I am trying to understand what happens to the zeros of the zeta function when you raise it to the 1/lnx power

@Fargle Looks like sometimes it's the ceiling and sometimes it's the floor. For $n=3$, the chromatic number is 1. For $n = 5$, it's 3.

My suspicion is that $n = 3$ is an edge case.

But it might turn out that, like, $\overline{C_{275}}$ is 137-colorable...

@AkivaWeinberger $|z - 3 + 2i| \leq 4$, what's the range of |z|?

I'm getting stuck using triangle inequality I wrote$ | |z|−|3-2i| | \leq |z−(3-2i)|$ so $| |z| - \sqrt{13} | \leq |z−(3-2i)| \leq 4 $ or $-4 \leq |z| - \sqrt{13} \leq 4 $ which gives $\sqrt{13}-4 \leq |z| \leq \sqrt{13}+4$ but that lower bound is negative.. ? should I take the lower bound as 0 instead? When I drew a picture, I'm getting the lower limit for |z| to be $ 4 - \sqrt{13}$, what's going wrong here? I'm confused, why does triangle inequality give a different result?

In which case I throw up my hands in disgust.

@user76284 Do you know this theorem: if $D$ is a kernel-perfect orientation of $G$ and $f(x)=1+d_D^+(x)$ for all $x\in V(G)$, then $G$ is $f$-choosable

That looks similar to lemma 5.4.3 in Diestel's Graph Theory.

or rather; does an odd cycle $C_n$ have a kernel-perfect orientation? I would think that the cyclic orientation is kernel-perfect, but that can't be true, becuase $\chi(C_n)=3$

Does someone of you have an idea about my question about hypothesis testing? math.stackexchange.com/questions/2744569/…

ah I've heard of Diestel, but haven't read it

@Rick $|3-2i|=|(-z)+(z-3+2i)|\le|z|+|z-3+2i|$

Wait hold on

Never mind

$z=0$ satisfies $|0-3+2i|\le4$, doesn't it?

@Fargle @ShaVuklia math.stackexchange.com/questions/2744899/…

Because $\sqrt{13}\le\sqrt{16}=4$

holyy

you were on the right track!!

So it's possible to have $|z|=0$.

hmm yes it does

And $|z|$ can never be less than $0$, so that's the minimum.

Basically, the solution set is a disk that contains the origin

@user76284 Phew, turns out I had it.

>_>

The algebra is the worst. Just draw the disk and compute the minimum and maximum distances out (which are normals to the boundary circle).

Hey @Fargle, @Akiva

anyhow, any ideas why the cyclic orientation of an odd cycle isn't kernel-perfect? I see absolutely no difference with the cyclic orientation of an even cycle for the kernel-perfect-ness

Howdy @Balarka.

@BalarkaSen The disk contains the origin, so the minimum of $|z|$ is found there.

oh ok..yes

@Akiva Ah nice, I didn't pay attention to the equations that much

h

$|\!\cap$

ん

@ShaVuklia Not sure what this terminology means. What is an orientation of a graph, and what does it mean for an orientation to be kernel-perfect?

I like that the Cyrillic, Latin, and Greek alphabets all have a capital H-looking glyph but have completely different ideas for what the lowercase version should look like

@Fargle orientation means that our edges are 2-tupes, so they have a direction

Cyrillic just shrinks most of its letters, which makes the most sense I guess

so we're working with directed graphs

Alright.

I like that I don't know any language

I don't know if you're familiar with the kernel of a digraph

@AkivaWeinberger Thanks a lot! Got it now

hi

in any case, it's explained there

\translate[English][Bangla]{What?}

independent set means that it doesn't contain vertices that are adjacent to each other

@AkivaWeinberger I once wrote a massively fuckheaded short story by google translating gibberish from English to Bengali and then back again, and used that for my literature project in school.

It got an A+

in any case, for $C_n$ ($n$ is odd), I was thinking; any isolated points have a kernel, so we don't have to consider those. And we only really need to consider the components of our induced subgraph. Now such a component is a path of lengt $\leq n$. and I would think that each path has a kernel?

I think the Bengali script is really pretty looking

I'm back.

Jesus lord.

It's the fifth most widely used writing system according to wikipedia

o I was so stupid! I forgot the subgraph $C_n$ itself

so never mind, the cyclic orientation isn't kernel-perfect indeed

Hi. Could anyone explain how to find this limit?

Hello nerds

sin(πcos(x))/(cos(x)+1) as x approaches π

@ShaVuklia Why doesn't $C_n$ have a kernel? I guess the wording of that definition has me thrown off.

Sup @Daminark

@BalarkaSen I remember you said once you're an expert in contour integration. I have 2 methods by contour integration, but not sure if these options are the best. How about comparing what I have to what you have?

How you doing Fargle?

Alive! You?

Pretty sure I'm alive. Maybe

I love contour integration. Let's play a bit with contour integration.

I have around 150 integrals to post here (meant to be done by multiple ways belonging to the contour integration area).

limit sin(pi*cos(x))/(cos(x)+1) as x approaches pi

?

Let's begin with this one.

@Fargle because for any independent set of $C_n$, there will always be a point outside of the set that doesn't have a successor in the set

please explain in layman terms. no derivative or hospital

@ShaVuklia Ah, I guess I didn't see "independent".

right

@Waiting I think I just claimed I like contour integrals, but admittedly I'm pretty bad at it.

This isn't the case for an even cycle because you can pick n/2 staggered vertices, I guess?

I don't think I'll be of any help on this one

what is a contour integral

limit sin(pi*cos(x))/(cos(x)+1) as x approaches pi =?

Whereas one can only pick $\lfloor n/2 \rfloor$ independent vertices in $C_n$ for $n$ odd, and each vertex has only one successor, and then counting finishes the proof.

I see.

@FarhadRouhbakhsh make the change of variable pi-x=y

yes exactly @Fargle

I really, REALLY should get more into graph theory. Seems right up my alley.

@Balarka so I dunno any geometry and thus my use of these words is gonna be sketchy at best, but if this makes any sense, is it true that a Riemannian manifold such that for any two points, there's a unique minimal geodesic between them, that the manifold is simply connected?

hahaha, yea it's great, I should read more about it too

@Fargle yoooooo combinatorics is the best

@Waiting i did that. then what?

if you want to think along, @Fargle, I have to find the choice number of an odd cycle

@Daminark I see that you have seen the light of the discrete. :)

@FarhadRouhbakhsh And what you got?

@ShaVuklia Sure.

Did you somehow mean "unique geodesic joining them", and not "unique minimal geodesic joining them"?

Oh, wait, antipodal points

I take that back.

#rekt

so we know it's 3 colourable

Yeah exactly

Hmmmmmmmm

HMMMMMMMMMM

What's the distinction between choosable and colorable?

In my mind the obstruction is that you can have two points and two paths between them achieve their distance

Or is there one?

=lim sin(pi*cos(y+pi))/(1+cos(y+pi))

Yeah I see your reasoning

(I'm guessing here based on having eavesdropped around that "minimal geodesic" means achieves the length)

@Fargle colourable just means that you can colour your graph with 3 colours

= -sin(pi*cosy)/(1-cosy)

$k$ choosable means that given any assignment of $k$ colours to each vertex separately, we can find a proper colouring

we are only allowed to choose colours from those lists

@Waiting so now what should i do?

-sin(pi*cosy)/(1-cosy) as y approaches 0

I see. So any k-colorable graph is automatically k-choosable?

@FarhadRouhbakhsh Use that $lim_{x\to0} (1-\cos(x))/x^2=1/2$ (prove that first)

@Daminark OK, if $M$ is not simply connected there's a nontrivial homotopy class. I think if $M$ is compact you can always choose a geodesic representative of that nontrivial homotopy class, and do the circle argument there.

@MikeMiller Help, is this the correct idea?

(or, at most k-choosable?)

at least $k$ choosable

math.harvard.edu/~lurie/937notes/937Lecture35.pdf

I am correct

$k$ colourable applies to the list assignment where each set is the same $k$ colours

I see.

Nifty

So there is a closed geodesic $\gamma$ in $M$ which is not null-homotopic. Mark two points $p$ and $q$ on $\gamma$ such that the two arcs from $p$ to $q$ in $\gamma$ that we get have equal length - you can always do this

This gives two points $p$ and $q$ in $M$ with two distinct - even homotopically distinct - minimal geodesics joining them, contradicting hypothesis on $M$

This should work

So, you want to find the minimal k such that $C_n$ is k-choosable? It seems like it's impossible for it to be 1- or 2-choosable--consider the list colorings where every node is the same color, or the same two colors.

@Waiting i know that. then i can divide the denominator to y/2 and get 1/2 but what i should i do with nominator then? -sin(pi*cosy)/(y^2)

@BalarkaSen The idea is that there is always an action minimizer, and the concern is that it's constant; you solve that by being homotopically nontrivial

yea, or you can use the fact that it's 3 colourable, so the choice number it at least 3

Yup

Right.

@FarhadRouhbakhsh First use this limit in your main limit.

I proved this in my big tome answer, I think. Or at least the same tools can do it.

Minimize the energy functional over a given nontrivial homotopy class

And it might be different from 3 because there might be non-trivial list colorings where there is no coloring.

@FarhadRouhbakhsh then you make the change of variable pi cos(x)-pi = t

t will go to 0

The rest is a piece of cake.

@Waiting you mean use this limit by "you this limit"?

@FarhadRouhbakhsh aren't you done yet?

how can i use this limit in my main limit?

@Fargle that is correct

the problem is, there aren't many theorems on list colourings in my book

Nice question though, @Daminark. I should have thought about it before.

@FarhadRouhbakhsh -sin(picosy)/(1-cosy) can be written as -sin(picosy)/x^2 * x^2/(1-cosy) and then split the last limit

@ShaVuklia Hrm. My suspicion is that the choice number is always 3, but I don't know how to prove this. It seems like it's far easier to exhibit counterexamples than to demonstrate k-choosability.

haha yea, it's my suspicion too:p

I guess I will ask on the main site, and leave it for now

I wonder what else can be said about compact Riemannian manifolds with any pair of points having a unique minimal geodesic between them. Not all simply connected manifolds appear as such; eg $S^2$ obviously doesn't

What are the other topological obstructions for such manifolds?

(cc @MikeMiller)

@ShaVuklia Still in uni?

@Waiting I dont understand. after changing variable to t you get -sin(t)/ y^2. what should i do with y^2

Even for C_3, there are lots of cases to check--where all lists are disjoint (000), where one pair of vertices share one color in the list but no other overlaps (100), and then (extending this notation) 110, 111, 200, 210, 211, 221, 222, 300, 311, 322, 333.

@Waiting I do study still, yes:p but I am at home now

@FarhadRouhbakhsh you make the change of variable here -sin(picosy)/y^2

@Fargle o yea, checking by cases is impossible:p

Sorry, my notation above failed

It was -sin(picosy)/y^2 * y^2/(1-cosy)

in a previous exercise I thought I was supposed to show that this graph is 2-choosable

@BalarkaSen Equivalent to saying a geodesic has a Jacobi field, right?

(it was in fact not two-choosable)

and the cases were endless:p

@FarhadRouhbakhsh Then make the change of variable pi cos(y)-pi=t for sin(picos(y))/y^2

@MikeMiller Ohhhh

So if you ever have a 1-parameter family of isometries fixing two points, then you're boned

This is what happens with $S^2$, we have a circle of rotations

@ShaVuklia Nice ;)

Yeah fantastic point. Fuck me I have forgotten math

what if you fix two points and let the homotopic curve vary with one parameter

@Waiting the answer would be sin(t+pi)/y^2. then what?

@MikeMiller Really, we're requiring that the exponential map has no singularities

Singularities as in, many-to-one points.

@FarhadRouhbakhsh I cannot tell you more than that. You should be able to do it alone from here. Just try to finish it alone.

Hm, wait, but it's a compact manifold. It has to have those.

@BalarkaSen You can definitely have \emph{recurrent} geodesics, but you're trying to demand they be minimal

@ShaVuklia That's downright diabolical.

Ah, hm.

@Fargle lol:p

I think it's special to the 2-sphere that the boundary of the convex hull which exponentiates to an open dense subset is a norm-1 thing

Special to the sphere, I should say

o damn, I linked you a post, but I thought it was an answer to my question on the odd cycle

but it was in fact an answer to that previous exercise, which was already solved

so never mind:p

lol, I noticed.

Ah-hah, I see.

I think on the genus 2 surface, the exponential map should look like some sort of pointy (hyperbolic-regular) polygon

I had literally posted it minutes ago, so I was a bit confused at how rapid the response was:p

or rather this hull

So, what's an example of a compact manifold such that there are no pairs of points admit different minimal geodesics between them?

I don't think I have one

@Waiting im really confused as i dont see any relation between t+pi and y^2. i dont know derivative or hospital rule or other things.

@FarhadRouhbakhsh Or simpler. For sin(pi cos(x))/x^2 you make the change of variable pi/2-x=y. And then you exploit the simple fact that lim_(x->0) sin(x)/x=1.

@MikeMiller That sounds about right. It's the universal cover after all.

@BalarkaSen $\Bbb R$? Or have I misunderstood your question

So has a fundamental domain

@Fargle Should have specified compact, thanks

Ah, sorry. That's a much trickier question.

@Waiting no. first explain the first solution please

mm seems to be

@FarhadRouhbakhsh Finalize the problem first, and the last suggested way is straightforward.

finalize?

@BalarkaSen So the question is about how long the geodesics are, mostly

You want enough isometries (preferably for them to act transitively) so that you only need to check one point, but not too many (or it will force new geodesics to exist)

That's a good strategy

So we should look for some simply connected homogeneous spaces, or some such?

@Waiting in your last solution i will get sin(picos(pi/2-y))/((pi/2-y)^2) should i expand that?

@BalarkaSen Ah, I was going to try T^2 because I forgot about the nontrivial homotopy classes

Yup

I think T^2 will still apply as the geodesics that recur are not minimal

I guess it depends what you mean

They're minimal as maps from $S^1$ but not $[0,1]$. The first is good if you don't have specified points, the second is good if you do

I don't understand precisely what you're giving T^2 as an example of

Can you reiterate? I'm slow

@Waiting =sin(pi*sin(y))/((pi/2-y)^2)

"For every $p, q \in M$, there is a unique geodesic passing from $p$ to $q$ with minimal possible length"

But take any meridian and subdivide it into two equal arcs

oh

damnit

the T^2 part of the isometry group doesn't force new geodesics

but the O(2,Z) part does

@Waiting where are you?

help me. actullay i have been thinking on this problem about 5 hours

Aha

@FarhadRouhbakhsh But you want to do it with more restrictions. I read now what you wrote above: no derivative or hospital

No Taylor series?

our teacher didnt tell us taylor series.

Hm, I think this works for simply connected manifolds of negative curvature

@FarhadRouhbakhsh Give me 5 min. Don't disturb me for 5 min.

@Waiting im 11th grade student

Eh, those are not compact

I'm pretty sure that 11th grade students are compact.

('Cuz compact negatively curved manifolds are covered by R^n (with some bad metric), so are aspherical. Simply connected just says they are topologically R^n)

@XanderHenderson By Cartan-Hadamard theorem, negatively curved 11th grade students students are all aspherical

@BalarkaSen compact?

@Fargle apparently we have that the choice number it at most* the maximum degree +1, so that concludes the proof

@BalarkaSen those?

@ShaVuklia We already knew it was at least, though. Do you mean at most?

yea indeed

Ah! That does it then.

OK LITTLE BOY

I'm done!

@Fargle come to Washington

its a nice place to be in

State or DC? I can't either way, but yeah

@FarhadRouhbakhsh we go in the initial limit, that is

limit sin(pi*cos(x))/(cos(x)+1) as x approaches pi

and then make the change of variable pi cos(x) + pi = y

Q.E.D.

LHS=sin(y-pi)/(cos(x)+1)

@Fargle Lol @ "State or DC"

Aw well

@FarhadRouhbakhsh With that replacement you actually get -pi lim_(y->0) sin(y)/y = -pi

can you write it more clear?

@FarhadRouhbakhsh What is the unclear part? :-)

you mean as y approaches 0, lim sin(y-pi) = -pi?

In the numerator you use that pi cos(x) = y - pi and in denominator you use that cos(x) = (y - pi)/pi. That simple!

Now, with that replacement y->0. So, once again, you finally have -pi lim_(y->0) sin(y)/y = -pi.

is dy/dx = x equivalent to dy=x dx due to the way dy/dx is defined.... I know it may seem like an elementary question

@BalarkaSen b-b-b-but why

In the sense of $x=\frac{dy}{dx} \implies y=\int dy=\int x\,dx$, yes

@Fargle I meant Discord you silly goose

'tis a silly code :P

I may be able to later. I actually have to run for now. Group project on TSP. :|

\emph{Eating peanuts with salt.}

@Fargle Bye!

?

@waiting how do you reach the final step -pi lim_(y->0) sin(y)/y = -pi. where do you get sin(y)/y from?

@FarhadRouhbakhsh what you wrote can be reduced to what I wrote above.

@FarhadRouhbakhsh ^^^

yes i know that siny/y = 1. i dont know where should i find it in the fraction?

after changing variables where can i find siny/y?

@FarhadRouhbakhsh lim_(y->0) (sin(y))/y = 1. You may see solutions at the link above.

The issue im having is... $\frac{ds}{d\theta}=\sqrt{r^{2}+(\frac{dr}{d\theta})^{2}}$ ... integrating both sides just gives me $s=f(\theta)$ where in fact im looking to get the summation of ds to give L and then I can form a Riemann sum

honestly Im trying to get my head around the derivation of the arc length of a polar curve

quite sure im making things more complicated

\emph{Keep eating peanuts with salt.}

@Waiting Hey little man, i told you that i know limit siny/y is equal to 1. im just asking you after changing variables and simplifying limits where do you find siny/y

@FarhadRouhbakhsh I have no idea what you're talking about there.

after changing variables you will get sin(y-pi)/(1+(y-pi)/pi) and there is no siny/y in it.

@FarhadRouhbakhsh Do you know the formula sin(a-b)?

yes.

also, simplify your denominator.

@FarhadRouhbakhsh sin(y-pi) = - sin(y). That simple.

yes i know that

where is y?

@FarhadRouhbakhsh simplify the denominator

1+(y-pi)/pi=y/pi.

ok

thats right

thank you

@FarhadRouhbakhsh You need many hours of practice with simple algebra. Just do practice.

oh, smbc, how I love thee: smbc-comics.com/comic/geometry

Hi @Ted

Hi demonic @Alessandro

hi yall

Yo

If I received some help from the chat or from the math stacksexchange site. How shall I acknowledge it in my work?

Oh my god tomorrow's 4/20

@quallenjäger "This paper would not exist without the help of my mom, my advisor, and xXx_username69_xXx on StackExchange"

I dunno

I think it depends on what sort of work you mean.

"bigcock69 from mathstackexchange has helped me considerably ..."

A+ idea

If you remove a point from a manifold, it cant be compact anymore. Why?