Mmm how is x^2+1 irreducible over Q adjoined root 2?

@Corellian perhaps the better question is how would it be possibly reducible?

@anakhronizein what is $\beta$

@0celo7 $\beta$ is induced by $\xi$. Since $\xi$ is locally defined as the kernel of a 1-form, $\alpha$.

I don't know what "induced" means in this context

I assumed it meant the 1-form arising from $\alpha$ defined only on the surface.

Let me double check Geiges.

It's the restriction of $\alpha$ to $S$.

ok

Given that I don't know what a characteristic foliation is, I doubt I can answer

I describe it at the start of the question.

What is given there is effectively the definition and everything in between from Geiges/Giroux.

So assuming they themselves are not skimping on details, the question should be more or less complete.

I would give that a 3/10 on an exam if I asked you for a definition

Okay, thanks for your help.

¯_(ツ)_/¯

I don't know what "via looking at" is supposed to tell me, in precise terms

It's fine.

Thanks.

@anakhronizein Balarka would know

I think Balarka maybe has attempted to help before.

Or that was with another thing about the characteristic foliation.

@AkivaWeinberger, please look at this:

I will ask him again anyways. That would have been before I made the question. Been struggling with understanding it for a year now.

Maybe I should email Geiges?? Or maybe I should get better at math.

One or the other.

What's the title of the book

An introduction to contact topology.

pages 77-79 is the definition, statement and proof.

Ok, I'll read that part

What is the deep result of this book

Why read it?

Well it's got a few cool results.

@Silent If $f$ were positive everywhere, we would have $0\le\int_0^1xf(x)dx\le\int_0^1f(x)dx$

One is convex surfaces and overtwisted contact structures.

and thus $\int_1^1f(x)dx\le\int_0^1xf(x)dx\le\int_0^1f(x)dx$

Question: Is $A = [A \cap (B-C)] \cup [A \cap (C-B)]$?

and then we could use the intermediate value theorem

Then there is stuff about symplectic fillings I have yet to look at.

Clearly $A$ is contained in the one set, but the other inclusion appears to be false.

Also the legendrian knot stuff is really neat.

@user193319 What if $A=B=C$

@Corellian a degree 2 polynomial is irreducible iff it doesn't have roots in the field

@AlessandroCodenotti same goes to degree 3

Too many people with A in the chat right now

Please leave

@anakhronizein What does $\perp$ mean in this context?

In his definition of the foliation?

perpendicular in which sense

@LeakyNun I know, but he has a degree 2 polynomial

In his definition, or where are you reading this?

So iow, $(a+b\sqrt{2})^2+1$ is never zero for rational $a,b$?

the correct answer is this:

yes, in the defintion of the char foliation

"because the centre of mass exists and is inside the solid"

@AkivaWeinberger Hmm...How could $\int_A |1_B - 1_C| = m(B-C) + m(C-B)$ if that set equality doesn't hold!? Perhaps this is an error in my book.'

It's with respect to the symplectic form $d\alpha\vert_\xi$.

@anakhronizein Ah, ok.

If you want I can provide a definition with respect to vector fields.

Actually no, I can't.

It would boil down to the exact same thing.

@user193319 $A=[A\cap(B-C)]\cup[A\cap(C-B)]\cup[A\cap(B\cap C)]$

Note that $|1_B-1_C|$ is zero on $A\cap(B\cap C)$

(The parentheses there^ are unnecessary to be honest)

@AkivaWeinberger Ah! Thank you!

Hi. Let $A$ be a square, symmetric, positive definite real matrix, and $B$ be one of its square roots. If, in addition to symmetry, some elements of $A$ coincide (e.g., $A_{1,2}=A_{1,3}$), then what about the corresponding elements of $B$? I believe that in general $B_{1,2}\neq B_{1,3}$, but is there anything else I can say for why that's the case?

@Julius "because."?

@LeakyNun Whoa!

@Corellian it's even not zero for real a,b :P

@LeakyNun @Silent But that still only works for $f$ positive

but with that physical intuition, I think we can find situations where it's false for $f$ not everywhere positive

@anakhronizein Isn't he showing that $\ker X=\ker\beta$?

Interpreted appropriately since these are different objects

Hello everyone. I need help in probability.

Numerical experimentation shows that it is false for $f(x)=\sin(5x)$ @LeakyNun @Silent

(which is not positive everywhere)

nice

where is its centre of mass?

@RahulJain just ask; don't ask to ask

@AkivaWeinberger I am so thankful

@0celo7 could be.

@AkivaWeinberger Did you use some software to find this? or just your brain?

Desmos

I tried $\sin(ax)$ and let $a$ vary

ok

the frequency with which this airplane clip comes to mind is a sign of how bad my sleep schedule is:

@LeakyNun @Silent WAIT! I messed up

lol

I was comparing $\int_0^1xf(x)dx$ to $\int_0^cf(x)dx$, not $\int_c^1f(x)dx$

I probably need to listen to some good wake-the-f***-up music

nah, just stay outa here for awhile :P

lol

if i'm not here, then I'll be tempted to go take a nap

which i'm not sure is an improvement right now

power naps work, sometimes

yeah, but i need to actually get stuff done atm

then your YouTube clip is the only choice, pal

@AkivaWeinberger surely you can just invert it?

I tried. The problem is the $x$ doesn't invert with it

$\int_0^c$ measures stuff on the more squished side. $\int_c^1$ measures stuff on the less squished side.

Never mind; I plugged it in wrong

@Semiclassical after you get some work done; you can go downstairs and flip those tables over :P

lol

So I'm beginning to seriously conjecture that it's true for all $f:[0,1]\to\Bbb R$

no, if i go down there i really will tamper with the signs

i think $|\int_0^1 xf | \le |\int_0^1 f|$

@LeakyNun That's false. Take nearly any $f$ with $\int_0^1fdx=0$.

ok

I think Cauchy–Schwartz bounds it by a constant multiple of $\sqrt{\int_0^1f^2dx}$, don't know if that's relevant

hmm. it's equivalent to $\int_0^1 (1-x)f(x)\,dx=\int_0^c f(x)\,dx$ for some $c$

I guess one thing would be to find $c_1,c_2$ such that $\int_{c_1}^1 f(x)\,dx<\int_0^1 x f(x)\,dx<\int_{c_2}^1 f(x)\,dx$

and then argue by continuity

Maybe write $\int_0^cf(x)dx$ as $\int_0^11_{[0,c]}f(x)dx$

or $\int_c^1f(x)dx$ as $\int_0^11_{[c,1]}f(x)dx$

the most obvious thing in that case would be the fact that, when $c_1=1$, the first one is just $0$

and $\int_0^1 xf(x)\,dx\geq 0$ since $xf(x)\geq 0$

is it true that $\int_0^1 x f(x)\,dx <\int_0^1 f(x)\,dx$? That'd finish it

@Semiclassical We don't know $xf(x)\ge0$

bah, you're right

If $f$ is positive everywhere, it's easy

@Semi! can I quickly ask you something?

@AkivaWeinberger hmph

sure

how should I parametrize $x^4+y^4\leq 1$, $z=0$? Should I send $(\theta,r)$ to $(\sqrt{r\cos\theta},\sqrt{r\sin\theta},0)$?

@ShaVuklia why not

as in, $x=\sqrt{r\cos\theta}$ etc?

yes

the problem is

I need to calculate the pull back

$c^*\omega$

that sounds reasonable, but sorta painful

so the derivatives get annoying

maybe it's easier to think of it as a composition of two mappings: $x=\sqrt{u},y=\sqrt{v}$ and $u=r\cos\theta,v=r\sin\theta$

right, so first we parametrize it with the roots, and then we do the usual parametrisation with the circle

right

so that will be 2 pull backs

i'm not sure that's a silver bullet but it lets you separate the difficulty

yea it's reasonable, like when you have to parametrize an ellipsoid

first turn it into a circle, then usual parametrisation

@AkivaWeinberger Random: It's a pain in the butt to prove that symmetric difference is associative

Maybe you know a slick way to do it

is it possible to prove that sine is cts at 0 using elementary methods

Symmetric difference being $[A,B]=AB-BA$?

what do you mean by "elementary", no diffs?

@Akiva No no the set thing

Oh, wait, you mean with sets?

Ya

(sniped)

@BalarkaSen Yeah I know a slick way

@Secret elementary meaning it doesn't go like "it's a power series so it's cts"

@AkivaWeinberger tech me senpai

but still using epsilon-delta

@Semi but will that thing be integrable? I will manage to do the separate pull backs, but the integrand get's horrible. Though we do get an ordinary integral, so maybe applying change-of-variable again back in the opposite direction could do the trick

@LeakyNun Little little arc goes sloop sloop to height

Remember the Iverson bracket? $[P]=1$ if $P$ is true and $[P]=0$ otherwise

Mhm

What form does the square root of ax^2 + bx + c take, where a, b, and c are constants? My initial thought was Ax + B, where A and B are constants, which would mean (Ax + B)^2 = ax^2 + bx + c, but this offers one less degrees of freedom than is necessary. When the left is expanded to (A^2)(x^2) + 2ABx + B^2, and therefore A^2 = a and B^2 = c, 2AB is determined, and very unlikely to equal b. Would Ax + B + Cx^-1 be a better starting point, or some other x term?

For example, $[3\le5]=1$ and $[\sqrt2\in\Bbb Q]=0$

@ShaVuklia not sure. but it can't be worse than doing it all at once by hand

So, basically, the defining thing of $A\triangle B$ is,

true

$[x\in A\triangle B]\equiv[x\in A]+[x\in B]\pmod2$

... Huh

Oh I see, yes

@user10478 the square root of ax^2+bx+c is sqrt(ax^2+bx+c). in general, you can't do simpler than that.

Wow that's slick.

Wowowowow

@BalarkaSen pass it to the boolean algebra

So $[x\in(A\triangle B)\triangle C]$ is $([x\in A]+[x\in B])+[x\in C]$ mod $2$

there are some cases when you can write ax^2+bx+c as the square of a linear polynomial, but that's not generic

(it's $0$ if that's even and $1$ if that's odd)

and that's clearly associative

Dam son

Mind is blown

(unless you're working in some abstract algebra setting. things are hinkier there)

@Semiclassical Shouldn't the arc length between two points on a parabola, for example, which involves sqrt'ing a quadratic polynomial, always have a real answer?

@user10478 sure. that answer is sqrt(ax^2+bx+c).

The Iverson bracket is just a notation trick. The point is that $\triangle$ on $\mathcal P(X)$ is isomorphic to ${\Bbb Z_2}^X$ @BalarkaSen

(if you want to take the square root of that)

I mean, it's like asking what the square root of 5 is

Since the latter is a group, so is the former

it's sqrt(5).

@AkivaWeinberger Right, I see.

there's not going to be a nicer representation of that in terms of rationals

Okay, so you just evaluate the integral without evaluating the sqrt?

pretty much, yes. (you may be able to simplify the integrand in that specific situation, I don't know off the top of my head)

in the case of points on an ellipse, for instance, the arc length formula leads to the definition of the elliptic integrals

(well, at least the elliptic integral of the second kind)

For the sake of curiosity, if one were to consider complex answers, wouldn't the form be Ax + B + iCx + iD, which would be one TOO many degrees of freedom?

still won't be enough

you've really got ux+v with complex u,v

Ahh, right

Okay, thanks for the help

np

Does any have any idea about this or can you help with it? math.stackexchange.com/q/2762937/542737

How did I get in here? I'm in a meeting now, so busy.

The answer: Chat is a black hole of unproductivity

@Semiclassical kinkier :P

@Secret Yes, it ruins you mathematically speaking (with some little exceptions).

@IceInkberry 1 lol

uh..., this function can never have a denominator of zero, thus top and bottom cancels out

You should be able to + e^x - e^x the numerator to write it in 1 + some fraction form, which will make the analysis easier

After that my advice is to check all stationary points and also behavior near -+infinity

one way to rewrite that is (e^x-1+e^-x)/(e^x+1+e^-x)

the main nice thing about that form is that it's obvious f(-x)=f(x)

so you only have to worry about x>=0

you've also go f(0) = 1 and f(infty) =1, so as secret suggests it's a good idea to consider f(x)-1

@IceInkberry yep. which you can in turn write as $1-(1+e^{-x}+e^{-2x})^{-1}$

at which point the question goes from finding the range of f(x) to finding the range of $1+e^{-x}+e^{-2x}$ which is much easier to do by hand

derp, typo

should be $1-(e^x+1+e^{-x})^{-1}$

so one wants the range of $1+e^x+e^{-x}$

no

e^(-x)->0 as x->+infty

while e^x->0 as x->-infty

I have posted a question, but I think in this form, it doesn't sound interesting. Any ideas how to improve it?

so exactly one of those terms will blow up at infinty

@IceInkberry not sure where you’re getting that from

Mmkay

The upper bound is right

Woops

Yeah, you’re right

Got myself thinking of it as 1+cosh(x) not 1+2cosh(x)

Yeah

1-2/3 = ?

And there you go

B/c it’s 1-1/(stuff)

Where stuff ranges from 3 to infinity

what's this notation? context is log-derivatives

$\sqrt[\mathrm dz]{\mathrm dP}$

whatever it is, I don't like it

Yes, ie 1-1/(stuff)

I don’t see where you’re getting tripped up

Yes. That’s the range of stuff

And then f(x) = 1 - (stuff)^{-1}$

Oh.

Hmm

Oh, typo earlier

Should be 1-2/stuff

[Random] when things just don't work:

where do you see a cat

eh, i see it

blue eyes, black ears

and a blue tail?

it's a rather misshapen cat, but I see it

that's a molecule electron density map thing, but it should not give those weird black regions. Have been spending 3 hours trying to figure out what's wrong so I can prepare for my meeting on fri

ugh, at least they give the answer to you: use the Bromwich integral to solve $\mathcal L^{-1} \{ (s \cosh )s^{1/2})^{-1} \}$

the answer is an ugly power series

oof

I think you've got a paren out of place

$(s \cosh)s$

$1 + 4/\pi \displaystyle \sum_{n=1)^\infty (-1)^n/(2n-1) \exp(-1/4(2n-1)^2 \pi^2 t)$

$$1+\frac{4}{\pi}\sum_{n=1}^\infty \frac{(-1)^n}{2n-1}\exp\left(-\frac{1}{4}(2n-1)^2 \pi^2 t\right)$$

anyone see the problem in the latex

lol

oh, you had {n=1) not {n=1}

aah thankis

so nice of them to give the answer lol

lol

yeah a parens is out of place, $\dfrac{1}{s \cosh(s^{1/2})}$

figured as much

well luckily the powers of $\cosh$ are all even

not too strange to be getting an ugly power series in that case

not fun, but not surprising

I guess I have to do the residue theorem on a finite halfdisk and then take the limit

ugh im skipping this and doing the rest of the assignment and coming back if i have time

I don't even know how to do that, use a residue theorem on a periodic function in that way

well, maybe at least write down the bromwich integral

okay, let's see

my guess is that you'll end up needing to do a $z=\sqrt{z}$ substitution

and then use periodicity

with the main weirdness being the integration contour

oh youre saying not to use the residue theorem?

no, you will use the residue theorem

hmm

i guess my point is that it may be better to do residues in terms of $z=\sqrt{s}$ (typo above) rather than $s$ itself

I'm getting $\dfrac {-1}{2z^3 \cosh z} \mathrm dz$

okay

where does cosh have zeros?

what's the formula, $\cosh z = \cos i z$?

right

alternatively, you want e^z+e^-z = 0 ->e^(2z)=-1

$i \pi/2 + \pi i n$

yeah, or if you start at n=1: $i(2n-1)\pi/2 $

so we already see that 2n-1 aspect showing up

the other nice thing is that $\cosh(z-2\pi i)=\cosh z$

so if you know the residue of $1/\cosh z$ at $z=i \pi$, you know the residue at any other such pole

Could you give me a hint how we could find the area of the shaded part?

one thing you should also note: does your integrand have a residue at z=0?

As a function of the distance of AB

figure out the area of the big circle, then work out the area of the small circles

It is the are of the black circle - areas of the two circle, right?

wait, back upo

yus

But how can we use the distance of AB ? @Semiclassical

well, I don't actually see A in your diagram

so...

It is on the vertical line that passes through B. AB is tangent to bith inner circles. @Semiclassical

regardless: if you know the radii of two of the three circles, you can infer the third radius as well

this is periodic in $2\pi i$ so it's periodic for $2 \pi i n$ but that's not exactly what we want because we want odd integers

ah

not even

that seems...unpleasant

you're getting odd integers

Let the radius be r, a, b. Then we have A - A1 - A2 = πr^2 - πa^2 - π*b^2 @Semiclassical

How can we use here the distance AB?

you've got poles at $i\pi/2$, $i3\pi/2$ etc

of the integrand right

you're going to need to use a better image, one which actually shows AB

@GFauxPas right

oh so we can just consider one inverval of length $2\ pi$ at a time you're saying

going in the imaginary directions

one pole at a time, sure

note that the zeros are actually spaced by $ i \pi$ not $2\pi i$

same as the zeros of cos being spaced by pi not 2pi

so we need to consider 2 poles in each interval?

or doesn't matter

I’d just go one pole at a time

You’ll probably also need to consider poles in the lower half plane

here is an other image: @Semiclassical

I’m not entirely sure what the contour looks like tbh

Do you have an idea? @Semiclassical

isn't it symmetric

wel let's see, let's put in $-iz$

wait no just $iz$

no not quite

According to archimedes there is a formula $A=\frac{\pi\cdot t^2}{8}$ where t is the distance AB. But could we get to that formula. Do you have an idea? @Semiclassical

ugh Im gonna come back to this later, if I have time. but thanks Semi

ok, so the lower bound of the colour range need to be set to 0 because the data only span from 0 to 0.01

Fixed for now

okay this looks more fun, Schwarz-Christoffell transformations

mapping the upper half plane onto a polygon

at least I can now RIP

Zzzzz

@MaryStar I'm dubious that you can find the area just by knowing the length AB. Seems like one should be able to draw more than one picture for a given AB

maybe if they told you that the big circle had radius 1?