@Huy Or does it depend on something else?

@VincenzoOliva: I would only use it if it is absolutely clear by the context for the targeted audience which theorem/lemma you mean.

@Huy Perfect then, thanks.

If $\tau \subset \tau'$ we say that $\tau'$ is finer than $\tau$. If $\tau'$ properly contains $\tau$, we say that $\tau'$ is strictly finer than $\tau$. What?

@Huy I now need to explain why $\displaystyle \int_{|z-1|=2} \frac{1}{(z-2)^3} dz = 0$. I think I can use Taylors theorem for this, that is $f^{(k)}(z) =\displaystyle \frac{k!}{2\pi i}\displaystyle \int_{\gamma} \frac{f(w)}{(w-z)^{k+1}} dw$, but i'm struggling to think of a function f to use. I tried $f(z)=z-2$ which almost gets me the result I need, but not quite.

Darn it. I miss my complex analysis.

@GustavoMontano: $\tau \subset \tau'$ allows them to be equal. $\tau \subsetneq \tau'$ doesn't, thus you say strictly.

Oh wait, I can use z=2

What @Huy said

I'm okay now

I think

So $\subseteq$ and $\subset$ are the same in Munkres?

@GustavoMontano: They're pretty much the same everywhere, from my experience.

@Gustavo Well, just be careful whenever you see $\subset$

Oh no.

This is the first.

Same everywhere except in Australia.

That's really annoying!

I've almost exclusively seen the use of $\subsetneq$ explicitly, for when it can't be the same set.

Will have to keep that in mind! Thanks @Huy, @BalarkaSen.

@user112495: Are you sure that statement is correct? I'm a bit rusty but it looks like the circle contains the pole this time, then the integral shouldn't be zero...

I usually don't keep a fixed convention about $\subset$ and $\subseteq$ in mind. Whenever I stumble across something which says $\subset$ but actually means strict inclusion (i.e., otherwise it ain't remains true), I just add a $\not -$ below the inclusion sign along the way... :P

I had a deep feeling that would be the case. But it's a first experience for me, so I check with my MSE buddies :).

Another pain in the neck is about whether or not $0$ is in $\Bbb N$...

@Huy What, about the integral being 0?

It's a tremendous pain in the neck if you're studying number theory from different textbooks, especially doing exercises

@Huy I'm fairly sure it's correct. I'm not using Cauchy's integral formula this time though.

@user112495: Yes. @BalarkaSen: Can you look at it quickly? I haven't done complex analysis in years.

There is a theorem that talks about singularities and integrals.

@Huy Me neither, but what is it about?

@BalarkaSen: chat.stackexchange.com/transcript/message/19346094#19346094

@BalarkaSen: I would have guessed from first impression that the integral isn't zero because the circle contains the pole, but I don't know for sure.

@Huy I'm using this result here en.wikipedia.org/wiki/…

@Huy The circle contains $z = 2$?

Where?

@Huy I only need $f(z)=z-2$ to be complex differentiable I think

It's just a straightforward application of Cauchy's theorem.

@BalarkaSen: Isn't that the circle with radius $2$ with its center at $1$?

Oh, yikes, I misread.

Then you have to residue out of it by bending the contour.

Is @user112495 familiar with residues?

@BalarkaSen I don't think so...

@BalarkaSen Can I not just use Taylor's theorem? en.wikipedia.org/wiki/…

@GustavoMontano: I just realised by plugging in. I always get that mixed up.

What are you talking about :p ?

@GustavoMontano: Either way the pole was inside of the circle, so it didn't really matter.

Indeed.

@user112495 Maybe you can.

@BalarkaSen: But I don't think the residue of that pole is $0$, so how can the integral become $0$?

@Huy I haven't computed the residue.

@BalarkaSen: Neither have I.

If I let $f(z)=z-2$ and consider the third derivative, I have $f^{(3)}(2) = \displaystyle\frac{2!}{2 \pi i} \int_{|z-1|=2} \frac{w-2}{(w-2)^4} dw$. So

That looks right, @user112495

Go on.

Does the Cauchy Integral formula yield the same answer? I believe it does.

Let $f(z) = 1$. Is it right to say that it is analytic everywhere? If so, you have that $z = 2$ lies in the circle and therefore you are more than able to apply that formula.

Well, @Huy, turns out the residue is indeed $0$.

And then $f^{(3)} = 0$, so $\displaystyle \int_{|z-1|=2} \frac{1}{(w-2)^3} dw$

@BalarkaSen: I just noticed. Surprising.

@GustavoMontano Isn't Cauchy's integral formula only for $\frac{f(z)}{z-z_0}$?

@BalarkaSen: So if I have $\frac{1}{(z-c)^n}$ for any $n > 1$ the integral around a circle containing $c$ vanishes, right?

The formula I have posted is an "extension".

@Huy Yes.

Hmm

@BalarkaSen: Wow. I did not remember that.

If we use the Cauchy Integral formula, then you also get zero. Since the second derivative of 1 is 0.

I'm hungry. Nutella.

@Huy Interesting problem : Can we explicitly classify functions $f(z)$ such that for a given a Jordan contour $\gamma$, $\oint_{\gamma} f(z) dz = 0$, where $\gamma$ contains finitely many poles of $f$?

I sure can't right now. :D

I haven't thought about it, but looks interesting.

Maybe we can bound $f$ or something.

shrugs back to comalg. sometimes soon, I'd need to study Remmert

gah

@TedShifrin I managed to work out the bit with similar triangles but I am still stuck

you're Lieing, @Alizter?

No @BalarkaSen my book was confiscated

I must physics instead

But.. how?

I have to do well in my regular studies so my parents decided it was best

after that rant, I guess?

yup

lol

Ted gave me a 'simple' physics problem that I am still stuck on

oh well

@Alizter: Can you show me the problem?

I just ate a cake that tasted like animal piss.

@BalarkaSen: Your statement implies you've tasted animal piss before.

I somehow knew you'd said that.

@BalarkaSen: You're a very smart boy.

I personally think the contrary.

OK, I gotta run.

Morning, @TedShifrin.

@Alizter: I'm heading to the office to get ready for the semester. But remember that you always do two basic things with elementary mechanics problems. You draw a force diagram and you think about conservation of energy. In this case, you also have motion along a circle, so you know something about circular motion, too.

Hi @Huy.

@TedShifrin I will read a bit about circular motion incase I am missing something

@TedShifrin: Let's just use Lagrangian mechanics. :3

Well, of course you know some calculus, so it's even easier. You need to know about centripetal acceleration, of course.

Well, sure, @Huy. That's the link between the first problem and the second problem :P

@TedShifrin: I haven't seen the problems. Can you show me?

@Huy: Usual gravity in both. (1) A particle slides frictionlessly along a spherical ball which is sitting on the ground (with gravity pulling downward). If it starts not at the very top, at what height does it fly off the ball? (2) A ladder with mass leans against a wall, starts to slide. Ignoring friction, at what height does it fly away from the wall?

@Huy Is the superadditivity of the geometric mean a "mainstream" result? In other words, is it necessary to reference/outline a proof of it?

I've seen the first one. I'm a bit confused about the second one. How does it slide? @TedShifrin.

@VincenzoOliva: Depends on your audience.

There's no friction, so it just slips down the wall and along the floor, right?

Yes, @TedShifrin.

OK, I'm headed to the office. Bubye for now.

@Huy Well, mathematicians. I meant in a paper.

@VincenzoOliva: I wouldn't know what you mean.

@Huy ?

@VincenzoOliva: I don't know what superadditivity is and I'd have to look up geometric mean.

@Huy Oh, alright, I'll look for a reference.

@VincenzoOliva: Also, "mathematicians" isn't really an audience, unless you're writing a novel or some kind of introduction textbook.

@Huy How should I have pointed out the general audience of a paper?

@VincenzoOliva: Why would you need to point it out?

@Huy To you, that is. Didn't you say "mathematicians" is not an audience?

@VincenzoOliva: Usually, if you write a paper, you expect a certain knowledge from the audience. It would be rather cumbersome to define terms like "injective, continuous, differentiable" in every paper again and again.

@Huy Indeed, I agree. Nonetheless your advice is to provide a proof/reference of Mahler's inequality, correct?

How can we conclude from the fact that $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding that $\mathbb{Z}$ is a subset of $\mathbb{Z}_p$ ?

@VincenzoOliva: As I said: It depends on your audience.

@evinda: It is not literally a subset, but can be thought of as one, because an embedding is an injective map which preserves some sort of structure.

Could you explain it further to me? :/ @Huy

@evinda: How do you construct such an embedding anyways?

@Huy Mh. Well, providing it will certainly do no harm.

It is : $\epsilon_p(x)=(x \mod p, x \mod{p^2}, x \mod{p^3}, \dots)$ where $x \in \mathbb{Z}$ @Huy

@VincenzoOliva: For example, look at this paper: arxiv.org/pdf/1311.7608.pdf For the author and anyone who's remotely interested in reading it, the words "semiabelian variety", "Zariski dense", "Frobenius morphism" etc. are very clear, whereas I personally have no idea what they mean.

@evinda: I don't quite understand your notion. What do the commas indicate?

Or what is $\mathbb{Z}_p$ in your notation, @evinda?

And what is $\mathbb{Z}_p$?

@Huy It is the set of integer p-adics

Oh, I see.

@Huy I understand, yes. I think now the key is how much the issue is related to the topic of the paper.

@evinda: Again: An embedding is defined as an injection which preserves some kind of structure. Which structure it preserves depends on the context. Now, if you have an embedding $f: X \to Y$, since it is injective, the cardinality of $X$ is less or equal than $Y$'s, and since it preserves a structure, you can think of $X$ as some kind of subset in $Y$ (instead of looking at $f(X)$).

@Huy Which is what you said in other words.

Hello!! Is there someone that can help me at the calculation of the degree of a field extension??

@evinda: $f(X)$ and $X$ have the same structure and equal size (due to injectivity), and often it is easier to look at $X$ instead of $f(X)$. $f(X) \subset Y$, thus sometimes formally one says that $X \subset Y$ (but don't take this literally!).

@Huy Becase of the fact that $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is injective, we conclude that the cardinality of $\mathbb{Z}$ is less or equal than $\mathbb{Z}_p$'s, right? What do we take as $f(X)$ in this case?

@evinda: $\epsilon_p(\mathbb{Z})$

@Huy Why do we know that $f(X)$ and $X$ have the same structure?

@evinda: Because $f$ is an embedding, and an embedding, by definition, is an injective, structure-preserving map.

@evinda: I would never write something like that, because $\mathbb{Z}$ is not a subset of $\mathbb{Z}_p$. It is a common notion, but in my opinion a rather confusing one (as you can see).

@Huy How else could I write it?

@evinda: I don't really know what you are trying to do. All you're writing are tautologies anyways. :P

@evinda: Maybe something like this: $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding, so we can identify $\mathbb{Z}$ with its image $\epsilon_p(\mathbb{Z})$.

@evinda: However, anyone who knows what an embedding is would leave out the last part of the previous phrase.

@Huy I am looking at a proof of a sentence for the lecture I will give.. One of the sentences is this one: Die Abbildung $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ ist eine Einbettung. Man fasst damit $\mathbb{Z}$ als Teilmenge von $\mathbb{Z}_p$ auf.

@evinda: And you prove that? It's really redundant to prove that.

@Huy You mean the second sentence?

@evinda: It's like saying $f$ is an injection, therefore it is an injection.

@evinda: Kind of. You can say it, but I don't know what you'd do to prove it.

@evinda: Maybe you can briefly remind them of what an embedding is and explain orally why you can identify its domain with its image in that case, just like we just did, but I wouldn't waste more than 15 seconds with it.

@Huy A ok.. From this: we conclude that $\mathbb{Z}$ and $\epsilon_p(\mathbb{Z})$ have the same structure and size can we conclude that $\mathbb{Z} \cong \epsilon_p(\mathbb{Z})$?

@evinda: Depends on what an embedding is in your context, but if it preserves the same structure a homomorphism is supposed to preserve, then yes.

(assuming you mean "is isomorphic to" with your tilde over the equal sign)

@Huy So could we conclude from the isomorphism the second sentence?

@Pedro @robjohn (and other moderators) Is it possible that any comment flags I raised in the past days were declined more or less accidentally (perhaps by the new mods)?

@evinda: I don't understand your question.

@Lord_Farin It is possible that flags get declined accidentally. There is no "undo" for marking a flag as "helpful" or "declined"

@robjohn But there has been no update in flag handling policy in the past year or so? I've been away for a while.

@Lord_Farin It used to be that we could only evaluate flags in a group, but now, it is possible to evaluate each flag individually, but that requires an extra step that can easily be forgotten.

@Huy The second sentence was this: Man fasst damit $\mathbb{Z}$ als Teilmenge von $\mathbb{Z}_p$ auf. After saying that "Since $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding, we conclude that $\mathbb{Z}$ and $\epsilon_p(\mathbb{Z})$ have the same structure and size." could we say that "Therefore \mathbb{Z} \cong \epsilon_p(\mathbb{Z}) and so we can identify $\mathbb{Z}$ with its image $\epsilon_p(\mathbb{Z})$" ?

@evinda: Was ist eine Einbettung in eurem Kontext? Ein injektiver Homomorphismus?

@robjohn E.g. this comment I flagged as obsolete, then again after it was declined; it has been declined again. What purpose will retaining this comment ever serve?

@Lord_Farin: I think you're tagging the wrong person.

@Huy Ja, eine Einbettung ist ein injektiver Homomorphismus

@Huy Sorry, I noticed.

@Lord_Farin there are some general guidelines, but different moderators can handle flags slightly differently. If you have any questions, you can bring it up here, on meta, or contact the community managers by emailing team@stackexchange.com

@evinda: Wenn du nur das Bild betrachtest, folgt direkt Surjektivität, und ein injektiver und surjektiver Homomorphismus ist bijektiv, also ein Isomorphismus.

@evinda: Es kommt ziemlich auf deine Audienz an, wie sehr du da ins Detail gehen willst. Ich persönlich würde das alles weglassen und einfach sagen, dass $\epsilon_p$ eine Einbettung ist und darum $\epsilon_p(\mathbb{Z}) \cong \mathbb{Z}$.

@Lord_Farin That is indeed obsolete... I am inquiring.

@robjohn: Do you know what $\operatorname{sh}(x)$ could stand for? Possibly $\sinh$?

@Huy it would depend on context

@robjohn: It's really difficult to explain the context tbh. :D

@robjohn: Have you seen other uses?

@Huy nope... you weren't planning on evaluating at $x=it$, were you? that would be $i\sin(t)$

@robjohn: We're looking at different charts on spheres, planes and hyperboloids, and in spherical coordinates we have $$g_0 = \frac{1}{1-k(r/R_0)^2} dr^2 + r^2(d \theta^2 + \sin^2 \theta d\varphi^2).$$ Then it is claimed, that we can obtain a variant thereof by replacing $r$ with $\chi$ according to $$\frac{r}{R_0} = \begin{cases} \sin \chi & \chi \in [0,\pi], k=1\\ \chi & \chi \in [0, \infty), k=0\\ \operatorname{sh} \chi & \chi \in [0, \infty), k = -1 \end{cases}.$$

@robjohn: The $k$s indicate which manifold we're on, $+1$ for sphere, $0$ for plane, $-1$ for hyperboloid.

@Huy Have you tried plugging in $\sinh(\chi)$ for $\frac{r}{R_0}$ and see if it works?

@evinda: Das kann man direkt wenn man weiss, dass $\epsilon_p$ eine Einbettung ist. ^^

@robjohn: No, I was hoping someone was familiar with the notion, which is why I asked first, but I will, in that case.

@Huy In that context, I would assume $\operatorname{sh}(\chi)=\sinh(\chi)$, but I would definitely check it before proceeding.

@evinda: Das wichtige ist, dass du weisst, warum was stimmt. Aber so "simple" Sachen würde ich echt nicht extra erwähnen, ausser halt er fragt dann explizit danach.

@Lord_Farin I can delete the comment, or if you want, you can flag it again so that you get an accepted flag (if that matters to you).

@Huy Ok, aber wenn er fragt könnte ich dann so erklären?

@Lord_Farin Ah, I see that the comment has already been deleted.

@robjohn It has already been deleted, it seems. But thanks for looking into it and confirming that it was indeed obsolete; that comforts my mind more than the number of helpful flags (I can't even get a badge anymore, so it's for bragging rights only).

@evinda: Jo. Kannst es mir auch gerne nochmals erklären, und ich sag dir, ob das so klingt als ob dus verstanden hast oder eher nicht ^^

@Lord_Farin It was a mistake. I apologize for that. Thanks for bringing it up, it will probably keep this from happening in the future.

@robjohn Ok, thanks for the quick resolution :).

@Huy Eine Einbettung ist ein injektiver Homomorphismus. Wenn wir nur das Bild von $\epsilon_p$ betrachten, haben wir dass $\epsilon_p: \mathbb{Z} \to \epsilon_p(\mathbb{Z})$ surjektiv ist... Also ist $\epsilon_p: \mathbb{Z} \to \epsilon_p(\mathbb{Z})$ ein Isomorphismus, also $ \mathbb{Z} \cong \epsilon_p(\mathbb{Z})$. Also fasst man $\mathbb{Z}$ als Teilmenge von $\mathbb{Z}_p$ auf.

@Huy Und?

Ich würde "wenn wir nur das Bild von $\epsilon_p$ betrachten" weglassen und als letzten Satz stattdessen eher sowas wie "somit können wir Z als Teilmenge von Zp auffassen"

One of the nicest limits I received today $$\lim_{x\to 0} \frac{1}{x^2} \left(\frac{1}{2}\log(\cos \alpha) + \sum_{n=1 }^{\infty} \frac{(-1)^{n+1}}{n}\frac{\sin^2(n x)}{(n x)^2} \sin^2(n \alpha) \right), \space 0<\alpha<\pi/2$$

@evinda: Weil die beiden Mengen gleich gross sind.

@robjohn: Do we really need to occupy the space for 2 posts for the permanent pins on the star-board? Wouldn't it be better if those links were together and in one-line?

But I guess the only way to get all that onto one line would be to drop proper grammar and talk like a caveman inorder to convey all that information directly.

Ok, here's the caveman's pin :

Read Rules | Math in $\LaTeX$ | Be Pin on Map (how?)

Doesn't it look better? Please star it and put it up. pweez. pwetty pweez with sugar ontop.

@Huy Sind die beiden Mengen gleich gross weil $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$? Oder gibt es einen anderen Grund?

@evinda: Sie sind gleich gross, weil $f$ injektiv ist, ja.

I think there's a gross glitch in here. Oh no, wait. That's Zherman, Ya? Ah, never mind. I'm a poor comic.

@Huy Ist aber nicht auch $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ injektiv? :/

@evinda: Doch, es ist ja eine Einbettung

@Huy @alizter: Any progress? :)

@TedShifrin No

@TedShifrin: I've been working on GR, mainly.

I have literally nothing left to try

@Alizter: Well, I gave you three ingredients. Let's see what you did with them.

@TedShifrin Wait a few minutes. I will need to retrieve my notes.

ok ... I'm working on letters of recommendation simultaneously.

@Huy Warum kommt man dann zum Ergebnis dass die Mengen gleich gross sind von der Tatsache dass $\epsilon_p: \mathbb{Z} \to \epsilon_p(\mathbb{Z}_p)$ injektiv ist? $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ ist auch injektiv aber die Mengen sind nicht gleich groß... :/ Oder habe ich es falsch verstanden?

@evinda: weil du beim einen nur das bild betrachtest und beim anderen ganz $Z_p$

@evinda: $f: X \to f(X)$ injektiv ist immer auch surjektiv

@Huy Achso... Wie könnte man das beweisen?