That's a cyclic group and thus can be embedded into $S^1$ as roots of unity

So, now the idea is that you consider $\sum_{\chi\in\hat{G}} \chi(h)\chi_*(h)$

Wait, I'm confused, is $f$ from above a homomorphism?

If so, $\Bbb C$ is additive but $S^1$ is multiplicative

Now, $\hat{G}$ forms a group as well (easy to check), so in particular, this above sum gives $\sum_{\chi\in\hat{G}} \chi(h) = \sum_{\chi\in\hat{G}} \chi(h)\chi_*(h) = \chi_*(h)\sum_{\chi\in\hat{G}}$, which makes that sum 0

Nope, we're just considering functions, period

OK, that makes more sense

For reference, you can think about $\hat{G} = Hom(G,\mathbb{C}^x)$, since finiteness forces it to lie on the circle

@AkivaWeinberger no, I mean how does your proof fail

Right, yeah

of course I know that ord(A) can be n

@LeakyNun The contradiction was that we had $n+1$ linearly independent vectors in $\Bbb R^n$.

But anyway, so thanks to what I said, we know that $\delta_1(h) = \frac{1}{|\hat{G}|}\sum_{\chi\in\hat{G}} \chi(h)$, but then of course you can express $\delta_g(h) = \delta_1(gh^{-1})$, so in fact $\hat{G}$ spans $\{\delta_g\}$, which then spans $\mathbb{C}\langle G\rangle$

hmm...

If we replace $n$ with $n-1$, there's nothing wrong with having $n$ linearly independant vectors in $\Bbb R^n$

@Daminark Cool, I see

So now, I can't remember how linear independence goes exactly

$\Bbb C\langle G\rangle$ is a finite-dimensional vector space, right?

An algebra, as it turns out

How many basis vectors do you have?

If it's the same as the dimension of the space, and the (proposed) basis vectors span it, they must be linearly independent

The problem is that we don't know that $|G| = |\hat{G}|$

The dimension of the space is $|G|$

Knowing this would prove that and there was a way to do this which at the time was good

It was like

So clearly it starts with "suppose $\sum c_n\rm thingy=0$"

So now I'll number the elements of $\hat{G}$ by $\chi_{1,\ldots,k}$

And then yeah, suppose $\sum_{i=1}^k a_i\chi_i = 0$

And that $a_1 \ne 0$

Then you can multiply through by $\chi_1^{-1}$

So you have $a_1 + \sum_{i = 2}^k a_i\chi_i\chi_1^{-1}$

Somehow you could derive another one so as to kinda cancel the $a_i$, and then keep going

Well, hold on

Call the trivial map $g\mapsto1$, $\chi_0$

Sure

Now I'm not actually sure

But if we prove $a_0=0$, we're essentially done, right?

@Daminark regarding your starred comment...

Because then we can multiply through by $\chi_n^{-1}$ like you said

@Semiclassical what information do I need to add to keep it from being a research question? I thought I explained all the prerequisites given in the paper.

Speaking of the starboard, what does Leaky's Greek mean?

@AkivaWeinberger When I asked about it, he said it was trolling and left it at that.

@Daminark Maybe something about how $\Sigma\{x:x^n=1\}=0$ unless $n=1$

By which I mean the sum of the elements in the set

@heather Well, your question ultimately comes down to "What is a 'max support flux vector sequence'?"

(which probably isn't an actual notation used by anyone, but the alternative is $\sum_{\{x:x^n=1\}}1$, which puts the important part in the subscript and thus is bad)

It might help to put that more front and center. As it stands you're burying the lede as it were

@Semi I'll say as much as that we also believed he was wrong

@AkivaWeinberger @Semiclassical (vote on the starboard lol)

What did I just post

Nothing, nothing at all

Yeah, that's the disadvantage of what I linked.

Lmao

"Background messages inside"?

(Google Translate)

That's what Google Translate says

@Semiclassical put it front and center. Better?

Google said something like, all those who don't know geometry, don't enter (or something like that)

That may help, yeah.

@Semiclassical oh...

I misunderstood

I didn't think you were talking about actual Greek

Hi guys, I just have a small question. I am solving a problem involving construction of confidence intervals for the parameter of a uniform distribution using some form of the max and min of 3 iid RVs with that uniform distribution as the pivot

ohhh

I guess we misunderstood each other, then.

I think I solved it already for the case of the maximum but my answer for the minimum seems problematic

wow, if I never scrolled up

@heather I'll say, though, that that phrase seems pretty specialized.

@Daminark tecepe.com.br/nav/vrtool/ageometretos.htm

googling "max support flux vector sequence" brings up your question as the first hit, for instance

seriously?

huh.

i tried to provide all relevant information the paper gave.

and the fourth result for me is the paper I'm referencing =P

@Daminark What is $\sum_{g\in G}\chi(g)$?

Yeah.

For the trivial map, it's $|G|$.

On that note, though, you'd probably be better off linking to the arxiv preprint.

Conjecture: For all other maps, it's $0$.

So... what do you say? @AkivaWeinberger @Semiclassical

Otherwise people are more likely to see it go to a paywall and not bother.

So X1, X2, X3 are uniform (0.5theta, 1.5theta). I want to get a confidence interval for theta using the minimum. First, I used (Xmin-0.5theta)/theta as my pivot. Is that correct?

@LeakyNun I don't mind it, but I'd rather avoid clutter if we don't need it.

added arxiv link.

@LeakyNun I am very bad at Euclidean geometry. So, uh, I vote it down.

@Semiclassical because Balarka was accusing me of trolling

Ohhhh.

Then you say "I'll bite" meaning I'll continue interrogating you

oh. No, I didn't mean it like that.

@Daminark, do you still exist?

...though I guess it would count as trolling in the original meaning of the word, heh. You put that on the starboard, I asked what it meant.

Yeah, I'm thinking

then what does "I'll bite" mean?

I just thought that it'd be good to have a quote in the description, and this even has historical significance @AkivaWeinberger

I guess it's a bit idiomatic.

I see that my English needs improving :p

@LeakyNun It's like, I'll ask the obvious question or something

@AkivaWeinberger I see

I mean, it does go along with the underlying image of 'trolling.'

But not the connotations that word has picked up.

Oh @Akiva I think I see why that's true

Choose $b$ such that $\chi(b)\ne 1$

The old literal meaning of trolling is: fish by trailing a baited line along behind a boat.

well, I've done as you've suggested @Semiclassical - any other big things to fix?

Not really. My main concern, though (and I don't have an easy answer to this) is that the answer may come down to "read the paper more carefully."

And it's going to be hard to get people interested in that.

Then $\chi(h)\sum_{g\in G} \chi(g) = \sum_{g\in G} \chi(hg) = \sum_{g\in G}\chi(g)$

So yeah that conjecture works (how'd you think of that?)

It seemed like a generalization of that^

The sum of roots of unity is 0? Huh?

which was the only relevant fact about roots of unity that I knew

Well I mean I guess in hindsight yeah sure

@Daminark Is 0.

@LeakyNun So the idea being that the starboard presented a tempting question---what does that mean---and my saying "I'll bite" is to say "I know it's obvious, but I'll ask anyways."

something like that.

@Akiva that's what I said

@Semiclassical alright

Because multiplying by any root of unity doesn't change it (it just rotates them around)

@Daminark You lying ******* :P

Alternatively, they're the roots of $x^n-1$, right? @Daminark

Oh now that you put it that way it's obvious

Also it follows from the generalization :P

So their sum is negative the $n-1$-eth coefficient, which is $0$. It's an old puzzle, I don't remember where I first heard that solution

Alternatively: This is literally how you solve for the coefficients of a Fourier series.

Of a complex Fourier series, rather.

Lol, the only Fourier analysis I know right now is over finite abelian groups :p

Well, you use $\int_{-\pi}^\pi e^{inx}dx=0$ unless $n=0$, which is a continuous version.

So maybe not "literally."

But it's the same idea :P

@akiva eh, same difference upon using $z=e^{ix}$ to turn that into a contour integral

Right, right, yes, yes

Balarka and I (and MeowMix?) have discussed this several times

Maybe that's why it stuck in my mind :P

just like how PPCG chat (used to) have a quote in the description

well, we also have "just ask; don't ask to ask" here, but I don't really think that's a quote

"Read Euler, read Euler; he is the master of us all."

--Akiva Hamburger

I mean, Weinberger

I didn't come up with that quote. (But I agree with it!)

Also, nothing about "wine"?

whining burger?

Quit whining, it was a good pun

Sniped, in a sense

And the first time I've heard it /s :D

once upon a time, there was a dog named Columbus

dog ate my whining burger

My first name sounds like "Aquí va," which is Spanish for "Here he comes"

and "Ki va," which is Hebrew for "For he comes"

"va" is "come" in Hebrew?

I thought it means "and"

Yeah, it's the same work in both languages, it's weird

@LeakyNun No, that's the prefix v-

@AkivaWeinberger oh, v- + ha- => va-

Right, yeah, va- is "and the".

can you write down the Hebrew for "come"?

But it's also a conjugation of "lavo"

@LeakyNun בא

(Infinitive: לבוא)

oh, it isn't Biblical Hebrew lol

wikt says it's ba not va

בָּא

@LeakyNun Essentially interchangeable

Denis Villeneuve might be one of the best directors post-2000 who has striked the perfect balance between Hollywoodish style and artistic qualities.

@AkivaWeinberger hmm..

Biblical Hebrew would use va more often than Modern Hebrew, in fact.

"Ki va" is part of a passage from Psalms.

@AkivaWeinberger I thought בָּא isn't a Biblical Hebrew word

nvm, I'm wrong

biblehub.com/hebrew/ba_935.htm

@AkivaWeinberger how would you translate the Shema into modern Hebrew?

Just for curiosity

@LeakyNun Looking through the examples, it looks like it's "va" iff the previous word ends in an "-i" sound

@LeakyNun The first sentence or the whole thing?

@AkivaWeinberger shema yishrael, adonai elohim, adonai ehad

IIRC

Maybe change "Shema" to "Hekshev" or something? But it seems pretty modern-sounding already anyway

@LeakyNun *eloheinu

@AkivaWeinberger right, our god

*ekhad

@LeakyNun I didn't know there were Jewish asians

He got the Shema wrong, I don't know if he's Jewish

@Zee I'm hardly Jewish

@AkivaWeinberger you see, I'm hardly Jewish

WAIT HOLD IT I THINK I UNDERSTAND

(caps ftw)

@Akiva

So, choose some $h$ such that $\chi_n(h) \ne 1$, and then, say, choose 1

@AkivaWeinberger what is the "–" in biblehub.com/interlinear/deuteronomy/6-4.htm

(Ve'ahavta et adonai eloheikha, bekhol levavekha uvekhol nafshekha uvekhol me'odekha…)

@LeakyNun There's some more "v-" (and "u-")

@AkivaWeinberger what is this?

never mind, it's just the next verse

If you apply our $a_1 + a_2\chi_2\chi_1^{-1} + \ldots + a_n\chi_n\chi_1^{-1}$ to $h$, we know this will be different in at least one term than if we apply to to $1$

It's the next sentence of the Shema. @LeakyNun

Not sure what the dash is

But now we can cancel out the $a_1$

Okay we're done

What's $h$?

And I would have thought "apply it to every element of the group and add them all up to cancel out the coefficient of $\chi_0$" works already

(and multiply on the left by $\chi_n^{-1}$ and do the same thing, for the other coefficients)

@AkivaWeinberger are you a Jew?

That's also possible, I think, I hadn't pursued that in particular much

And wrt h, it's just some group element such that $\chi_n(h)\ne 1$

@LeakyNun Yes

@AkivaWeinberger oh...

I mean, follower of Judaism

…Yes

interesting

I eat only kosher and don't use electronics on Saturdays and learn Talmud in school and all that jazz

I suppose I shouldn't be asking you more questions in regard to that :p

out of fear of tuning this chatroom into a religious firestorm

@Daminark I don't understand, how are we canceling out $a_1$?

We have $a_1(h) + a_2\chi_2\chi_1^{-1}(h) + \ldots + a_n\chi_n\chi_1^{-1}(h) = 0$

And $a_1(1) + a_2\chi_2\chi_1^{-1}(1) + \ldots + a_n\chi_n\chi_1^{-1}(1) = 0$

The first term is just $a_1$, right?

You subtract

That'll give 0 on the right side, and something which doesn't include $a_1$ but is still non-trivial on the left

Oh, by "cancel out $a_1$" I thought you meant "prove that $a_1=0$"

I mean I think this gives you that $a_1 = 0$

We know that the sum of the $a_i$ has to be 0

Because you apply the element 1, that should still be 0 by assumption

So now, if you subtract, you get $\sum_{i=2}^n a_i(\chi_i\chi_1^{-1}(h) - \chi_i\chi_1^{-1}(1)) = 0$

But that's, in turn, equal to just $\sum_{i=2}^n a_i\chi_i\chi_1^{-1}(h)$

Say I had a polynomial with coefficients in $\mathbb{Z}_{p}$. How would I go about converting it into a polynomial with coefficients in $\mathbb{Z}$?

So we know that sum is 0, and that the sum plus $a_1$ is 0, so that should give $a_1 = 0$

And from there you continue and should be done

@Daminark We don't know that $\sum_{i=\color{Red}2}^na_i(\chi_i\chi_1^{-1}(1))=0$ yet!

We know the sum of the $a_i$ is zero

but that includes $a_1$

@ALannister Well, each coefficient is still an integer. So that would seem to give an integer polynomial automatically, though not in a canonical way (since adding $p$ to a coefficient doesn't change it as a poly in Z/p)

Oh right shit

@Daminark Just apply it to all the elements and sum them

@Daminark GET REKT

Using the thing you proved, that becomes $a_1|G|=0$.

($a_1$ the coefficient of the trivial map.)

Wait, hold on

@Semiclassical but let's say $f = g_{1}g_{2}\cdots g_{k}$ where $f$ and all the $g_{i}$ are thought to be polynomials in $\mathbb{Z}_{p}$ but $h$ is any divisor of $f$ in $\mathbb{Z}$. If my goal were to show that some $g_{i}$is a divisor of $h$ over $\mathbb{Z}_{p}$ wouldn't I have to do some conversions somewhere?

Why do the $\chi$ form a group, again? @Daminark

It does seem a bit weird.

Oh, just multiplicatively.

Yee

I guess what I was getting at was that $2x^3+x^2+x^1$, for instance, can be understand as a polynomial with coefficients in either Z/3 or Z.

We talk about reduction modulo $p$, but what about increasing modulo $p$? Or undoing reduction modulo $p$?

OK, yeah, then for the coefficient of an arbitrary $\chi$ just multiply through first by $\chi^{-1}$ and the do the same.

@Balarka hopes and dreams were crushed today

Though I think I'm buying what you're saying @Akiva

@AkivaWeinberger should I ask you more questions on that?

On what, my Judaism?

Sure

But I could equally go with $2x^3+4x^2+x^1$; it's the same polynomial in (Z/3)[x], but not in Z[x].

@Semiclassical that's true. But in $\mathbb{Z}$, we wouldn't ssay that polynomial is equal to $5x^{3} + x^{2} + x$

Exactly.

Yeah, $\Bbb Z\to\Bbb Z_3$ is a many-to-one function. The reverse would have to be one-to-many, which isn't a function

So I dunno.

@AkivaWeinberger I think my first question would be that how do you know God exists

Anyway, I was about to post something on the main MSE page. This is supposed to be an intermediary step on the way to proving Eisenstein's criterion for monic polynomials

@LeakyNun That's kind of personal

@AkivaWeinberger well, I can take it back if you don't want to answer it

I don't see how an impersonal question could be asked about your Judaism, though.

I believe in contrarianism. The fundamental principle of contrarianism is nothing is positive, everything is contrary.

That is to say, neither contrarianism is I nor I am contrary.

Not is (not)

Half the real numbers tho

@Daminark You mean, Unreal mbers

You mean, infinity-topoi

That is just a special case of (infinity, n)-topoi

Might as well generalize

Gnhh i should be asleep

It's, uh, 6:50am?

Well, it is summer

yep

I have school, I am just bunking it

Aren't you applying to a college at some point

Not that there's anything wrong with not going, but I think you mentioned a specific place

Next year

So presumably skipping all your classes is not conducive to that

College was nice

But It was such a waste of time

Nice typo there

The ONLY good thing about college is meeting people, professors and students

(Also, when you do get to college, if you sleep through all your classes there, too, what's the point of going in the first place?)

@Akiva Er, I don't think my high school attendance has anything to do with admission on college

What's the point of going to lectures even if you don't sleep?

College is about personal connections, all else you can do for free in your local library

@AkivaWeinberger vOv. if I get admitted, I probably will try to fix my sleep schedule

I'm genuinely scared about the admission exam

@BalarkaSen Your high school grades do, at least

And, they don't look at attendance? Weird.

@AkivaWeinberger Well, not majorly, but yeah

Why do you have school on 28 June?

It was vacations for me back then

Hell if I know man

HI

@Avantgarde Have you heard of the greatest rap of all-time?

@ChrisNguyen Hey

@BalarkaSen I've heard some good rap. What've you got?

pls not that guy again

I need help on a problem

Here is the legend

Nah this guy's much better

Can someone help

"Just ask, don't ask to ask."

@BalarkaSen Disgusting

need to wash my ears

@Avantgarde Here's a killer review of the music video

What is the radius of convergence of $\sum_{n=1}^{\infty} \frac{(xn)^n}{n!}$

Which convergence tests do you know?

I'm not sure if that rap is a joke or not

I think this should be nice for limsup type things, no?

The review was very insightful for me.

@ChrisNguyen $1/e$, I think.

Like, $\frac{1}{\limsup \frac{n}{\sqrt[n]{n!}}}$

@Semiclassical I know the ratio test

Not sure the ratio test helps here, but worth checking.

@ Akiva Weinberger can you explain

That'd be $\dfrac{a_{n+1}}{a_n}=\dfrac{(x(n+1))^{n+1}/(n+1)!}{(xn)^n/n!}$

siiigh

Oh wait a sec

finally

Here's my link on the main MSE page

So we're trying to calculate $\limsup \frac{n}{\sqrt[n]{n!}}$

That simplifies to $x\cdot \left(1+\frac{1}{n}\right)^n$.

oh hey, that should look familiar.

@Semiclassical That simplifies to $x(1+\frac1n)^n$, no?

So the ratio test will work here.

DAMMIT

Sniped

looool

I was scrolled up, I didn't see what you wrote!

How come I'm not seeing latex

I think that's less "sniped" and more "perfect guard"

I'm new here so plz help

@ChrisNguyen Look at the room description on the top-right

Do you see where it says "LaTeX in chat"?

Now, I think we know that $\sqrt[n]{n!}$ grows like $\sqrt[2n]{2\pi n}\frac{n}{e}$

Yes

So then $n$ over that is gonna be $e\sqrt[2n]{2\pi n}$

Which should go to $e$

What next @AkivaWeinberger

It's next to a link with some instructions

tinyurl.com/cfqcvpc

If we know the sequence is increasing this should make that the limsup

^That link, specifically

@Daminark, we already got this

I already started my idea so I wanted to finish no matter what

Ok i clicked startChatjax but nothing happened

Did you click it while in this room?

You have to add it as a bookmark, like it says

(If you're on mobile, there's a separate thing on the bottom)