopy, I meant $C^0$-small. It's small but increases length of the circle as much as it wants

(There exists such isotopies)

I know what you meant.

Just make it wiggle.

OK great

Go on

Yep

So consider $(-ydx + xdy)/(x^2 + y^2) - \omega$

This integrates to zero along $h_1(S^1)$, because both of those forms integrate to $2\pi$

So it's an exact form in an nbhd $V_{h_t}$ of $h_1(S^1)$

Let's say it's $d\psi$. Fix a definite nbhd $U$ of $S^1$ and bump it up to $d\rho\psi$ for some bump function $\rho$ on $V_{h_t}$

Why is the integral of omega 2pi? Wouldn’t it be something involving N?

It should be the length of the perturbed circle, no?

Hmm, you're prolly right

well that's no good

I wanted $\omega$ to integrate to $2\pi$ but go to 0 in the C^0 norm

I know. Maybe you can arrange it so that the other form has the same integral along the perturbed circle.

$N$ varies though

I wanted to take $N \to \infty$

I guess it's not entirely clear to me what form PV chose. I messaged him a while ago, let's see if he gets back

Is the actual proof of the approximation in EM too complicated to put into practice?

What is this btw?

Like what type of maths?

It's a corrugation trick @0celo7. I don't think it's too complicated but it seems it's only theoretically interesting

@Phase my water cooler is hot

This game is going to melt my PC

stop playing it then ez

They interpolate between 2-plane distribution by a zig zag ladder trick

But no, realistically if you stop overclocking and just eat the shit it'll probably not melt your PC

It’s pretty damn good though

balarka pls wat is it I wanna google

H principle

Oh ive watched this before

@BalarkaSen yeah I saw the neat pictures

ages ago

About like, the sum of certain properties being constant right?

Like the amount of 'going to the right' and 'going to the left'. It's been ages so that's probably far from what it's actually about.

Is this topology?

well its about turning the spheres inside out

yes it is

@0celo7 it gives things like existence of contact structures on open manifolds

sp00ks

My knowledge of topology goes as far as that really primative formula thing

V - E + F

Euler characteristic is very good

Maths professor made a mini-puzzle thing once about the Topology of a level on Sonic

the one where you get all the big rings

and you go to the space where all the coloured balls are