Howdy folks, I was looking at this answer about the smoothness of the power map from the circle to itself and it seems like the answer isn't quite right. At least, $\arccos{u}$ seems like it should be $\arccos{\frac{2u}{1+u^2}}$ and even then, I graphed the resulting coordinate representation and its not obviously smooth

ANyone have any insight on if I'm right here, or I'm overlooking something about the answer?

@Steamy can I ask you something real quick? it's about something in my syllabus (in Dutch)

Sure, go ahead

When I read (c), I actually interpreted it as (e). I thought that in (c), they meant that we had $f\colon(G,X)\to X\colon(g,h\circ S)\mapsto gh\circ S$. But apparently that’s (e). So any idea what they mean at (c)?

@ShaVuklia Isn't (c) just the group acting on the set of all its subsets?

yes that's indeed what they say

oh

hm

yea

I think I'm seeing the difference now

oh yea, I see it. thanks

Yup, that's pretty much it

also, "ondergroep" :O

lol

yeppp

our teacher is a belgian

because deelgroep is Dutch, and ondergroep is Vlaams, right?

like, technically

The other way around afaik

oh huh

yea apparently

ohhhh yea

our teacher complained how it's ondergroep in Dutch :P

What's Vlaams??

the Dutch they speak in Belgium

Oh, Flemish?

lol yea

Ah, I hadn't heard the other name before

lol well it's Dutch for Flemish

And the best Flemish dialect is West-Flemish :P

Which even has its own wikipedia: vls.wikipedia.org/wiki/Voorblad

lol you probably live there

Maaaaaaybe?

looooooooll. don't worry, nobody here in this chat will stalk you:P

pfff, people already can :P

Not sure what good it would do them

@AkivaWeinberger puedo preguntarte una cuestion sobre topologia?

Claro

have you heard of cantor's leaky tent?

Yeah

I don't know how exactly one proves it's connected (with the top)

well I saw the outline of the proof

is that you assume the contrary

and argue that the top must belong to either set

and since the set is open, so must its neighbourhood

and somehow expand that to the whole tent

@AkivaWeinberger who do you think will know?

piensas que quien quizas sabe?

Like, the one that first introduces it

Maybe you could find a paper on it

(Those got sent in the reverse order^ ^^)

I even found a proof online

but that doesn't mean I can read it...

Guys, I have a question.

@ALannister just ask

How do I use the Lagrange error bound to prove that $P_{n}(0.5) \to \ln(1.5) $?

I suppose we're using the function $\ln(1+x)$ here

So, once I have the Taylor polynomial for $\ln(1+x)$, then what?

I'm assuming $P_n$ is the $n$th partial sum of that Taylor series?

Yes @AkivaWeinberger

I forget what the Lagrange error looks like

Oh, I found it math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/…

Right, so $|P_n(0.5)-\ln(1.5)|\le\dfrac M{(n+1)!}x^{n+1}$

where $M$ is the maximum of $|\frac{d^{n+1}}{dx^{n+1}}\ln(1+x)|$ on the interval between $0$ and $1.5$

Right.

\So, then what?

Well, we can actually calculate the $n+1$-eth derivative of $\ln(1+x)$

right, but don't we need the thing on the right to go to zero?

The first derivative is $(1+x)^{-1}$, the second is $-(1+x)^{-2}$, the third is $2(1+x)^{-3}$...

@ALannister Yeah, but we can't show that unless we figure out what $M$ is

or at least bound it

Right, so what you're gonna want to do is find a pattern for the $n+1$-eth derivative of $\ln(1+x)$, find a bound for its (er, for its absolute value), and show that the thing above goes to zero

When I wrote $\dfrac M{(n+1)!}x^{n+1}$ I guess I meant $\dfrac M{(n+1)!}(0.5)^{n+1}$

since $x$ is $0.5$ here

The pattern for the $n+1$st derivative of $\ln(1+x)$ is $|f^{(n+1)}(x)| = n!|(1+x)^{-(n+1)}|$

No critical points on $(0,1.5)$

Right, and we want to find a bound for it when $x$ is between $0$ and $0.5$, right?

Yes

$1+x$ is thus between $1$ and $1.5$

Still no critical points there.

So, its max would need to be at an endpoint

$(1+x)^{-1}$ is between $1.5^{-1}=\frac23$ and $1$?

still no critical points there.

Only critical point is x=-1

And then $(1+x)^{-n-1}$ would be between $(\frac23)^{n+1}$ and $1$

The point is that its maximum would be $1$

Ah. Yes.

So, M=1

Well we're trying to bound $n!|(1+x)^{-n-1}|$

That would be between $(\frac23)^{n+1}n!$ and $n!$

So, M=n!

Seems like it

Better.

Right, so we have $|P_n(0.5)-\ln(1.5)|\le\dfrac M{(n+1)!}0.5^{n+1}$, right?

@LeakyNun \o

Or, er, $\dfrac{n!}{(n+1)!}0.5^{n+1}$

@CowperKettle hi

\o/

And that fraction simplifies

Hm, what would /o\ mean

 \ /
  |
 /o\
¯¯¯¯¯

Handstand

It simplifies to $\displaystyle \frac{0.5^{n+1}}{n+1}$

@AkivaWeinberger spider

though i guess that'd be ////o\\\\

Or, even better: $\frac{1}{2^{n+1}(n+1)}$

Which of course, goes to zero as $n \to \infty$

/╲/\╭ºoꍘoº╮/\╱\

@Semiclassical

@ALannister Yup

Assuming we haven't made any mistakes

aghghghhh

@AkivaWeinberger is that a spider?

too many eyes

in the creeps-me-out sense, not the anatomical sense

óó_ÒÒ

Spiders have a bunch of eyes

They've got a bunch of everything.

+262 rep lol

sure

You see that black spot on the ceiling

That's not a spider

That's the only part of the ceiling that's not spider

Screams

Do they have an even or an odd number of eyes?

several eyes blink out of sync

Maybe the number of eyes isn't integral? :^)

hmm, from google: "Overall, 99% of all spiders have 8 eyes and of the remaining 1% nearly all have 6, but there are a few exceptions. Sometimes there can even be a varying number of eyes in the same spider family. For example, there are spiders in Cybaeidae that have eight eyes, six eyes, and two eyes."

Thanks @AkivaWeinberger for the help; not so much for the spiders.

regarding odd numbers of eyes: quora.com/What-animals-have-an-odd-number-of-eyes

(Why does no one ever thank me for the spiders $\ddot\frown$)

"Starfish have an eye at the end of each leg."

wuuut

phenomena.nationalgeographic.com/2014/01/08/…

to be fair, calling them eyes is misleading in that they're not like human eyes

"However, its vision is rather poor. It’s colour-blind, and sees the world only in shades of light and dark. Its light-detecting cells work very slowly, so fast-moving objects are invisible to it. And it has poor spatial resolution, so it can’t see fine detail."

still enough to be useful, but not anywhere near the ability of human eyes

Dammit I thought I was original

(You don't need to know what that is, the point is that I thought of it and then later it turns out it already exists)

> first described by Leonhard Euler

yeah, I heard of that at a math conference once

Why has he done everything

b/c Euler was the man

This is pretty cool

(More of that grid thing, with an actual song)

Hey, at least Euler published his findings

Gauss was this giant troll who discovered stuff and then deemed it "not worth publishing"

he said that to a lot of people

So when someone else sent him a letter about their discovery, he could reply with "yeah I'd like to praise you for this, but I'd just be praising myself because I already did this years ago"

lolyes

"nice job finding what I did too!"

Kronecker, Lobachevsky, numerous other people I can't recall

Well tbh if it did turn out that music grid thing was original, I probably would never get around to publishing it

@BalarkaSen could I ask you about topology?

no

and if someone came up to me with that idea later, I'd be like

"Oh cool I came up with that a while ago too"

you can only ask me about higher topos theory

Why is Cantor's leaky tent connected (with the top on)?

Gauss was also one who never published until he was really satisfied with what he'd done.

yeah no I definitely don't care about that

I think his Latin motto translates to "Few, but ripe."

you can probably find proofs online

@AkivaWeinberger I have already thought about it and have written a short paper on it but it's on the drawer there somewhere.

@BalarkaSen :O

a paper

I didn't publish it

@Steamy can I ask you one last thing? it's so cumbersome to translate all this Dutch stuff:< and I just want to check something small

Sure

I have a paper, I think it's 8.5x11

If we’re going to apply the idea of the proof to $V_4$, is it correct that $f\colon V_4\to S(V_4)\colon x\mapsto \epsilon_x$, where technically $\epsilon_x=x$? So technically $f=\operatorname{id}$?

@ShaVuklia Cayley's theorem right

yea

yes, $f=\operatorname{id}$

It pretty much says Cayley's theorem on top :P

hahah

I can read like 50% by my knowledge of German

oh that explains it

another 50% by my knowledge of English (those terms there)

Is $V_4$ your notation for the cyclic group of order $4$ ?

@SteamyRoot no, the klein-4 group

@LeakyNun I can read it all from my knowledge of German and Danish

what Leaky says

oh

never seen anyone write that with a $V$ o.O

German rocks. Dutch sucks. West-Flemmish even more.

@SteamyRoot V for vier (German of 4)

@SteamyRoot it's not infrequent to refer specifically to the Klein 4-group as $V_4$

lol, what leaky says, again

Well, that explains it I guess

@MikeMiller Which does seem somewhat redundant since the $V$ also stands for "four"

loll

that's true :P

I just stick with $\mathbb{Z}_2 \times \mathbb{Z}_2$ :P

$4_4$

you basically have to look at the group table @ShaVuklia

no that's not my question @Leaky

I specifically wanted to use the theorem:P

but alright, you already confirmed

so that's good

So $a$ is the permutation that sends $(e,a,b,c)$ to $(a,e,c,b)$

i.e. $(1,2)(3,4)$

yea yea I know that :P

@ShaVuklia Whether $f$ is the identity depends on how you have defined the Klein 4 group

If it is as that subset of $S_4$ then yes

yea that's what they do

they say call $e$ $1$, and so forth

V_4 is a pretty interesting group

or actually

I don't know

@ShaVuklia No, they are just renaming elements here, not defining the group

It's the simplest group which is not cyclic.

@TobiasKildetoft yea true

that's like so amazing to me

@BalarkaSen smallest?

so technically, it's not the identity then

I'm not really sure what you mean with $f$ being the identity. We're trying to prove $V_4$ is isomorphic to a subgroup of $S_4$

right, it isn't the identity

So assuming this isomorphic-ness off the bat is just a circular argument.

yea I thought that the elements of $V_4$ were already considered permutations of 1,2,3,4. But that's actually what they're showing :P

@SteamyRoot yep, I just realised that :P

@ShaVuklia $f$ is de groepmorfisme van $V_4$ aan $S(4)$

@LeakyNun The answer to that question is yes, if even there is a question.

If $1 = (0,0), 2 = (1,0), 3 = (0,1), 4 = (1,1)$, then you map, say, $2$ to $(1 2)(3 4)$ because $2 \cdot 1 = (0,0) + (1,0) = (2,0) = 2$, $2 \cdot 2 = (1,0) + (1,0) = (0,0) = 1$, $2 \cdot 3 = (1,0) + (0,1) = (1,1) = 4$, $2 \cdot 4 = (1,0) + (1,1) = (0,1) = 3$

@SteamyRoot thanks. I somehow got it, but I also didn't get it. But thanks for writing it out, that helped.

and @Leaky I know you were saying the same:P but the coin fell just now.

@ShaVuklia alright

@LeakyNun Sure, but I think it's a lot more obvious/clear by writing out the elements as elements of $\mathbb{Z}_2 \times \mathbb{Z}_2$.

@SteamyRoot I've always thought of $e,a,b,c$ as the standard representation @_@

@ShaVuklia to test your understanding: what does the Cayley theorem's algorithm give you for $G=S_3$?

let me think (no hints pls)

At my universit, people don't even learn that representation or the name "Klein Four group". They just learn what $\mathbb{Z}_2 \times \mathbb{Z}_2$ is :P

@SteamyRoot then you're the only one here who doesn't know that $V_4$ stands for the klein-four group lol

I guess

I still hate that belgian book by Igodt :P

I'll tell him :^)

hahahahahaha

Oh, by the way, what's the name of the Belgian guy teaching you algebra?

Raf Bocklandt

Don't know him, must be from Gent

what is a field in abstract algebra?

yea

@Leaky $S_3$ is isomorphic with a subgroup of $S_6$?

Fix $3$ of the $6$ elements :P

@KasmirKhaan It is a commutative ring where all non-zero elements are invertible

@SteamyRoot Except that is not what Cayley gives you

@ShaVuklia right

lol that's a great discovery then :P

@ShaVuklia which subgroup?

dude I know that

what Steamy said

Oh, are we insisting on using Cayley?

sort of, I guess.

@TobiasKildetoft thanks :)

Well, technically, even if you use cayley, you'll identify $S_3$ with $3$ elements in a set of $6$ elements, and then study the action of $S_3$ on those first $3$ elements.

It still fixes the final $3$ elements :P

@SteamyRoot No, the specific subgroup given by Cayley will have transitive action

Oh, wait, yeah

Woops, brainfart

I should probably stop switching between my own research and the chat, it's clearly confusing the hell out of me >.<

Also, though unrelated, all elements of the same order become conjugate in $S_n$ withthe Cayley embedding. Though for $S_3$ they already were.

I gave up writing any more for tonight. Paper is looking to be somewhat short as a first version, but I will just need to add some more results later, as I really need this part to be on the arXiv no later than September 15th

Why the deadline?

@SteamyRoot I am applying for a Marie Curie fellowship with that deadline, and I refer to the paper I am working on in the project description

Ah, right.

So I need to have the parts of the paper referred to ready by then. I can then extend with some further things I realized recently later on, which will substantially improve the paper.

(OMG we're finally learning what fields and rings are)

they mention that stuff throughout the entire first year

@ShaVuklia congratulations

hahahah

thx

@ShaVuklia so what is de ondergroep?

you mean what it translates to? subgroup.

what is the subgroup of $S_6$ that Cayley gives you for $S_3$

the one where you fix 4,5,6

@ShaVuklia No, it really is not

Does the boundary of a domain in the plane characterize it?

The elements of $S_3$ are the $6$ points

you need to understand/recall what Cayley embedding does, firstly

@Emolga Well, uh, given any Jordan curve there are two domains in the plane which has that as a boundary

so I don't think I understand your question

@BalarkaSen Yes, of course, but e.g. is there exactly one domain whose boundary is the same as a given annulus?

Yeah, well, complement of an annulus on the plane is disconnected and the closure of each component contains exactly one of the boundary components.

But the same is not true if you replace "annulus" by "disk" is all what I mean