(I'm referring to exercise 11)

Incidentally, exercise 10 is essentially the same thing as Gauss's lemma, and note how it requires a symmetric connection

("(Local) geodesic frame" means that you have vector fields $E_i$ in a neighborhood $U$ of $p$ such that they're orthonormal everywhere in $U$ and such that $\nabla_{E_i}{E_j}=0$.)

gah

what is this garbage

how do I get the parentheses to not be stupid

$\displaystyle\left(\sup_{M\setminus B_1}G\right)^{q-q_i}$

$\displaystyle(\sup_{M\setminus B_1}G)^{q-q_i}$

Another question: There is a cup product on cohomology with compact support for an oriented manifold $M$, right? I.e. a bilinear map $H_c^p(M) \times H_c^{q}(M) \to H_c^{p+q}(M)$.

I see the problem @0celo7

$\displaystyle\big(\sup_{M\setminus B_1}G\big)^{q-q_i}$

Hrm

The parentheses are gonna be symmetrical, so if they reach the subscript they're gonna also go really high

yeah

I think I can just put the power on $G$

$\displaystyle\left(\sup_{M\setminus B_1}G\right){}^{q-q_i}$

lol

Just brought down the exponent

this works just as well

$\displaystyle\left(\sup_{M\setminus B_1}G \vphantom{\begin{matrix}\\\\\\\\\\\\\\\\\\\\\end{matrix} }\right)^{q-q_i}$

I think I solved your problem, Ocelot

I would like to be paid for my services

ok

@AkivaWeinberger by definition of $d$

there you go

Thank

@AkivaWeinberger But more seriously, do you not know differential forms?

Not in a way that's connected to the Lie bracket

I am starring that insanely stretched out bracket, because any insanely streched out object is hilarious for hilarious reason that it is strectched out

prose-ception

hi chat

Given that that's a big sup, I can only conclude that the proper pronunciation of such is "wassuuup"

I think I've completed the main proof of Chapter 4. But now I have to make sure all the times I divide by zero are ok...

> no, you cannot divide by zero

:P

you can divide by zero on a set of measure zero

eh???????????? mindsplosion

Hmm... I guess I don't really understood measure zero sets enough

@Secret you only need things to be defined up to sets of measure zero most of the time

I know how measure zero sets don't contribute to integrals, but I never heard about them being involved in some notion of divide by zero. Does it mean we can write indeterminates in them and not affecting the result because it is measure zero?

Yep

For instance, $\int_{-1}^1 x^{-1/2}$ makes sense

rigorously, you mean the integral of the function that equals $x^{-1/2}$ for $x\ne 0$, and is simply undefined at 0

@AkivaWeinberger Look up the Cartan's magic formula

it makes sense in the principal value sense, no?

@Semiclassical no, that function is really just integrable

so that means when we compute that, we are actually computing $\lim_{a \to 0}\int_{-1}^{a} x^{-1/2} + \int_{a}^{1} x^{-1/2}$ and $\{0\}$ is the measure zero set thus it does not contribue?

ah, you're right.

the antiderivative is $\sqrt x$, so.

@Secret no

you're really just defining $0^{-1/2}$ to be whatever

and the integral does not depend on what this whatever is

perhaps $\left(\lim_{a\to 0} \int_{-1}^a x^{-1/2}\,dx\right)+\left(\lim_{b\to 0} \int_b^1 x^{-1/2}\,dx\right)$ is a better way to put it?

difference being that in the principal value sense you assume that the two limits are taken in the same way

whereas here you really don't need that assumption

no, it really isn't

@Semiclassical that's not how Lebesgue integration works

0celo: ah, I guess that's because $\int_0^0 stuff =0$ so whatever stuff is (as long it is bounded) will not matter

@Secret even if it's unbounded

@abenthy Yes.

$\int_0^0\infty=0$, but that's pretty shit writing

@0celo7 well, that'd explain why I'm not able to say things right :P

How does a set being measure zero prevent unbounded stuff from leaking out into the result?

because I will thought something like the indeterminate form $0 \cdot \infty$ will pop up when the integral is evaluated?

I suspect one has to be careful about the distinction between functions and distributions here

No

really?

I think we should do a reading of Halmos' measure theory

ugh

pass

I'm just going to stick "almost everywhere" on these dividing by zero equations

hehehe

@AkivaWeinberger Now go back to my energy answer and see that exercise 10 gives variations which are generated by elements of the kernel of $d\mathcal{E}$

These are called Jacobi fields

you prankster you

if $f$ is analytic and i know that there is $z_n \to z$ with $|f(z_n)| < a_n$ over a ball around z. can i deduce from that $|f(w)| < a_n$ for $w$ in the boundary?

Real analytic or complex analytic?

complex

for example $|f(1/n) | < e ^ {-n}$

can i deduce from that $|f(w) | < e \ ^ {-n}$ for $|w| = r $ for some $r$ ?

There's more work to be done with this, actually. One needs to apply the Maximum Principle

I hate this proof

i know it has something to do with the max. princ.

no, my stuff

but im not sure if what im saying is even true..

but you might need it too!

ah, didnt see what you wrote :P

lol coincidence

im asking this question regarding my question here:math.stackexchange.com/questions/2689315/…

i want to show $f ^ {n} (0) = 0$

using Cauchy's integral formula

and i need to bound $|f(w)| $ over a circle..

so... I got that Halmos book, which pages contains the relevant info to derive "$\int_a^a \infty = 0$"?

page 104

I think I have to use a distributional derivative. What a complete PITA.

hmm...

how will $|f_n-f|$ make sense at $\{a\}$ if $\lim_{x\to a}f(a)$ is unbounded

I am convinced by the rest of the argument, except this step

moreover, if $f$ is unbounded at $\{a\}$ then even if $n\to \infty$, no matter how large $n$ is $|f_n-f|$ will still be a positive unbounded value, and hence unable to shrink to zero?

I guess another way to ask this question is: Why does the indeterminate "$\infty - \infty$" not arise here?

@BalarkaSen So using the Poincaré isomorphism $D \colon H_c^*(M) \to H_{n-*}(M)$ the cup product on cohomology with compact support gives a ring structure on $H_*(M)$ via $\alpha \beta = D(D^{-1} \alpha \cup D^{-1} \beta)$?

That annoying thing about h bar when you have a question that no SE site can ask about, but it is related to physics and what-ifs, thus you have no choice but to ask the chat, but then the people who hold the answers are terribly uresponsive. RIP me

There are many things about weird people, but that one trait that I value in them is their higher probability to spontaneously respond to indirect questions

@Secret Why don't you ask on Physics.SE if its related to Physics?

It's too speculative, it will be instantly closed for "not mainstream physics"

I am basically asking about whether some notion of inheritance can be supported by quantum mechanical rules

so that we can have quantum systems that evolve like organisms

@Secret Ah, Physics Overflow? Also, if that's the case then you can even suggest a new site on Area51.

There's already one, but it is not beta yet

@Secret I see. I'll follow it too then.

This part of it:

$||z_1|-|z_2|| = |z_1+z_2|$, this one (this equality case) isn't obvious to me.

It's usually helpful in this kind of problem to square both sides

And then express the absolute values in terms of $|z|^2=z\overline{z}$

That expression has so many absolute signs, I am not sure if it can ever be interpreted geometrically

Pretty sure there's at least a geometric interpretation of $||z_1|-|z_2||\leq |z_1-z_2|$ (reverse triangle inequality)

note that in that case $|z_1|,|z_2|,|z_1-z_2|$ correspond to the distances between the origin and the points $z_1,z_2$

and that has a direct interpretation vis a vis triangles

(draw it and see what I mean)

@Secret It can be.

I think that for this the condition is that $\text{Im(z)<0}$, do you agree @Semiclassical ?

It's also worth noting that $||z_1|-|z_2||=|z_1+z_2|$ becomes $||z_1|-|z_2||=|z_1-z_2|$ if you replace $z_2$ with $-z_2$

So if you understand the condition of equality for one of the statements, you get the condition of equality for the oother

I think you mean Im(z) < 0 ?

@Semiclassical yes

I want to agree, but what is z in relation to z1, z2?

Ahhh

$\text{Im(z})_2<0$

I don't think that's going to work.

I suspect what you want to do is divide both sides by $|z_1|$ to get $\left| 1-\left|\frac{z_2}{z_1}\right|\right| = \left|1+\frac{z_2}{z_1}\right|$

in which case it reduces down to the case of $|1-|z|| = |1+z|$

For that, I could believe Im(z)<0 is the condition

(though I haven't checked it)

Abcd: (Unrelated PS Slereah is shitposting again. He is one reason why I have less inhibition to spam h bar due to the persistent low weirdness (maths has roughly 3 nonweirds as of 2018, while h bar still has 5 if I recall))

(Alessandro is interesting in that he is semi-weird, because his display of weird and non weird behaviour is so far statistical, thus he fit into neither categories)

I think the distinction between weird vs. nonweird is a bit too sharp at times

I'm not sure where I land, for instance

(probably depends a bit on the time of day for me, lol)

@Semiclassical (Imagine (using vectors)). Consider two points z_1 and z_2 on different sides of the origin i.e. $Im(z_2)<0$. Also let $|z_1|>|z_2|$...Equality of the above inequality only holds when they are collinear. $\implies $||z_1|-|z_2||= |z_1+z_2|$$

@Abcd What if Im(z1)<0 as well?

@Semiclassical then the equality is for this: $||z_1|+|z_2||= |z_1+z_2|$

(geometrically)

hmm

I think according to the statistics, you are weird, because like most weird people, you like to explore interesting things in a domain and when people ask questions or discuss things that does not sound making sense at first glance, but it is actually sensible, weirds tend to contribute rapidly, weeding out any errors, and suggesting further ways to optimise the issue in question

e.g. recall you investigations on the lambert W function branches

@Secret "to spam h bar"? You want to spam h bar?

One more question please (related to De Moivre's theorem)

@Secret I think weird vs. not-weird is a bit of an odd description for that, then. but i've actually seen this distinction elsewhere

one descriptor being that hot vs. cold as an analogy to statistical mechanics

If $n\in Z $, $(\cos\theta+i \sin\theta)^n = \cos n\theta + i \sin n \theta$. I understood this.

But lemme find a source rather than try to do it off the top of my head

@Secret haven't found hot/cold yet, but this is one version of what I had in mind: ncbi.nlm.nih.gov/pmc/articles/PMC4410143

I don't get how.

But in some way, nonweirds also have an advantage, for examples, they are less suceptible from being **** by trolls because they will quickly ignore them as questions are more likely to make no sense to them if the relevant jargon and culture is not in use

and trolls are bad at jargons obviously

right, the temperature bit comes by way of analogy with how the Boltzmann distribution works in stat mech

at very high temps, all states are equally likely regardless of temperature and so are equally probable

at very low temperatures, only those states with minimum energy are probable

Basically, why does DeMoivre's theorem fail for non Integers and how to generalise it?

@Abcd Simplest example will be exponent 1/2

What would DeMoivre say about the case of $\theta=0$, for instance?

@Abcd oh yeah, I am a hothead, a bit insane and also unstable at times, but I will try my best to mitgate frustrations anyway so that chats won't be spammed

Anyway, I don' wanna distract this discussion anymore, thus I am gonna stay silent for some time

@Semiclassical $1^{1/2}= \cos 0= 1$

Right. What about $\theta=2\pi$?

$1= -1$

Closer.

The LHS is just (cos(theta)+i sin(theta))^(1/2)

and what's cos(theta)+i sin(theta) here?

Edited

$1^{1/2}=-1$

Right.

But you already had $1^{1/2}=1$.

So? The principal argument is $0$ not $2\pi$

Sure. But does your LHS know about that?

All your LHS knows about is that you've got cos(theta)+i sin(theta)=1 and you raise it to the 1/2.

The right-hand side of your equation knows more than that, but the left-hand side doesn't.

It's not obvious... Both sides of the equation have same info IMO.

How? When $\theta=0$, you have $(1)^{1/2}$. When $\theta=2\pi$, you again have $1^{1/2}$.

but $\theta$, the principal $\arg$ is $0$ It cant be $2\pi$

Since when are angles required to be between 0 and 2pi?

cosine and sine are just functions. They don't care about the fact that you're talking about complex numbers.

and for cosine/sine, you can plug in any angle you want. doesn't have to be less than 2pi

yes.

The more general point is that, as written, $(\cos\theta+i\sin\theta)^{1/2}$ is $2\pi$-periodic whereas $\cos \frac{\theta}{2}+i\sin\frac{\theta}{2}$ is $4\pi$-periodic.

Consequently you'll have issues the moment you try to look past the 2pi window

What's $\int_0^1\delta dx$? $~\frac12$?

@AkivaWeinberger I presume you mean a Dirac delta centered at zero?

Yeah

It's probably undefined

It depends on the context, from what I know.

Hm

@Abcd part of the issue is with how the mapping $z\mapsto z^p$ is defined in the first place

@Semiclassical then? How to generalise DM theorem?

well, note that you've got $1^{1/2}=1,-1$ from the above. But you won't get any more values by choosing larger values of $\theta$; it'll just reduce two one of the two cases above.

I guess one way to put it is that while $$(\cos \theta+i\sin\theta)^{1/2}=\cos \frac{\theta}{2}+i\sin\frac{\theta}{2}$$ no longer makes sense, it's still true that $$\cos\theta+i\sin\theta =\left(\cos \frac{\theta}{2}+i\sin\frac{\theta}{2}\right)^2$$

the second is DeMoivre's theorem with $n=2$ and angle $\theta/2$.

What you can then note is that this remains true if you replace $\theta$ by $\theta+2\pi$

yes

More generally, suppose you have exponent $1/p$

Then $$\cos\theta+i\sin\theta=\left (\cos\frac{\theta}{p}+i\sin\frac{\theta}{p}\right)^p$$

And this will remain true if you increase $\theta$ by $2\pi$, and you can repeat this

so you get more generally that $$\cos\theta+i\sin \theta=\left(\cos\frac{\theta+2\pi k}{p}+i \sin \frac{\theta+2\pi k}{p}\right)^p$$ for any integer $k$.

Hi, could you show me how this holds?

I know that $\frac{f(x+h)-f(x)}{h}\rightarrow f'(x) $ as $h\rightarrow 0$ and similarly $\frac{f(x)-f(x-h)}{h}\rightarrow f'(x) $ as $h\rightarrow 0$ but don't know what happens to $\frac{f(x+h)-f(x)}{h^2}$ when $h\rightarrow 0$

The next point is that, while this works for any integer $k$, you'll eventually find that the RHS will repeat itself if $k$ keeps increasing

@Abcd so, if I start with k=0 and increase by one each time, at what point will I begin to repeat?

@Semiclassical $k = \dfrac \theta {2\pi}$

That's not an integer, though.

What you want is the value of $k$ for which you get the same thing as you had initially

so $\cos\left(\frac{\theta+2\pi k}{p}\right)+i\sin\left(\frac{\theta+2\pi k}{p}\right)=\cos\theta+i\sin \theta$

$\displaystyle\frac{\frac{F(x+h)-F(x)}{h} - \frac{F(x) - F(x-h)}{h}}{h}$

@Semiclassical how to get that value of k?

$k = \dfrac{\theta(p-1)}{2\pi}$?

well, you want $\cos((\theta+2\pi k)/p ) = \cos(\theta)$ and $\sin((\theta+2\pi k)/p) = \sin(\theta)$

Again, no. Theta is arbitrary.

Think in terms of the unit circle. How much do you have to increase $\theta$ to be sure you're going to have the same value you started with?

oh, blah

Bad typo in what i had above.

What I meant to say was $$\cos\left(\frac{\theta+2\pi k}{p}\right)+i\sin\left(\frac{\theta+2\pi k}{p}\right)=\cos \left( \frac{\theta}{p} \right ) + i \sin \left( \frac{\theta}{p}\right )$$

@Semiclassical $\pi$

@Rick and?

there we go, finally

@Abcd Not quite. Is $\cos(0)$ the same as $\cos(\pi)$?

@Leyla Alkan You can see this by writting $F(x+h) = F(x) + hF'(x) + {h^2\over 2}F''(x) + o(h^2)$

@Semiclassical $2\pi$

Right.

So we need the argument of cosine to increase by $2\pi$

:43380393 What have I become ...

Which means we need $\frac{\theta+2\pi k}{p}=\frac{\theta}{p}+2\pi$

Which value of $k$ will give that?

Guys, according to my book, a kernel of a digraph is an independent set $S$ having a successor of every vertex outside $S$. Any ideas on what a successor is? It means that every vertex is connected to another vertex outside $S$?

also hi @Astyx, long time no see

Wow I almost didn't recognize you with that profile pic

How are you ?

hahah:p

I'm fine, how are you

I'm good

Why is $\int_{-\infty}^{x} \theta(y-a) dy = \int_{a}^{x} dy = x-a$ not true? I know I should get $\theta(x-a)(x-a)$

@ShaVuklia From wikipedia's article on digraphs: "An arrow (x, y) is considered to be directed from x to y; y is called the head and x is called the tail of the arrow; y is said to be a direct successor of x and x is said to be a direct predecessor of y. If a path leads from x to y, then y is said to be a successor of x and reachable from x, and x is said to be a predecessor of y. The arrow (y, x) is called the inverted arrow of (x, y)."

@Semiclassical o oops, you caught me on laziness:p

thanks tho!

@Lozansky $\theta(y)=1$ if $y>0$ and zero otherwise, right?

Oh. What happens when $x<a$?

@Semiclassical Yes

We don't care

uh

Or maybe we do

Yeah, you'd better. If $x>a$ then $\theta(x-a)$ is pretty boring!

stressful music intensifies

Meh, I assumed $x>a$ :P

(It's also boring if $x<a$ is allowed but it's boring in a different way)

This was a nice check, thanks Semi

"You're not even interesting, you're just differently boring."

I would have preferred "oddly boring :P"

lol

goo.gl/images/Rzgan6

And in swedish, a drill is called "borr"

yeah, in english it's just "bore"

@Semiclassical k=p

@Abcd yep.

So you go through k=0,1,2,...,p-1 and these are all different

but once you take k=p you're back where you started.

So there's $p$ statements, one for each of k=0 through p-1

@Semiclassical Just a minute

The idea now is that each of these will be a pth root of $\cos\theta+i\sin \theta$

kk

to reiterate: $$\cos\theta+i\sin \theta=\left[\cos\left(\frac{\theta+2\pi k}{p}\right)+i\sin\left(\frac{\theta+2\pi k}{p}\right)\right]^p$$ is true for any integer $k$ but it's only necessary to consider $k=0,1,2,...p-1$.

Why did you do this?

Because it's true by DeMoivre's theorem (take the power to be p and the angle to be theta/p)

and you should note that it's exactly what I just wrote in the case of $k=0$ (as well as the case $k=p$)

The upshot is that DeMoivre gives p different versions of this statement, one for each of k=0 through p-1

@Semiclassical Got your method, thanks. But why does De Moivre fail for non integers?

For the above (the reiterate part) it doesn't. $p$ at this point is just some positive integer, so DeMoivre works just fine in that particular case.

But now you've got $$\cos\theta+i\sin \theta=\left[\cos\frac{\theta}{p}+i \sin\frac{\theta}{p}\right]^p=\left[\cos\frac{\theta+2\pi }{p}+i \sin\frac{\theta+2\pi}{p}\right]^p=\cdots=\left[\cos\frac{\theta+2\pi k}{p}+i \sin\frac{\theta+2\pi k}{p}\right]^p=\cdots$$

@Semiclassical yeah each "inside" value is a root

What that means is that we can't say $(\cos+i\sin\theta)^{1/p}=\cos\frac{\theta}{p}+i\sin\frac{\theta}{p}$

That could be true.

But it could instead be true that it equals the second root

or the third, etc

all the way up to k=p-1

I see.

Got it!

Consequently, the most we can say is that the $p$th roots of $\cos \theta+i\sin\theta$ are the numbers $\cos\frac{\theta+2\pi k}{p}+i\sin\frac{\theta+2\pi k}{p}$ for $k=0$ up to $p-1$.

As a check on this, let's go back to the very first case.

namely, $\theta=0$ and $p=2$

@LeylaAlkan I think they mean $\displaystyle\frac{\displaystyle\frac{F(x+h)-F(x)}{h} - \bigg(\frac{F((x-h)+h) - F(x-h)}{h}\bigg)}{h} = \frac{F'(x)-F'(x-h)}{h} = F''(x)$

What we then conclude that is that the square roots of 1 are $\cos 0+i\sin 0=1$ and $\cos\pi+i\sin \pi=-1$

Which is exactly what it should be.

yeah

So you do get a generalization of DeMoivre's theorem here, in that there are $p$ possible values for $z^{1/p}$

Oh okay, thanks! @Rick

Which really shouldn't come as a surprise: The equation $w^p=z$ has $p$ roots in $w$.

This comes up a lot when you discuss the $\theta=0$ case

in which case the values $\cos\frac{2\pi k}{p}+i\sin \frac{2\pi k}{p}$ for $k=0$ to $p-1$ are called the pth roots of unity.

(b/c unity = 1)

A similar idea works if you want to instead do $z^{q/p}$. In fact, the only thing that changes is that the angle is now $q\theta$ throughout instead of just $\theta$.

@Rick but the result seems to be $F''(x-h)$

On the other hand, this idea fails if you want $z^r$ with irrational $r$. In that case, you don't get that nice "everything repeats once k=p" property

so instead you need every integer $k$ and therefore any generalization of DeMoivre would involve an infinite number of roots...at which point one stops trying

Ok

You get some weird stuff if you go down that route

For instance, $i=\cos\frac{\pi}{2}+i \sin\frac{\pi}{2}$. So if you trust DeMoivre you have $i^i =(\cos(\pi/2)+i\sin(\pi/2))^i = \cos(i\pi/2)+i\sin(i\pi/2)$

But you can show that this last line equals $e^{-\pi/2}$

So $i^i = e^{-\pi /2}$...which is weird

One usually just stays away from monsters like that.

For the most part, one just worries about stuff like $z\mapsto z^{p/q}$ so that there's only a finite amount of weirdness

hello hello hello

i need help finding an atlas for the ellipsoid

i see it is a 2d manifold by the regular value theorem

Guys , how do I differentiate this wrt x so as to to get the derivative independent of $a$ : $\sqrt{1-x^2} + \sqrt{1-y^2}=a(x-y)$ .

differentiate wrt what?

wrt $x$.

@LeylaAlkan In my final step, substitute h as -t, you'll get $F''(x)$

@Tanuj

i think bringing x-y to the left side might work

@HerrWarum Uhm , I still can't get rid of that $a$.

@Tanuj if you divide by x-y, isn't a a constant that will give zero on differentiating?

@HerrWarum ah you meant that , will I not have to apply quotient rule then ? I think it will take long to solve then

@Tanuj long but easy

@HerrWarum yup , any other alternatives ?

I cant think of any off the top of my head

@HerrWarum okay , no worries. Thanks :)

$e^x=(e^{2\pi i+1})^x=e^{2\pi xi+x}$

(not actually)

@Tanuj, did you get up till $\displaystyle\frac{\text{d}y}{\text{d}x} = \sqrt{\frac{1-y^2}{1+x^2}}\cdot \frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2} +y}$ ?

If you substitute $a$ as $\displaystyle\frac{\sqrt{1-x^2} + \sqrt{1-y^2}}{x-y}$, it'll simplify to $\displaystyle\sqrt{\frac{1-y^2}{1+x^2}}$

I don't want to have PDE exam tomorrow :(

I hope I get hit by a car on the way

@Lozansky can relate, I've got functional which includes some PDE and this isn't fun