@Blue @Blue Sorry, I'm new here and perhaps shouldn't interfere, but I saw you posting the Extended Real Numbers definition and just wanted to mention that this definition is logically flawed. Real numbers are defined as an infinite set. Then infinity is defined as larger than any real number. Therefore the definition is circular. Real numbers are defined using infinity before infinity is defined using real numbers. There are other inconsistencies there as well.

(lol)

@safesphere That's not true, at least not using standard set theory and the usual constructions of the real numbers e.g. by Dedekind cuts.

@ACuriousMind it's b8

@0ßelö7 Part of Being Nice is assuming good faith.

@ACuriousMind you assume good faith for me?

@0ßelö7 Unless prior experience indicates it would be foolish to, yes.

complete ordered field is not a construction, hence worthless

@ACuriousMind I don't want to start argument, but IMHO the ERN definition is logically flawed on so many levels that it cannot be taken seriously. This doesn't affect most results though, because people use the intuitive definition instead.

@Blue You can show that there is a unique such field, but that doesn't prove existence. You need to do an actual construction like: natural numbers (Peano) -> integers (adding additive inverses) -> rationals (fractions) -> reals (e.g. Dedekind cuts)

Or, of course, you can just assume the reals exist. It won't make a difference unless you're interested in foundations :P

@Blue I wouldn't be so sure. Putting different labels on the same thing doesn't change its meaning. Sorry again, I'm not a mathematician and there are many concepts that I don't know. However, what I do know i try to understand really well and I can see no logic when it's not there.

if you're not interested in foundations, you have no business doing math

@0ßelö7 I want to say the majority of mathematicians disagree :P

@ACuriousMind says the physicist

@0ßelö7 I know a lot of mathematicians

ACM just told you how to prove existence

@Blue The proof of their existence is precisely given by constructing the reals

You first prove "all complete ordered fields are isomorphic". Then you prove at least one exists, namely the reals by whichever construction you like

Wildberger thinks the Dedekind cut thing is wrong

the video is too long, I ain't gonna watch it

it isn't

@Blue This isn't number theory

Mhhh...analysis?

"foundations of analysis"

number theory is primes and stuff

@0ßelö7 I prefer Cauchy sequences anyway :P

@ACuriousMind I think there's issues with that because you need limits or something

You can probably just define $\Bbb Q$-limits and be fine though

(you need limits to define the equivalence relation)

@0ßelö7 You can't wave the magic wand called "completion" because that presupposes the reals but you can just mimic all the standard constructions - since there are arbitrarily small rationals, there's no need for the $\epsilon$ in the limits to be allowed irrational

that's literally what I just said

It is!

@ACuriousMind All spaces are over $\Bbb R^n$, $n\ge 2$. The claim is that given a vector field $v\in L^2\cap C^\infty$, there is a unique orthogonal splitting $v=w+\nabla q,$ $\nabla\cdot w=0$ such that both summands are in $C^\infty\cap L^2$.

the usual claim is for some bounded set, then there are boundary conditions one can use

@ACuriousMind From Wiki: "A Dedekind cut is а method of construction of the real numbers. It is a partition of the rational numbers into two non-empty sets." These sets are infinite. Therefore this definition of real numbers uses infinity.

but for this global thing I see no good solution. The proof seems to indicate one should use a plateau function $\rho$ on $\Bbb R$ to write $v_n=v\rho(|\cdot|/n)$, so that $v_n\in C^\infty_0$ and $v_n\to v$ in $L^2$

Then one solves $\Delta q_n=\nabla\cdot v_n$ by convolution, then define $w_n=v_n-\nabla q_n$. Then somehow $w_n$ and $q_n$ converge to what I need

But I have no clue what space $q_n$ is even in

ideas?

@safesphere Well, apparently you do want to start an argument ;P Sure, the sets are of infinite - even uncountable - cardinality. That's not a problem. You construct the reals $\mathbb{R}$ that way and then you add two additional elements $A$ and $B$ to this set and define that $A < x$ and $x < B$ for all reals $x$

there might be a typo and $v\in H^1\cap C^\infty$ is right

having the derivative be nonintegrable seems like a really bad idea

And then we replace $A$ and $B$ by the more common symbols $-\infty$ and $\infty$. These have nothing to do with infinite cardinality, and "infinity" is just a name for these elements.

@0ßelö7 I don't really know what you're talking about. Not a functional analyst, remember? :P

Here is the time to save a question to the archive.org ...

@ACuriousMind what part is unclear

@Blue Tomahto - tomaito ;)

@ACuriousMind I was invited to give talks at the new geometry seminar about regularity theory btw

there seems to be agreement that I'm the first one crazy enough to write all the details down

@0ßelö7 Not so much unclear as that I don't have an intuition for any of this

@AlexKChen You can do similar simulations in Matlab. I didn't know about vpython, but as I see it, it seems quicker than Matlab

@Blue No, he just restated the ERN definition with all its missing logic. For example, "all reals $x$" is an infinite set that cannot be consistently used in the definition of infinity.

@ACuriousMind No Sir, I really don't want to start an argument. I'll just shut up and only suggest you giving this a thought on your own. Aside from this definition being obviously circular, it also arbitrary uses the "larger" relation outside of the number line where it is defined. It puzzles me that people don't see this self evident lack of logic. Alright, I'll shut up now :)

hmm pdfs.semanticscholar.org/ecfa/…

Hi, everybody.

@DanielSank ¡Hola!

@Mithrandir24601 Holo. Gotta go to meeting.

@DanielSank Enjoy... Maybe meetings at Google are enjoyable?

They often are, largely because of the meeting manifesto:

Item #1 is wonderful.

If you have "group meeting" but no specific thing to discuss, you cancel it.

Item #2 is also wonderful.

i.e. "no shame in leaving if you're not useful in the meeting".

@DanielSank Nice! One of the groups in Bristol (QComm) implements 2) a lot and it's a really good rule. Those seem to be really good... I'll bear that in mind, thanks :)

Are polar coordinates ever more useful than Cartesian coordinates?

@SirCumference Yes. So much yes

@Mithrandir24601 Really? :O

The fact is that when it's initially taught, teachers don't explain the advantages or usefulness of it

amazon.com/…

1018 pages!!!!

$10^{18}$ pages??

@SirCumference Every time you want to do a lot of rotations. Simply adding to the angle is much easier than rotating a Cartesian vector.

@0ßelö7 I've just (about a week or two ago) finished the third Malazan book of the Fallen. Very nearly 1200 pages. It's not a textbook, but still

I might need this book just so I can throw it at intruders

@ACuriousMind aha

reads

Hmm. This seems wrong too

Well, all of physics is wrong

@ACuriousMind I can't into math. Is $1/|x|^{n-1}$ in $L^2$ (assuming we ignore the origin)? $1/|x|^{2n-2}$ seems pretty integrable.

$n\ge 2$

If $n$ is the dimension, it's false for $n=1$.

:P

@ACuriousMind the Jacobian gives $n-1$ factors of $|x|$, right?

And it's also not integrable for $n \geq 2$, you just get an integration of 1 from 0 to infty in polar coordinates

you get $2n-2$ factors of $r$ from the Jacobian?

it seems you get n-1 factors

as I thought

(just use dimensional analysis on $\sqrt g dx$)

So you get $1/r^{n-1}$

that's integrable for $n\ge 3$