I believe one would start with observing that the factor 1/(t^2-1)=sum_n t^(-2n)

Is it true to say that the volume of $$\Omega \subset \mathbb{R}^n $$ only depends on $$ \partial \Omega $$ ?

When $$ \Omega $$ is a bounded open connected set?

If $\Omega$ is reasonable, yes.

I don't know why I feel this is true but I got this ichy ich that I need to be careful when saying it :P

Well, if the boundary is a non-measurable subset, I suppose it could be questionable.

In what context is this coming up?

Variational calculus

So aren't you looking at manifolds with boundary in that context?

Yes. Lipschitz domain

Yeah, the boundary is measurable

If you're doing variational stuff, you're going to want to make sense of mean curvature of the boundary almost everywhere or something like that.

Ok. So far mean curvature is not in the material. But we are half way to the course

If you're trying, for example, to get minimal surface area of the boundary for a given volume enclosed, you will have a hypersurface of constant mean curvature.

Ain't that meaning that my surface is like a sphere or ... not sure what is the opposite of a sphere

or flat surface?

Yes, a sphere is an example. But it's not the only example.

Soap bubbles are examples; when they attach to one another, they're no longer spheres.

There are interesting non-compact surfaces of revolution that have constant mean curvature.

Oh this is cool!

Thank you @TedShifrin

Sure thing.

Sorry to ask, but conventionally does raising an ideal to the power of zero yield the unit ideal?

E.g. Wikipedia's page on nilpotent ideals mentions that ideals can be raised to any natural power

To most of the world, natural numbers do not include $0$.

We're back to this arcane European tradition of including $0$, when it causes so many difficulties.

@TedShifrin Huh, really?

I suppose one could define the $0$th power to be the unit ideal, just as we observe that $a^0=1$ for all nonzero $a$ and $0!$ is defined to be $1$.

I just wanted to make sure since e.g. the page on nilpotent elements specifically says positive integer powers, unlike with nilpotent ideals

i love it when wikipedia pages are plainly related in a way that makes it natural that someone would bounce between them, but have discrepancies like that because of different editors, different times.

someday an AI will harmonize everything.

I don't see any reason why one cannot set $I^0 = R$, but my complaint about the language still stands.

Gotcha, thanks

I dunno, including zero seems like the only way to make "natural number" meaningfully different from "positive number"

It's still a matter of notation. How do you take sums of $0$ products of elements of the ideal?

"Positive" is just easier to say and less ambiguous so I assume the use of "natural" is always intentional

It's easier to write $\Bbb N$ than to write $\Bbb Z^+$.

@TedShifrin Yeah, I agree it's probably just convention if the 0th power of an ideal exists

@TedShifrin Eh it's not that much easier imo, at least when you consider how much easier $\mathbb{Z}^+$ is for the reader to understand

in some contexts it would also be weird to say 'positive,' e.g. if you are only clearly only doing arithmetic with positive integers. there's no "negative" so you wouldn't necessarily want to imply that there is.

I guess that power series could be written as $\sum_{n\in\Bbb N} a_n x^n$ if we include $0$, but I've never once tried to write the summation like this anyhow.

@vitamind I'd be surprised, because both these sums appear in formula (18) of the paper and if they were equal, the formula should simplify a lot. in any case, there are two options since we both agree that the RHS involving the sum over the Möbius function diverges: 1. you are mistaken about the sum on the LHS converging 2. the paper is wrong. I'm not qualified to judge which is the case, but if you want to find out, 1. is where you should doubt yourself.

I guess if I ever see someone explicitly write the word "natural" instead of "positive" I assume it's intentional

things that are "well-known" and can "probably be found in <textbook>" are where the most mistakes happen

Since they're both the same word length

In my 40 years of teaching, I frequently introduced $\Bbb N$ and mathematical induction before ever discussing $\Bbb Z$ and ordering (i.e., a notion of positivity).

Word length is stupid.

It's a matter of mathematical content, actually.

positive looks like it might take up more ink when printed out.

in some fonts, anyway. i'll go and check.

I feel like there ought to be a committee/organization that comes up with the standard notation

rolls $\pi^3 + e^{1/2}$ eyes

At least for situations like this

what a continental spirit.

There are plenty of other conventions in France, for example, that appear nowhere else.

The use of $]1,2[$ for an open interval has been discussed here for hours upon end.

i was just going to say, maybe the academie francaise could assert jurisdiction over this issue.

the zero-th power of an ideal should definitely be the entire ring

I will be among the first to admit that $(1,2)$ can be ambiguous, as it can be an interval or an ordered pair. Just deal with it.

or the ICJ in the Hague.

it's generated by all empty products after all

@Thor: But that is convention, not the usual definition. That's just my point.

Why is not the empty product generating nothingness?

@TedShifrin I guess I just expected there's be a lot of incentive for people to come up with a standard notation

Especially for a field like math where precise communication is so valuable

It's for the some logic that gives $a^0=1$ for the usual addition law of exponents to work.

no, it is part of the usual definition

ah ok, your point is that the empty product itself is convention, I still disagree

I disagree with your disagreement, of course.

@Thorgott It's the most famous result in analytic number theory isn't it? So it's well known. Of course my first thought was and is that I am mistaken, that's why I asked. Another option might be that their formula only holds for N<\infty (even if this does not make much sense since the mentioned the case N\to\infty for epsilon.) I'll try to figure it out. Thanks for writing your thoughts.

This should be a good one for Leslie. Does he come down on the side of agreeing with Ted again?

you won't like my justification in terms of categorical products

i decline to answer the question. 5th amendment

@vitamind $N$ is always less than infinity! A limit doesn't ever have an infinite $N$.

@leslietownes We've done away with the Bill of Rights. Deal with it.

@TedShifrin Yes thanks but I think it's clear what I'm trying to say. I did not want to write a whole limit expression, but of course you are correct.

So you mean that it holds for finite $N$ but that it does not hold in the limit.

@TedShifrin I don't know. This problem seems a lot harder than I imagined it to be.

maybe we are talking past each other, but the sums in question $\sum\frac{\mu(n)}{n}\mathrm{li}(x^{1/n})$ and $\sum\frac{\mu(n)}{n}(\int_{x^{1/n}}^{\infty}\frac{dt}{(t^2-1)t\log(t)}-\log2)$, right? if those two being equal is the famous result in analytic number theory, I haven't heard of it.

I'm not paying any attention to the actual mathematics here, because it is antipodal to my knowledge.

@Thorgott I'm talking about the explicit formula for pi(x). We can deduce from it that the LHS converges. If it was divergent, the most famous theorem in analytic number theory is wrong. (Sorry for not being precise.)

I don't see what the ugly sum we care about has to do with the explicit formula for $\pi(x)$

@Thorgott Google Play has a (free) pre-version of Edwards book. See page 34 eq3.

Set $x\to x^{1/n}$ and if you take the Möbius inversion of that expression you'll get the explicit formula for pi(x). (Note that log2 is a constant and that sum mu(n)/n)=0.)

the $\sum\frac{\mu(n)}{n}\sum_{\rho}\mathrm{li}(x^{\rho/n})$ converges as well?

Yes, conditionally.

then I'm confused

It's not often that @Thor is conditionally confused.

i'm absolutely confused.

Well, that's less stressful.

it's almost pleasant

I'm tending in that direction. Whether I'll converge remains to be seen.

@Thorgott Looking at (19) and (20), it seems that the third option N \neq \infty might be correct. (Still, if this is the case, this would result in further questions.)

wdym N \neq \infty

if its valid for all N, we can take a limit

That is one of the many further questions. But in (19) they want to find an approximation, so they clearly want a finite N.

oh, I think I get it

all of this is only for $x<2^{N+1}$

so taking $N\rightarrow\infty$ doesn't work

That's 1. very nice, where did you find the result? 2. confusing

they state so in the third line of p.970

Hey

oh look, it's a nerd

@Thorgott Okay this answers one question but opens two others. Here (eq3) is the exact form. Two problems: 1. The result, which is stated is exactly equal to the -ln+arctan term but it should be an approximation (with epsilon). 2. I forgot. I hope I remember.

@Thorgott no u

*Correction: I don't know if it is exact, it is stated as exact but well Wikipedia does not reference this so...

@AminIdelhaj can't believe you'd publicly own me like that

These fists don't miss

But yeah what've you been up to?

Hi Demonark

Hey Ted!

nothing extraordinary, just some math

currently typing up some riemannian geometry homework

how've you been doing?

Sorry I was out. I'm wrapping stuff up for the semester, grading calc exams

Then I'm switching to summer mode. Data science bootcamp and summer school in microlocal analysis

@Thorgott pinging since I responded late

@robjohn your proof math.stackexchange.com/questions/75130/… is great

but It is now obvious how did mathematician immediately constructed the proof out of thin air

can you tell me the motivation about how someone get this inequality

Microlocal analysis, huh?

I can't see the relationship between these area squeezing and rate of change sin(x)/x

Yup, that'll be one of my big things this summer. My advisor said it's pretty useful to know about that area for the more "asymptotics of Laplace eigenfunctions" side of his work

Nothing to do with thin air. Just about relating areas and using the unit circle defn of trig.

@robjohn if you know it please tag me thank you

This proof has been around for centuries, @user863565

oh, sounds exciting

Should be yeah

@user863565 Do you mean $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ or $\frac{\sin(x)}x\le1\le\frac{\tan(x)}x$?

the first is a limit, but there are several inequlaties in that post.

and as Ted says, it is a limit that has been known for a long, long time.

got that sincing feeling

@TedShifrin My real-only proof of $\sum_{n\in\mathbb{Z}}\frac1{n+z}=\pi\cot(\pi z)$ was done at Mammoth Lakes, so it could have been affected by the thin air

@copper.hat that happens when you cosine a loan

now you are off on a tangent

people say that a lot, it makes me not feel normal

i'm stuck :-)

But if I walk against the normal direction, I find my center

Just thought I'd point that out

:-)

@robjohn yes I mean lim sinx/x to 0

I want to know about motivation

it isn't intutive to me how to derive the proof of it but it is obvious the proof is true

@copper.hat Finally!

how can one relate the area of that 2 triangke and one triangle plus arc and derive that inequality from it for sinx/x exactly