hey I was wondering if anyone could help me, ive been confused why would the answer not be $1/5sin^5(2x)+c$ and instead be $1/10sin^5(2x)+c$ I mean I get that the the derivative of sin was taken so that 1/2 would come out side but if that was the case why wasn't it $ 1/10cos^5(2x) +c$

Nvm I think I know what I did wrong

the differentiation has a mistake

its 2cos2x which means u add 1/2 to the integral thats how the 1/10 comes about

I have the following problem in an AP textbook. I was wondering about the second step where 2pi becomes pi and the boundaries become 2 and 0 from 4 and 0? Why does this happen?

Is it because xs are substituted for ys?

@Ted @Eric This can be done by the jiggling trick in holonomic approximation theorem.

Take any path from $(0, 0, 0)$ to $(x, y, z)$. You can wiggle it enough to be tangent to the given distribution.

In fact by a $C^0$-small isotopy.

Wow @Balarka

I like my approach just integrating better

Maybe it's a good exercise for me to write down an answer using that

I'll try it sometime today

Just woke up now

Im a simple non fancy man

I like non fancy answers

This reminds me when were proving in our algebra pset that there aren't any proper 2-sided ideals in $M_2(\mathbb{R})$ and were going back and forth on whether doing Jordan form is overkill

@Li357 That is correct

lol @Dami

proper 2-sided ideals of $M_n(R)$ are in 1-1 correspondence with proper 2-sided ideals of $R$ me thinks

In this case, that's a field, so balloon noises

iseewhatyoumean.jpg

@Eric Congratulates on the 3K

I cant spell

@Balarka so it seems

Come on dudes they are closing the question

I wanted to answer it

@Lozansky Okay but why does 2pi become pi?

I'll pull some strings later to get it opened but maybe I'll just start writing the answer

@Li357 They made some mistakes. The limits should be $0 \to 4$ not $0\to2$ and the constant should be $2\pi$ not $\pi$

@Lozansky That's what I thought and worked it out but the answer is different from the answer I got with the washer method (same problem)

Nevermind, I made a mistake in my work

I never realized until my topology class that breaking chalk is really easy to do, our class is taught so individuals go up to the board and write proofs and virtually everyone breaks the chalk in half instantly when they go up there.

I haven't seen a chalkboard/blackboard in some time. They replaced them all with whiteboards so that instead of breathing in chalk dust you breathe in marker fumes.

can any one tell the direction of gradient at any point of $z=x^2+y^2$

?

$(2x,2y)$?

@BalarkaSen thanks

Can gradient perpendicular to tangent plane?

The reason I shared the story is that I was so surprised and also amused because the way he propositioned me almost sounded like he was trying to get me on a get rich quick scheme

@Lozansky

The gradient is normal vector perpendicular to the tangent plane right? then, why do we plot the gradient field in 2-D space? for example $f(x,y)=x^2+y^2$ is paraboloid. the gradient at the certain point must be normal to the tangent plane containing the point $p$. Why do we plot the gradient field of the paraboloid in the plane? Can't we consider the paraboloid as $z-x^2-y^2$=0?

sorry!!! this is not exact field. we get similar figure. Right?

Those are level sets?

@EricSilva Funny how such things happen, right?

It turned out to be a good day for you

@ManeeshNarayanan in order to plot this on an actualy graph we would need to plot $R^2 $ on one axis ( the domain) and $R^2$ on the range axis, this would need a 4-dimentional piece of paper to do

Yeah I'm recovering from a cold tho so that sucks

Also I am no longer sure my holonomic approximation fuckery works. I think it only gives me a curve which is $\varepsilon$-close to being tangent to the distribution

Lol

For arbitrarily small choice of $\varepsilon$

we dont have a 4-d piece of paper so reduce the cases using level sects and vector field so we can actually graph the function for some kind of intrepretation

The gradient vector is orthogonal to the level curves if that is what you are asking

@EricSilva Ugh

its not a picture of the object in 4-d but its as close as our 3-d brains can manage

@Faust ok

@Lozansky yes. I could figure out the proof by using the chain rule.

@Balarka I guess if you tried to take a limit of a bunch of integrable curves that are almost tangent you might end up with fuckery in convergence

So maybe unsalvagble

I was confused by seeing this 2-D graph

Yeah I think this distribution is special

Yeah this property is weird

I mean obviously it'd fuck me up if the distribution was horizontal and $\gamma_\epsilon$ curves going from origin to $(1, 1, 1)$ staying $\epsilon$-flat

@ManeeshNarayanan they are rather confusing when they are first introduced especially if no one tells you why your doing it.

Somehow the limit should exist in this distribution.... maybe....

@EricSilva Strange, because this distribution is everywhere nonintegrable looool

it shouldn't have this property

integrability in codimension > 1 is weird

@Faust Gradient is normal to the surface, right?How can I interpret this?

thats the reason Gromov's thing works, eg. he broke integrability being a closed condition by wiggling in higher codimension

Yeah this guy is like maximally weird

I could figure out gradients are perpendicular to the level curves using chain rule.

I guess complete nonintegrability is exactly what I mean by weird

@Faust

@Eric The solution you guys came up with is just finding an area problem, yeah?

find a $\gamma$ such that $\gamma \cup \Gamma$ encloses a certain area?

How strange!

Yeah

@ManeeshNarayanan Gradient is a different idea that uses the same picture, its representation of the oritation of the surface...

That's a completely shit solution because it provides no visual information... damn you analysts

Very strange

(I mean shit very nonliterally, I really like this solution)

someone else would be more qualified to explain the diffrence ( i understand they are simliar pictures but they are representing completly diffrent things and the diffrence is kind of subtle)

I want a good pictorial reason for that solution tho

I just fucked with integrals

The property is pictorially obvious at least

Cause the distro is a very twisty boi

yea

But I mean like it's an interesting question when this property holds for a distribution

Does the dist automatically need to be nonintegrable if this holds for all paths $\gamma$?

stuff like that

Does this even hold for all everywhere nonintegrable dists?

I feel like yes for everywhere nonintegrable sides but I don't have a good reason

@Faust we use gradient for finding the normal vector, right?

I just feel like you gotta be real twisty to end up everywhere nonintegrable

thats the picture yup

@ManeeshNarayanan yes

How that coming? I don't understand

If there are 3 independent variable I can see

Twisty, huh

as the consequence of level curves

@Faust

@BalarkaSen i cant think of one that doesnt look like this twisty guy

@ManeeshNarayanan It's easy to prove the gradient of $\phi$ in the point $P$ is orthogonal to the level set $\phi = c$ through the point $P$. Just note that a small displacement $d \textbf{r}$ along the level set won't change the value of the scalar field $\phi$ (since the function is constant) and so $d\phi = \nabla\phi \cdot d\textbf{r} = 0$ which proves $\nabla\phi$ is orthogonal to each $d\textbf{r}$ in the level set which implies $\nabla\phi$ is orthogonal to the level set.

Where $\nabla$ is the gradient operator

@EricSilva $dz = xdy$ is the only standard one I know.... the "standard contact form"

right

hmm

well the gradient vector is orthogonal to the level curves yes.. but you calculate it in a diffrent way, the level curvers are a picture but you need not draw the picture to calculate the normal vector

is it a situation like in symplectic stuff where there is no real local theory

@ManeeshNarayanan

Also note that you do not have to use the gradient to find the normal vector

cause i cant picture a guy that would end up nonintegrable unless it just ended up looking twisty

(definatly not most of the time one can just read off the normal vector from the function or simply guess it with alot less effort)

I can't translate between the twist and the Frobenius condition

Hm

The latter says the space of sections of that distribution is bracket-closed

Maybe bracket-closed translates to something that is non-twisty?

How?

If you can parametrize say a surface by $\textbf{r} = \textbf{r}(u,v)$ then $\textbf{n} = \dfrac{\partial \textbf{r}}{\partial u} \times\dfrac{\partial \textbf{r}}{\partial v}$

I have very little pictorial intuition for the Lie bracket

@ManeeshNarayanan you should try your prof's office hours its a rather delicate topic and he would be more qualified to know what you would/ wouldnt understand explanation wise i am rather unqualfied to explain it

@Secret what exactly is a closed set of randomness? is it bounded?

@BalarkaSen bracket closed means that if you try to twist around in an infinitesimal parallelogram spanned by two vector fields it's twisting in the distro

cause the lie bracket is like the accelaration of the curve you get from flowing around a parallelogram

That's true

I kinda see what you mean

so it's not changing direction out of the distro too fast i think

literally "not too twisty to break free"

hmmmm!

@Faust I have completed my masters, Now I am revising the entire topic that was in the course.

maybe bullshit intuition tho? idk i think the picture makes a little sense

if you have a masters in any branch of mathematics your much more qualified then i am >.>

@Lozansky The figure must be level surface right?

How do you construct an ON-basis for $\mathbb{P}_2$ in $L^2$?

that is why it is normal. right?

@Balarka so like if you have a nonintegrable distro of codim 1 that means that if you take the one form that defines it and exteriorily differentiate you get a 2 form that is nondegenerate on the distro

this sounds symplectic-y

Because $d\omega(X, Y) = X\omega(Y) - Y \omega(X) - \omega([X, Y])$, yeah

@ManeeshNarayanan Yeah that's a level surface

$[X, Y]$ is not on the distro, so you'd get $d\omega \neq 0$

ok. Thank you very much.

mhm

@Lozansky

so you have a rank dim of your guy - 1 symplectic vector bundle

@ManeeshNarayanan There is a lot more to said about the gradient though :P

@Eric weeeird

I was too confused. Now It is more clearer.

@Lozansky

this makes me think that all these completely nonintegrable guys must be the same locally

because you can make a symplectic guy out of any of them and those are all the same locally

Darboux theorem?

yeah right?

i only vaguely know about this

me too

@ManeeshNarayanan Glad to hear :)

this is a fun idea

p interesting stuff

maybe its time for us to learn symplectic and contact geometry lol

lol

Eliashberg Mishachev chapter 9 is an intro

oh shit

trying to lure Eric to the dark side

the unspellable names

the motivation is piling on

@Lozansky small doubt is here. How to find the normal vector of paraboloid using gradiend? $z=x^2+y^2$. By considering the level surface of $f(x,y,z)=x^2+y^2-z$. right?

found a pdf

lol im supposed to be reading sociology rn

@Eric "the playful twisting of the distribution given by the contact form fills you with determination"

-Undertale

level surface corresponding to $f(x,y,z)=0$ right?

@ManeeshNarayanan Yeah that is correct

oh my god

goooooooold

Thank you. @Lozansky. Now it is very clear. Thank you for your help.

thanks@Faust

see u later.

peace

@ManeeshNarayanan Another way is parametrization. Let $x=u, y=v, z=u^2+v^2$. Then the origin vector or whatever it's called would be $\textbf{r} = (u,v,u^2+v^2)$

So the partial derivatives become $\partial \textbf{r}/ \partial u = (1,0,2u)$ and $\partial \textbf{r}/ \partial v = (0,1,2v)$

Thus the normal (i.e. gradient) is the cross product between these, and you get the same result

@Balarka ill check out chapter 9 if im free this weekend

neat! hit me up if you do

mmk

seems like an interesting tale

shit i guess i should read colding minicozzi thoroughly this quarter since im getting a job to learn minimal surfaces next quarter

lol

there's too much good geometry out there

2much2read

sniped

the pictures in chapter 1 of colding minicozzi give me life

I should learn Willmore theory

yeah

learn and read bryant with me

I think I'll spend the day looking at doCarmo today

need to revisit geometry

that reminds me @TedShifrin needs to learn me a thing about the Bryant proof

the basic boiz

back to basics 2: do Carmo boogaloo

I want to write a book and then title its sequel "electric boogaloo"

please dont

Or maybe write a paper

it's so weird that that's a meme but ive never met anyone else who's seen breakin

that meme is dead

Memes are only dead if you believe they're dead!

no

stop

dead meme is dead

you can't rickroll in 2018

Pshhh

rick rolling just like

isnt funny

it's just a chore for both sides

@EricSilva as the arbiter of humor I tentatively disagree so that I can hold it as an open option

if you were the arbiter of humor everyone would literally already be dead

+1

"Omae wa you shindeiru everyone"

no

otoh some 2018 memes are cancerous

so i prefer rickroll

did anyone else notice in there 3rd year math just got alot easier for some reason?

i dont understand the tide pods meme

yes, that is one

@Faust 3rd year means different thigns to different people in different places

so idk

people started joking about how it looks like a candy

and people posted videos of eating and vomiting it out

thus started the tide pod meme

i dunno i mean the math im doing now is harder than before but somehow it feels easier

why are there so many memes about eating something that is difficult and gross to eat

maybe im just understnading what the math is saying easier

theres also the knuckles meme

from the VR chat cancer

idek that one

better that way

i dont really follow new memes

or old memes

i just have hobbies instead

@EricSilva it all started with the infamous Filthy Frank TV video

about the hair cake

and it's successor the vomit cake

ive never watched a filthy frank thing

and it's successor the human cake

the famous trilogy

those sound like things i wouldnt watch

you know the SD cards for the switch have some kind of chemical coating on them so they taste really. so ofc the millennial generation has to try these gross tasting cartridges long story short a buncha of people started getting sick (ofc only adults) from ingesting too much of this chemical.

what does that really say about the world we live in?

@Faust that's just cause we're told not to do it

repositories.lib.utexas.edu/bitstream/handle/2152/35301/…

Good read

i mean i think it's dumb but i also respond to authority by rebellion

but it literally tastes disgusting... why would you suck on it for the hours necessary needed to ingest enough of the chemical to get sick. i can completely understand the curiosity to try it once...

no yeah i agree i am not going to like, put something disgusting in my mouth as a rule personally

unless there's a payoff

@Eric i don't think you would like filthy frank. his comedy is the most extreme form of racist, edgelordish, offensive kind. take south park and multiple it by 1 million

i would definitely not like it then

edgelordish

that sounds made up

or like some kind of wierd crustacean people thought was a fish

as a rule i think edgy comedy is self serving garbage that tries to make up for not being good by pushing you to the extreme for some cheap gag

I'm more open minded about them but FF is unimaginably bad. I think the badness is why it's funny

there are some B-movies which are so bad they gain a cult status

kinda like that

that stupid one with the sharks

The Room :p

i cant enjoy bad movies

they make me angry

i think there's also a difference between edgy comedy and comedy where the joke is layered in what's bad about the edginess

it's always sunny in philadelphia straddles the line there

The Room is brilliant

@Eric right that's true

the room is genuinely good and genuinely bad at the same time

Vox made a good video about Filthy Frank, I think it's worth watching

it contemplates the content of that channel from all aspects

eh i could also like

use my time to read eliahsfdkjhslkabdvfbsdferg mishvnsdfkljnvsbdflkjbchev

or C-M

much better

also accurate spelling there

yeah i tried my best

Marco

today is my birthday. i hope i won't have to deal with familial farce too much

Congrats your a year older lets celebrate that your an entire year closer to death.

happy birthday

@Faust that's right

@BalarkaSen maybe itll be tragedy instead of farce

never understood birthdays lol

maybe i'll die

tRaGeDy

thatd be fun

on the bright side your matter will be re-assimilated into to some new and unusual things

eventually you'll get swallowed up by the sun and the bugger off to anthor solar system

maybe meet some new life

interesting prospect

Your birthdays are different from mine

oh yeah whatcha do on your birthday?

or maybe youll meet an anti-balarka and annhilate

cry mostly

lmao

right into the cake

? why

stick it to the man

cuz it makes it taste better

hmm never would of thought to try that

maybe ill prepare a hair cake for myself

alright im gonna leave now

Cya!

yeah its 4am here

i have to read about religon

yikes

bye everyone

bye

aight me too see ya nerds

It is one thing i find fascinating where in the hell did all the antimatter go?

bubye

@Faust over there

classic gag

anyone know of an intresting paper on a topic related to number theory?

i need to read a research paper and then make a write up of what it says

but ill fall asleep if its boring

analitical number theory is fine too

Should probably summon Mathein, he might know

how do i fix my broken proof

its breaking my heart too

what broken proof?

@MatheinBoulomenos Summon spell

i had an idea of proof for something and it didn't work

ok...

@Balarka my heart got broken too since I think there's a geometric argument for finding the units of $\mathbb{Z}[\omega]$ where $\omega$ is a cube root of unity

geometry is the best

The vague idea is that you want to show that no elements of $\mathbb{Z}[\omega]$ have norm less than 1, so no units have norm greater than 1

And then you just narrow it down to those elements in $\mathbb{Z}[\omega]$ on the unit circle, which I think should just be the 6th roots of unity. So that just does it

Aha

My original argument for it was quite different but I found it rather fun

So, $(a+b\omega)(c+d\omega) = (ac - bd) + (ad + bc - bd)\omega$

Guess Mathein is currently our of the service area

Meaning you're trying to solve $ac - bd = 1$ and $ad + bc - bd = 0$

First off, you note in the back of your mind that this means $a$ and $b$ are coprime

Then you write it as a matrix equation

$\begin{pmatrix} 1 \\ 0\end{pmatrix} = \begin{pmatrix} a & -b \\ b & a-b \end{pmatrix}\begin{pmatrix} c \\ d\end{pmatrix}$

If the matrix isn't invertible, this means $a^2 - ab + b^2 = 0$

But then $x^2 - xb + b^2 = 0$ doesn't have real solutions

So you know the matrix is invertible, and you have a unique solution: $A^{-1}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}\frac{a-b}{a^2 - ab + b^2}\\ \frac{-b}{a^2 - ab + b^2}\end{pmatrix}$

But then we want these to be integer solutions, so $a^2 - ab + b^2 = det(A)$ divides $a-b$ and $-b$, so it must divide $a$ and $b$, which are coprime, so it has to be $\pm 1$

I like this solution

Right?

Then you just need to check that $x^2 - xb + b^2 + 1$ doesn't have integer solutions because discriminant, and that $x^2 - xb + b^2 - 1$ has discriminant $4 - 3b^2$

So that requires $b = -1, 0, 1$

Symmetrically, so must $a$

So now you just need to find what works

Right

> A mathematical theory is not to be considered complete until you have made it so clear that you can explain it to the first man whom you meet on the street

Quote by David Hilbert

But...

Does there exists maths so complicated that it simply cannot be expressed in simpler terms?

I wonder if such proposition is provable in first order logic...

I was suspecting that, but I am not sure whether that can be proved. My current thoughts is that if some maths has too much nesting in their definition, then it will be basically imcompressible to more daily life life language

but I am not sure if that is a good assumption to interpret precisely the phrase "made it so clear that you can explain it"

IUT requires a background in algebraic geometry, and that in turn requires a background in abstract algebra and geometry and a bit of analysis. IUT also requires category theory, so I will expect every definition will include terms that basically when unravelled completely, will be a highly nested sentence

I am not sure how many layers of unravelling is considered too nested for the public though...

There's also an extra complication of concrete vs abstract concepts. Soem concrete concepts have physical form that allows people to understood it easily by showing them how they behave, hence providing some kind of shortcut that bypass many layers of nestings, but otherwise when expressed purely in words, it will be almost as long as a chapter