There are two equivalent ways. One is to say that the union of the sets in the cover only have to contain $[0,1]$ rather than equal it

Every C^k manifold admits a C^infty atlas if k >= 1 in fact

didn't you only assume that the function was continuous on $[0,1]$?

so you're allowed to go outside of $[0,1]$.

@BalarkaSen ok that's very weird and far from my cup of tea so I won't delve deeper right now

The other way is to define: A set $S$ is "open on $[0,1]$" if there exists an open set $A$ such that $S=A\cap[0,1]$.

anyways the reason i asked about manifolds was because something kind of brought it up on that test question that got marked wrong @Balarka

This way, $[0,1/2)$, for example, is open (on $[0,1]$). This is called the subspace topology.

it said that the law of reflection only works on smooth surfaces

but i argued that all surfaces are "smooth" if you "zoom in" enough

@Alessandro I've never actually read the proof. It's some kind of smoothing argument.

You should read the proof at some point and tell it to me

@AkivaWeinberger Yeah i've heard of that

The two ideas are equivalent.

it's kind of like just considering the section of the standard topology intersecting $[0,1]$

@ZachHauk What was that?

@BalarkaSen that depends on how much patience you have :P

what, the law of reflection?

scienceworld.wolfram.com/physics/LawofReflection.html

@ZachHauk There are nowhere differentiable examples though

@AlessandroCodenotti in the real world?

@Alessandro I can wait.

What has math to do with the real world?

@AlessandroCodenotti It's not super hard. It uses two things: 1) convolution with a bump function smooths a function, and you can convolve with an approximation to the identity, so make something smooth without changing it too much; 2) diffeomorphisms are open in the set of C^k maps with the C^k topology when k>0.

the law of reflection!!!

it reflects over the normal of the surface

which i argue is the surface itself

@Zach One thing I always wondered was what happens if you try to reflect off a light beam on a perfect kink, like the graph of y = |x|

the light would go to either side of that probably :P

you would need to have an infinitesimally small light beam :P

Sure, technicalities aside, what'd be the result?

There's no tangent so tell bye-bye to law of reflection

do you know the answer?

Nah

The idea being you take charts, possibly with C^k transition functions, and convolve the transition functions.

break the universe?

idk

I'll read your message again after finding out what a convolution is, right now the first point doesn't mean much to me sadly

That'd be a fun result

i'm gonna take a nap

hey my math question has this E with a ^ on top of it at the end of a set of numbers in interval notation

nice talking to ya'll

1/8 , 1/16 , E.

anyone know what this i

s

@CausingUnderflowsEverywhere can you show us it?

@AlessandroCodenotti Fair enough.

did I say interval notation uh I think I said it wrong

I meant a sequence

my bad

post it

take a picture and post it

well clearly

there's a limit but what is E

i mean

you're on windows right?

and is the question in a book or online?

Ệ

sorry like I'd have to use my phone and import the image

it's on paper

oh

:/

tell you what

draw a picture on paint

then save it and post it on here

thanks

I'm going to have a nightmare tonight, with alkyl halides chasing me down a deserted lane

I can't even imagine what that'd be like but it's gonna happen

i'm going to have a nightmare tonight, with Ted chasing me down with quadratic forms while I procrastinate and play games

bizarre

games are fun and rewarding experiences

Hi @Astyx

Hi @Zach and everyone else

hey you didnt say hi to me

@CausingUnderflowsEverywhere draw your picture!

hi astyx

i want to finish helping you and take a nap! :P

take a nap

ill ask my prof

whatever it is, the limit is 0

anyways, i dont mind, just use paint

or im gonna be left hanging thinking what the hell that symbol was

:(

whatever

bye friend-os

Bye @Zach and everyone else

LOL

ILL BE BACK WITH A PICTURE @ZachHauk just hang tight, you'll get pinged and it'll end up in your notifications alright. it might take up to 2 business days

back beacuse

shithead brother is in room so i cant sleep

oh and hey @Akiva... don't you just like to slip in and out

Something like that

if we have 400 sqr feet area of a room and we know that its about 350 sqr a gallon, do we divide 350/400 or 400/350 to find how much gallon of paint is needed to paint the 400 sqr feet?

we have 400 sqr and 350 sqr / gallon

and we want to get gallons

So we would want to do 400 sqr / 350 (sqr / gallon)

because then the "Sqr"'s cancel out

and the gallon goes to the numerator

so we get it in gallons

so we do 400/350?

yep

is there like a logical explanation for this beside the math way

like why do we divide by 350?

i just said

I am havin a brain fart rn

we have 400 squares

and 350 squares per gallon

so with 1 gallon we can cover 350 squares, right?

yes

but we'll need a little bit more to cover that extra 50 squares

so it will be a bit more than 1 gallon, right?

yea

so, tell me

which one is larger than 1 gallon

400/350 or 350/400?

400/350

ohhhh

ok i get it now

Hi there.

Any idea of how to prove $ f(x) = 2x^2 - x\sin(x) - \cos^2(x)$ has exactly two roots?

@Topologicalife All roots are contained in $[-1,1]$, since it's bounded below by $2x^2-|x|-1$, which is positive for $|x|>1$.

Why all roots are contained in $[-1,1]$?

Ah, sorry.

Now, we just need to know that it's increasing for $x\in(0,1)$ (since it's even, so we can take advantage of the symmetry).

Its derivative is $f'(x)=4x-x\cos(x)-\sin(x)+\sin(2x)$, so we want to show that it's greater than zero in that range.

Note that the derivative is $0$ at zero. So we just need to show that the derivative is increasing, that is, that its derivative is positive in that range.

Yeah, but $f''$ is pretty ugly too.

$f''(x)=4+x\sin(x)-2\cos(x)+2\cos(2x)$

That's bounded below by $4+x\sin(x)-2-2=x\sin(x)$.

So all we need is that $x\sin(x)>0$ for $x\in(0,1)$, which is clearly true (both $x$ and $\sin(x)$ are positive in that range). So we're done.

Nice solution, thanks you.

Real ugly, though. Don't know if there's a better way.

so $|x| < x \sin{x} \quad \forall x\in \mathbb{R}$ ?

No, it's $>$. @Topologicalife

Uhm.

Because $|\pi/2|>0$, for example

@AkivaWeinberger they refuse to e-mail the people

How hard is it to email someone

Yeah, but if it is $>$, then why you can bound below with $|x|$

(let me think a little)

@Topologicalife I didn't. All I know is that, for $x\in(0,1)$, we have $x>0$ and $\sin(x)>0$.

The latter comes from the fact that $\sin$'s first positive root is $\pi/2>1$.

(Since $\pi>2$.)

My problem is the first statement: chat.stackexchange.com/transcript/message/35723403#35723403

that one

@Topologicalife Notice the minus signs

Ah, right :)

$|x|>x\sin(x)$, so $-|x|<-x\sin(x)$.

Okay, thank you, nice

@Topologicalife To show that that's positive for $|x|>1$, notice that we can assume $x$ is positive (by symmetry)

and that $2x^2-x-1=(2x+1)(x-1)$.

@AkivaWeinberger It shouldn't be $2x^2 - |x| + 1$?

$ \cos^2{x} \leq 1$ so $-\cos^2{x} \geq -1$

err :)

@Topologicalife You got it?

Yeah, I'm writting the solution so I can understand it better.

Reading it only I don't get all the details.

@Topologicalife I was just playing around with desmos, mostly

desmos.com

hi chat